Atomic Physics Chapter 1 Basic Properties of Atom
11/20/1309/03/2021 Jinniu Hu 1 1 What is an atom An atom is the smallest unchangeable component of a chemical element.
1. Unchangeable means in this case by chemical means
2. Moderate temperatures: kT < eV
Mass range: 1.67×10−27 to 4.52×10−25 kg Electric charge: zero (neutral), or ion charge Diameter range: 62 pm (He) to 520 pm (Cs) Components: Electrons and compact nucleus of protons and neutrons 09/03/2021 Jinniu Hu The mass of an atom
Atomic mass unit (AMU): 1u: 1/12 of the mass of a neutral carbon atom with nuclear charge 6 and mass number 12 Mass number (A): The total number of protons and neutrons in nucleus Mole (mol): 1 mol is the quantity of a substance that contains the same number of particles (atoms or molecules) as 0.012 kg of carbon 12C. 1 mol of atoms or molecules with atomic mass number A
AMU has a mass of A grams. 09/03/2021 Jinniu Hu The mass of an atom
The relation between 1u and NA
1 27 1u = =1.660539040(20) 10 kg NA ⇥ Electronvolt 19 1 eV = 1.602176565(35) 10 C 1V ⇥ ⇥ 19 =1.602176565(35) 10 J Mass–energy equivalence ⇥ E = mc2
1u transfer to eV 1 u = 931.478 106 eV/c2 ⇥ = 931.478 MeV/c2
09/03/2021 Jinniu Hu The mass of an atom The mass of electron: 31 m =9.10938356(11) 10 kg e ⇥ 4 =5.48579909070(16) 10 u ⇥ =0.5109989461(31) MeV The mass of proton: 27 m =1.672621898(21) 10 kg p ⇥ =1.007276466879(91) u = 938.2720813(58) MeV The mass of neutron: 27 m =1.674927471(21) 10 kg n ⇥ =1.00866491588(49) u = 939.5654133(58) MeV
09/03/2021 Jinniu Hu Avogadro’s Number
Avogadro’s number is a Bridge from macroscopic to microscopic physics.
1 mole of any substance contains the same number (NA ) of atoms (molecules) Mass of 1 mole of the substance N = A Mass of an atom 23 1 =6.02214078(18) 10 mol ⇥ 1. The Faraday constant and elementary charge F = NAe 2. Gas constant and Boltzmann constant R = kBNA 3. Molar volume and atomic volume Vm = VatomNA 09/03/2021 Jinniu Hu 14 2 The Concept of the Atom 14 2 The Concept of the Atom c) Determination of Avogadro’s Constant from c)Electrolysis Determination of Avogadro’s Constant from ElectrolysisAnother method for the determination of NA is based on Fara- Avogadro’sAnotherday’s method law Number for for electrolytic the measurements determination processes. of ItN statesA is based that the on Fara- electric day’scharge law for electrolytic processes. It states that the electric Thecharge Faraday’s constant F NA e 96, 485.3383(83) C/mol (2.15) = · = F NA e 96, 485.3383(83) C/mol (2.15) is the electric= charge· = transported to the electrode in a is transported to the electrode in an electrolytic cell, when a an electrolytic cell, when 1 mol of singly charged ions is transported1molofsinglychargedionswithmass to the electrode in an electrolyticmX cell,and elemen-when with1molofsinglychargedionswithmass masstary chargemX ande haselementary been deposited charge at the e electrode.hasmX andbeen elemen- Therefore, a/2 tary charge e has been deposited at the electrode. Therefore, depositedweighing at the the masselectrode. increase ∆m of the electrode after a charge a/2 weighingQ has thebeen mass transferred, increase yields:∆m of the electrode after a charge Therefore, weighing the mass increase m of the Q has been transferred, yields: electrode after a charge Q hasQ beenQ transferred,MX Fig. 2.9 Elementary cell of a cubic crystal ∆m Q mX Q M yields: = e = e XNA Fig. 2.9 Elementary cell of a cubic crystal ∆m mX = e Q M=X e NA NA (2.16a) Phase ⇒ =Q Me X∆m NA (2.16a) Phaseplanes ⇒ = e ∆m planes 09/03/2021where MX is the molarJinniu mass Hu of the ions with mass mX. where MX is the molar mass of the ions with mass mX. ϑ Example ϑ ExampleIn the electrolytic process In the electrolytic process d ϑ Crystal d Crystalplanes AgNO3 Ag+ NO3− d ϑ ↔ + d planes AgNO3 Ag+ NO3− of silver nitrate the transport↔ of+ charge F means a simulta- of silverneous nitrate deposition the transport of the molar of charge mass FMmeansNA a simulta-m(Ag) at = · neousthe deposition negative electrode, of the molar which mass can beM measuredNA bym(Ag weighing) at = · Fig. 2.10 Bragg-reflection of X-rays by two crystal planes thethe negative cathode electrode, before and which after can the be charge measured transport. by weighing With the theatomic cathode mass before number and after of silver the AM charge(Ag transport.) 107.89 With AMU the the Fig. 2.10 Bragg-reflection of X-rays by two crystal planes = wavelength λ,theinterferenceisconstructiveandtheampli- atomicAvogadro mass number number of silver AM(Ag) 107.89 AMU the = tude of the different partial waves add up. This is expressed Avogadro number wavelength λ,theinterferenceisconstructiveandtheampli- 107.89 AMU Q tudeby of the the Bragg different condition partial waves add up. This is expressed NA (2.16b) 107= .89 AMU∆m Q· e by the Bragg condition NA (2.16b) 2d sin ϑ m λ. (2.17) = ∆m · e · = · is obtained from the measured mass increase ∆m of the elec- 2d sin ϑ m λ. (2.17) is obtainedtrode and from the transported the measured charge massQ increase(∆m∆/mMof)N theA e elec-. At a given wavelength·λ one obtains= · maxima of intensity I (ϑ) = · trode and the transported charge Q (∆m/M)NA e. Atof a given the scattered wavelength radiationλ one only obtains for maxima those inclination of intensity anglesI (ϑ)ϑ, = · offor the which scattered (2.17 radiation)isfulfilled. only for those inclination angles ϑ, d) Determination of N from X-Ray Diffraction A for whichOne sees(2.17 from)isfulfilled. (2.17)thatform > 0thewavelength The most accurate method for the determination of N is d) Determination of NA from X-Ray Diffraction A One sees from (2.17)thatform > 0thewavelength based on X-ray diffraction or X-ray interferometry, which 2d The most accurate method for the determination of NA is λ sin ϑ < 2d basedare on used X-ray to measure diffraction the distances or X-ray between interferometry, atoms in awhich regular =2dm λ sin ϑ < 2d arecrystal used to [10 measure]. This theyields distances the total between number atoms of atoms in a per regular volume has to be smaller than= twicem the distance d between adjacent crystalif the [10 crystal]. This structure yields the is total known. number of atoms per volume crystal planes. For visible light λ d,butforX-raysof if the crystalLet us consider structure a is cubic known. crystal, where the atoms sit at the has to be smaller than twice the distance$ d between adjacent corners of small cubes with sidelength a (Fig. 2.9). When a crystalsufficient planes. energy Forλ visible< 2d can light beλ achievedd,butforX-raysof (see Sect. 7.6). Let us consider a cubic crystal, where the atoms sit at the $ cornersplane of wave small with cubes wavelength with sidelengthλ is incidenta (Fig. on the2.9 crystal). When under a sufficient energy λ < 2d can be achieved (see Sect. 7.6). Note planean waveangle withϑ against wavelength a crystalλ planeis incident (Fig. 2.10 on the)thepartialwaves crystal under In (2.17) ϑ is the angle of the incident radiation against the an anglescatteredϑ against by the a different crystal plane atoms (Fig. of adjacent2.10)thepartialwaves planes with dis- Note crystal planes not against the normal to the planes, different scatteredtance d byinterfere the different with each atoms other. of adjacent In the direction planes withϑ,which dis- In (2.17) ϑ is the angle of the incident radiation against the − from the conventional definition in optics. tancecorrespondsd interfere to with the direction each other. of In specular the direction reflection,ϑ,which their path crystal planes not against the normal to the planes, different difference is ∆s 2d sin ϑ.If∆s equals an integer− m of the from the conventional definition in optics. corresponds to the direction= of specular reflection, their path difference is ∆s 2d sin ϑ.If∆s equals an integer m of the = 2.2 Experimental and Theoretical Proofs for the Existence of Atoms 13 where M is the molar mass and κ Cp/Cv [9]. Since Fg (m "L Vp)g (2.11a) = = − · the2.2 acoustic Experimental losses and of Theoretical the spherical Proofs resonator for the Existence are low, of Atoms the 13 resonances are very sharp and the resonance frequencies is just compensated by the friction force f can be determined with high accuracy. Tuning the 0,nwhere M is the molar mass and κ Cp/Cv [9]. Since Fg (m "L Vp)g (2.11a) = = − · loudspeaker-frequencythe acoustic losses of through the spherical several resonator resonances are low, from the Ff 6πηrv, (2.11b) 2 = − n 1 to higher values the squares f0,n of the resonance is just compensated by the friction force =resonances are very sharp and the resonance frequencies frequencies can be plotted against n2 which gives a straight f0,n can be determined with high accuracy. Tuning the 2 which spherical particles of radius r experience when they fall lineloudspeaker-frequency with the slope RκT/(Mr through0 ).SinceMand several resonancesκ are known from Ff 6πηrv, (2.11b) 2 with the velocity v in a medium= − with viscosity η,thenetforce andn T and1 r to0 can higher be accurately values the measured, squares f0 the,n gasof the constant resonance R = canfrequencies be determined can [ be9]. plotted against n2 which gives a straight is zero (see Ref. [11]and[17]). The constant sink velocity is 2 then β) Measurementline with the slope of the Rκ BoltzmannT/(Mr0 ).SinceMand Constant κ are known which spherical particles of radius r experience when they fall and T and r0 can be accurately measured, the gas constant R with the velocity v in a medium with viscosity η,thenetforce The Boltzmann constant k was first determined in 1906 by (m "L Vp) g isv zero (see Ref. [11]and[17]). The constant4 π sink3. velocity is Jeancan Baptiste be determined Perrin (1870–1942). [9]. He observed the vertical g − · · where Vp 3 r (2.12) then = 6πηr = densityβ) Measurement distribution n of(z the) of Boltzmann small latex Constant particles in a liquid within a glass cylinder (Fig. 2.8). At equilibrium the Boltz- The downward flux of particles jg vg n creates a concen- The Boltzmann constant k was first determined in 1906 by (m "L Vp) g 4 3 vg − · · where= V·p π r . (2.12) mannJean distribution Baptiste Perrin (1870–1942). He observed the vertical tration gradient= dn6πη/dzr, which leads to an= upward3 diffusion density distribution n(z) of small latex particles in a liquid flux m∗gz/kT within a glass cylindern(z) n (Fig.(0) 2.8e− ). At equilibrium the(2.9) Boltz- The downward flux of particles jg vg n creates a concen- = · dn (m ="L · Vp)g mann distribution trationjdiff gradientD dn/dz,D whichn leads− to· an upward, diffusion(2.13) = − · dz = · · k T is obtained, where m∗g (m "L Vp)g is the effective flux · = − m· gz/kT weight of a particle withn(z volume) n(0V) p,(i.e.,itsrealweightminuse− ∗ (2.9) = · where D is the diffusiondn coefficient.(m "L Vp)g its buoyancy in the liquid with density "L). This gives the j D D n − · , (2.13) Finally,diff stationaryd conditionsz are reachedk T when both fluxes gradientis obtained, where m g (m "L Vp)g is the effective = − · = · · ∗ = − · just cancel. This means · weight of a particledn with volumemV∗,(i.e.,itsrealweightminusg n(z) p · , (2.10) where D is the diffusion coefficient. its buoyancy ind thez = liquid− with· k densityT "L). This gives the · Finally, stationary conditions are6πη reachedr D when both fluxes gradient jdiff jg 0 k · . (2.14) just cancel. This+ means= ⇒ = T The mass m of the particlesdn can be determinedm∗ g by measuring their size (volume) under a microscopen(z) and· , their density with(2.10) dz = − · k T 6πηr D standard techniques. · j j 0 k · . (2.14) Therefore, thediff Boltzmann+ g = constant⇒ = k canT be determined CountingThe mass m theof number the particles of n(z can) yields be determined dn/dz and by therefore measuring from the measurements of viscosity η,diffusioncoeffi- thetheir Boltzmann size (volume) constant under from a ( microscope2.6). The rather and their tedious density count- with cient D,temperatureT,andtheradiusr of the spherical standard techniques. ing can be avoided by the following consideration. Due to particles.Therefore, the Boltzmann constant k can be determined gravityCounting the particles the number sink down. of n If(z) theyields gravity dn/ forcedz and therefore from the measurements of viscosity η,diffusioncoeffi- the Boltzmann constant from (2.6). The rather tedious count- cient D,temperatureT,andtheradiusr of the spherical ing can be avoided by the following consideration. Due to Theparticles. most accurate method to measure k will be discussed gravity the particles sink down. If the gravity force in Sect. 2.3.1. b) DirectThe most Determination accurate method of Avogadro’s to measure Constantk will be discussed Fromin Sect. measurements2.3.1. of the absolute mass m of atoms X (see Sect.Avogadro’s2.7)andthemolarmass Number measurementsMX (i.e., the mass of a gas ofb) atoms Direct X Determination within the molar of volume Avogadro’sV Constant22.4dm3 under = normalFromFrom conditions measurements measurementsp and of Tof the)theAvogadroconstant the absolute absolute mass massm of m atoms of atoms X (see X
Sect.and 2.7the)andthemolarmass molar mass MX , the AvogadroMX (i.e., theconstant mass of a gas of atoms X within the molar volume V 22.4dm3 under NA MX/mX = normal conditions p and=T )theAvogadroconstant can be directly determined. can be directly determined. The molar mass for gasNA is MdefinedX/mX as the mass of a gas The molar mass MX can be= also obtained for nongaseous substancesof atoms from X within the definition the molar volume V = 22.4 dm3 under cannormal be directly conditions determined. p and T . The molar mass MX can be also12 obtained for nongaseous MX 0.012mX/m( C) kg substancesThe molar from mass the =can definition be also obtained for nongaseous substances from the definition 12 Fig. 2.8 Stationary distribution n(z) of small particles in a liquid when the absolute mass of the carbon atoms12 m( C) is mea- MX 0.012mX/m( C) kg sured (see Sect. 2.7). = 09/03/2021 Jinniu Hu 12 Fig. 2.8 Stationary distribution n(z) of small particles in a liquid when the absolute mass of the carbon atoms m( C) is mea- sured (see Sect. 2.7). 16 2 The Concept of the Atom Avogadro’s Number measurements Table 2.1 Different methods for the determination of Avogadro’s number Method Fundamental constant Avogadro’s number General gas equation Universal gas constant R
Barometric pressure formula (Perrin) Boltzmann’s constant k NA R/k =
Diffusion (Einstein) Torsionsal oscillations (Kappler)
Electrolysis Faraday’s constant F NA F/e = Millikan’s oil-drop experiment Elementary charge e X-ray diffraction and interferometry Distance d between crystal planes in a cubic N (V/a3) Mm for cubic primitive crystal A Mc crystal = Mm 3 Measurement of atom number N in a single NA N NA 4 Mm/!a for cubic face centered crystal = · Mc = crystal with mass Mc and molar mass Mm
The transmitted part of beam 4 now interferes with the Faraday constant F,theelementarychargee and Avogadro’s diffracted part 5 of beam 3 and the detector D2 monitors the number Na.Thevaluesoftheseconstants,whichareregarded total intensity, which depends on the phase difference between today as the most reliable ones according to the recommenda- the partial waves 4 and 5. Detector D1 measures the interfer- tion of the International Union of Pure and Applied Physics ence intensity of the superimposed transmitted beam 3 and IUPAP (CODATA 2016) are given on the inside cover of this the diffracted beam 6 of beam 4. book. When the slice S3, which can be moved against the oth- ers, is shifted into the z-direction by an amount ∆z the path difference ∆s between the interfering beams is changed by 2.2.4 The Importance of Kinetic Gas Theory for the Concept of Atoms 09/03/2021∆z Jinniu Hu δs 1 sin (90◦ 2ϑ) 2∆z sin ϑ. (2.22) = sin θ [ − − ]= · The first ideas of a possible relation between the internal energy U of a gas and the kinetic energies of its molecules The arrangement is similar to that of a Mach–Zehnder inter- were put forward in 1821 by John Herapath. Later in 1848 ferometer in optics. However, since the wavelength λ of X- James Prescott Joule (1818–1889) (Fig. 2.13)whohadread rays is about 104 times smaller than that of visible light, the the publication of Herapath found by accurate measurements accuracy of the device must be correspondingly higher. If the slice S3 is shifted continuously, the detectors monitor max- ima or minima of the interference intensity every time the path difference δs becomes an integer multiple of λ. The maxima are counted and its total number N at a total shift ∆z is
2∆z sin ϑ N · . (2.23) = λ The total shift ∆z is measured with a laser interferometer to within an uncertainty of ∆z/z 10 6 10 7 [12,13]. = − − − Example d 0.2nm, ∆z 1mm, ϑ 30 N 5 106,which = = = ◦ → = × allows an accuracy with a relative uncertainty of 2 10 7. × −
Instead of X-rays also neutrons can be used [14]. Table 2.1 compiles the different methods for the determi- nation of the gas constant R,theBoltzmannconstantk,the Fig. 2.13 James Prescott Joule Can One See Atoms?
Scattering of visible light by single atoms. Each image point corresponds2.3 Can One to See Atoms?one atom 19
Scattering particle
Laser beam
Scattered light
Lens
Image of the scattering CCD image plane microparticles Fig. 2.18 Schematic illustration of Brownian motion Fig. 2.17 Scattering of visibleJinniu light byHu single atoms. Each image point 09/03/2021 corresponds to one atom There is a nice demonstration that simulates Brownian about its size or structure can be obtained. One can only say: motion. A larger disk on an air table is hit by many small “It’s there.” discs, which simulate the air molecules. If the large disc car- There are several other methods that give similar infor- ries a small light bulb, its statistical path over the course of mation [13]. With computer graphics one can produce nice time can be photographed and the path lengths ξ between pictures of such “atom images” on the screen, which may be two successive collisions (the free path) can be measured impressive because they appear to give a magnified picture of (Fig. 2.19)[15]. the microworld of atoms and molecules. However, one should The basic theory of Brownian motion was developed inde- always keep in mind that such pictures are produced due to the pendently in 1905 by Albert Einstein (1879–1955) and Mar- interaction of light or particles with atoms. Only if this inter- ian Smoluchowski (1872–1917). The distribution n(ξ) and action is fully understood can the interpretation of the images the resulting mean free path Λ 1 ∞n(ξ)dξ.Itisclosely n0 give a true model of atoms or molecules. This will be illus- = 0 trated by the different techniques discussed in the following related to diffusion [15]. We will only briefly! outline the basic sections. ideas here. Assume particles in a gas show a small gradient dn/dx of their number density n,describedbythelinearrelation 2.3.1 Brownian Motion (Fig. 2.20)
The biologist and medical doctor Robert Brown (1773–1858) n(x) n(0) Gx. (2.36) discovered in 1827 that small particles suspended in liquids = − performed small irregular movements, which can be viewed Under the influence of mutual collisions the particles perform under a microscope. Although he first thought that these statistical movements with a probability distribution f (ξ) movements were caused by small living bacteria, he soon where ξ is the length of such a displacement in the x-direction found out that the movement could also be observed for inor- between two collisions. The number density of particles with ganic particles that are definitely not alive. movement ξ,isthen: The observation can be explained if one assumes that the particles are permanently hit by fast moving atoms or n(ξ)dξ nf(ξ)dξ with n n(ξ)dξ, (2.37) molecules coming from randomly distributed directions = = (Fig. 2.18). " The visualization of Brownian motion is very impressive. where the distribution function f (ξ) is defined as It is possible to demonstrate it to a large auditorium when 1 using cigarette smoke particles in air, illuminated by a laser f (ξ)dξ n(ξ)dξ. beam and viewed through a microscope with a video camera. = n Also here, the atoms are not directly seen but their impact ∞ f (ξ)dξ Λ on the smoke particle can be measured and, provided the mass ⇒ = of the smoke particle is known, the atomic momentum trans- "0 ferred to the particle, can be determined. Can One See Atoms?
Brownian Motion: small particles suspended in liquids performed small irregular movements, which can be
viewed under a microscope 20 2 The Concept of the Atom 2.3 Can One See Atoms? 19 (a) (b) Scattering particle
Laser beam
Scattered light
Lens
Image of the scattering CCD image plane microparticles Fig. 2.19 a Irregular path of a puck on an air table, which is hit statistically by smaller pucks (lecture demonstration of Brownian motion) Fig. 2.18 Schematic illustration of Brownian motion b Distribution of free path lengths ξ between two successive collisions and mean force path lengths Λ Fig. 2.17 Scattering of visible light by single atoms. Each image point corresponds to one atom Jinniu Hu 09/03/2021 Integration over all volume elements along the negative x-axis There is a nice demonstration that simulates Brownian yields with (2.36) about its size or structure can be obtained. One can only say: motion. A larger disk on an air table is hit by many small “It’s there.” discs, which simulate the air molecules. If the large disc car- 0 ries a small light bulb, its statistical path over the course of ∞ There are several other methods that give similar infor- N (n(O) Gx) f (ξ) dξ dx. (2.40a) mation [13]. With computer graphics one can produce nice time can be photographed and the path lengths ξ between + = − x $ ξ $ x pictures of such “atom images” on the screen, which may be two successive collisions (the free path) can be measured =−∞ =− impressive because they appear to give a magnified picture of (Fig. 2.19)[15]. Renaming the variable x x gives the microworld of atoms and molecules. However, one should The basic theory of Brownian motion was developed inde- = − # always keep in mind that such pictures are produced due to the pendently in 1905 by Albert Einstein (1879–1955) and Mar- interaction of light or particles with atoms. Only if this inter- ian Smoluchowski (1872–1917). The distribution n(ξ) and ∞ ∞ N (n(0) Gx#) f (ξ) dξ dx#. (2.40b) action is fully understood can the interpretation of the images the resulting mean free path Λ 1 ∞n(ξ)dξ.Itisclosely + = + = n0 x $ 0 ξ $x give a true model of atoms or molecules. This will be illus- 0 #= = # trated by the different techniques discussed in the following related to diffusion [15]. We will only briefly! outline the basic ideas here. In a similar way we obtain for the rate N of particles moving sections. Fig. 2.20 Illustrating drawings for the derivation of (2.45) − Assume particles in a gas show a small gradient dn/dx from right to left in Fig. 2.20 of their number density n,describedbythelinearrelation 2.3.1 Brownian Motion (Fig. 2.20) For a positive gradient G in (2.36)thenumberN of particles ∞ −∞ + moving through a unit area in the plane x 0intothe N (n(0) Gx) f (ξ) dξ dx, (2.41a) The biologist and medical doctor Robert Brown (1773–1858) = − = − n(x) n(0) Gx. (2.36) x-direction is larger than the corresponding number N in x$0 ξ $ x discovered in 1827 that small particles suspended in liquids = − + − = =− x-direction. Therefore, the net particle diffusion flux performed small irregular movements, which can be viewed − Under the influence of mutual collisions the particles perform through a unit area in the plane x 0is(Fig.2.20) which can be transformed by the substitution ξ ξ (note →− # under a microscope. Although he first thought that these statistical movements with a probability distribution f (ξ) = that the distribution function f (ξ) is symmetric and therefore movements were caused by small living bacteria, he soon where ξ is the length of such a displacement in the x-direction N N f ( ξ) f (ξ)) into j + − − ex . (2.38) − = found out that the movement could also be observed for inor- between two collisions. The number density of particles with diff = ∆t ˆ ganic particles that are definitely not alive. movement ξ,isthen: The observation can be explained if one assumes that Out of all n(x)dx particles within the volume dV A dx ∞ ∞ = N (n(0) Gx) f (ξ #) dξ # dx. (2.41b) the particles are permanently hit by fast moving atoms or centered around the plane x xi with unit area A,only − = − n(ξ)dξ nf(ξ)dξ with n n(ξ)dξ, (2.37) = − x$0 ξ $ x molecules coming from randomly distributed directions = = those particles with an elongation ξ > x1 can pass through = #= (Fig. 2.18). " the plane x 0. Their number is = Since the name of a variable is irrelevant, we can rename The visualization of Brownian motion is very impressive. where the distribution function f (ξ) is defined as x x in (2.40b)andξ ξ in (2.41b). Subtracting (2.41b) It is possible to demonstrate it to a large auditorium when # → # → 1 ∞ from (2.40b)weobtainthedifference using cigarette smoke particles in air, illuminated by a laser f (ξ)dξ n(ξ)dξ. dN n(x) f (ξ)dξ dx. (2.39) = n + = beam and viewed through a microscope with a video camera. ξ $ x =− Also here, the atoms are not directly seen but their impact ∞ f (ξ)dξ Λ on the smoke particle can be measured and, provided the mass ⇒ = of the smoke particle is known, the atomic momentum trans- "0 ferred to the particle, can be determined. Can One See Atoms? Cloud Chamber: Incident particles with sufficient kinetic energy can ionize the atoms or molecules in the cloud chamber, which is filled with supersaturated 2.3 Can One See Atoms? 23 water vapor. + (a)
Conductive layer Heater Barium supply R
− Anode Tungsten tip ZnS screen Electric field lines
(b) Enlarged image > of the tip r r~ 10 nm
How(c) to Make(d) a Cloud Chamber https://www.thoughtco.com/how-to-make-a-cloud-chamber-4153805 Fig. 2.23 Cloud chamber tracks of α particles (= He nuclei), which are emitted from a source below the lower edge of the photograph. One α particle collides with a (not visible) nitrogen nucleus at the crossing 09/03/2021 17 Jinniu Hu point of the two arrows, forming an 8 Onucleusandaproton.TheO nucleus flies towards 11 o’clock, the proton towards 4 o’clock (from W. Finkelnburg: Einführung in die Atomphysik, Springer, Berlin Hei- delberg New York, 1976)
single atoms “visible.” Since their basic understanding Fig. 2.24 a Basic concept of the field emission microscope. b Enlarged demands the knowledge of atomic physics and solid state view of the tungsten tip. c Image of the tungsten surface around the physics, they can only be explained here in a more qualitative tip, 107-fold enlarged on the screen of the field emission microscope. d way while for their quantitative description the reader is Visualization of Ba atoms on the tungsten tip referred to the literature [18,19]. the tungsten tip is already E 1011 V/m. Such high electric a) Field Emission Microscope ≥ fields exceed the internal atomic fields (see Sect. 3.5)andare The oldest of these devices is the field emission electron sufficiently large to release electrons from the metal surface microscope (Fig. 2.24)developedbyErnst Müller in 1937 (field emission, see Sect. 2.5.3). These electrons are acceler- [20]. A very sharp tip at the end of a thin tungsten wire serves ated by the electric field, follow the electric field lines, and as a cathode in the middle of an evacuated glass bulb. The impinge on the fluorescent screen at the anode where every anode has the form of a sphere and is covered on the inside electron causes a small light flash, similar to the situation at with a fluorescent layer (like a television screen). When a the screen of an oscilloscope. Most of the electrons are emit- voltage V of several kilovolts is applied between cathode and ted from places at the cathode surface where the work func- anode, the electric field strength at the cathode surface is tion (i.e., the necessary energy to release an electron) is mini- V mum. These spots are imaged by the electrons on the spherical E er , anode (radius R)withamagnificationfactorM R/r.With = r ·ˆ = R 10 cm and r 10 nm a magnification of M 107 is = = = where r is the radius of the nearly spherical tip of the tungsten achieved (Fig. 2.24). wire (Fig. 2.24b). With special etching techniques it is pos- Even with this device, only the locations of electron emis- sible to fabricate tips with r < 10 nm! This means that for a sion are measured but no direct information on the structure of moderate voltage V 1kVtheelectricfieldatthesurfaceof atoms is obtained. If other atoms with a small work functions = 24 2 The Concept of the Atom
2.3 Can One See Atoms? 25 are brought to the cathode surface (for example by evaporating The drawbacks of the transmission electron microscope The fluorescencebarium light canatoms be viewed from an through oven nearan optical the cathode) then the electron are the following: microscope or theemission secondary electrons,mainly comes emitted from from the these sur- atoms. One can now see face element dx dthesey of the atoms sample, and are theirextracted thermal by an electric motions on the cathode surface •Duetostrongabsorptionofelectronsbysolidmaterials, extraction field and imaged7 onto a detector where a signal the penetration depth is very small. One therefore has to S(x, y, t) is producedwith that 10 dependsfold magnification on the intensity (Fig. of the2.24d). prepare the sample as a thin sheet. secondary electrons emittedBut still from these the small magnified focal area spots dx d doy not give a direct image •Theelectronbeamhastobeintenseinordertoobtain around the point (ofx, y the),whichinturndependsonthecharac- size of the Ba-atoms but merely information on their sufficient image quality with a high contrast. This means a teristic propertieslocations of the sample and at their that location motions. [27,28]. larger current density j and total electron current I Aj, = d) Scanning Tunneling Microscope where A is the illuminated area. b) Transmission Electron Microscope The highest spatial resolution of structures on electrical con- •Theunavoidableabsorptionheatsthesampleup,which The electron microscope, first invented by Ernst Ruska in may change its characteristics or may even destroy parts ducting solid surfaces has so far been achieved with the scan- of the sample. This is particularly critical for biological ning tunneling microscope,1932 has invented meanwhile at the been research improved labora- so much that it reaches a samples. tories of IBM in Rüchlikon,spatial resolution Switzerland of [29 0.1,30 nm]in1984by [24–26]. The electrons are emit- Gerd Binning ( 1947) and Heinrich Rohrer ( 1933), who ∗ ted from a heated cathode∗ wire with a sharp kink (hair needle were awarded the Nobel Prize in 1986 for this invention [29]. Most of these drawbacks can be avoided with the scanning Can Onecathode) andSee are accelerated Atoms? by a high voltage (up to 500 kV). electron microscope. Similar to the electron field microscope a tungsten nee- dle with a very sharplyWith specially etched tip formed is used, electricwhich is orhow- magnetic fields, serving as c) Scanning Electron Microscope ever, not fixed butelectron is scanned optics in a controllable (see Sect. way2.6 at)theelectronsareimagedonto a very Fig. 2.26 Image of nerve cells in a thin undyed frozen slice taken In the scanning electron microscope (Fig. 2.28)theelectron small distance (a few tenths of a nanometer) over the surface with a transmission electron microscope (with kind permission of Zeiss, beam is focused onto the surface of the sample (which nowMicroscopes is the sample, whichwith is prepared Atomic as a thin foil (Fig. Resolution:2.25). While (Fig. 2.29). transmitting through the sample, the electrons are deflected Oberkochen) not necessarily a thin sheet), where it produces light emission If a small voltage of a few volts is applied between the tip by excitation of the sample molecules and secondary electronsField(cathode) andEmission the surfaceby elastic (anode) collisions the electronsMicroscope or can loose jump energy from by inelastic collisions. by impact ionization. The electron beam is scanned over the the needle into theThe surface transmitted by a process electrons called tunneling are imaged (see again onto a fluorescent surface of the sample by an appropriate deflection program Sect. 4.2.3). The electricscreen current where depends a magnified exponentially image on the of the absorption or scat- for the electron optics. This is quite similar to the situation in Transmissiondistance between tiptering and surface. centers WhenElectron in the the tip sample is scanned is overproduced Microscope (Fig. 2.26), which aTVtube[26]. the surface by piezo elements (these are ceramic cylinders can be viewed either through an optical microscope or with that change their length when an electric voltage is applied to Field emission V1 V0 them), any deviation of the surface in the z-direction from the tip Electron Scanning Electron Microscope source exact xy-plane results in a change of the tunnel current. Generally the tunnel current is kept constant by aHair controlled needle cathode Scanningmovement of the tip inTunneling vertical direction, which always Microscope keeps Electron it at the same distance ∆z from the real surface and therefore First condensor source First lens anode Atomic Force MicroscopeAperture y x Second Magnetic Second condensor anode y lens condensor lens x y Screen x z z Sample Lens for Electron beam imaging and Amplifier and scanning control feedback y Magnetic Fig. 2.27 Field-emission electron source where the electrons are emit- x-Piezo objective lens ted from a point-like tungsten tip and imaged by electrostatic lenses Electron Tunnel Field Emission Microscope Tip current detector
Sample aCCDcameraandanelectronicimageconvertingsystem. With electron microscope even small biological cells can be Fig. 2.28 Scanning electron microscope Scanning Tunneling Microscope Fig. 2.29 Scanning tunneling microscope Imaging made visible [22]. lens The spatial resolution of the electron microscope increases 09/03/2021 Jinniu Hu with decreasing size of the electron source. A nearly point- like source can be realized with field emission from a sharp edged tungsten tip (Fig. 2.27) like that in the field emission microscope. The emitted electrons can than be imaged by the Fluorescence electron optics to form a nearly parallel beam that traverses screen the sample. Each point of the sample is then imaged with a large magnification onto the screen. Fig. 2.25 Principle setup of the transmission electron microscope The size of atom
Assume that the masses of 1 mole atoms is A, and the atom is spherical 4 A ⇡r3N = 3 A ⇢
The radius of atom The density of substance
The radius of atom 1 3A 3 r = 4⇡⇢N ✓ A ◆ The units for the radius of atom 9 10 1 nm = 10 m, 1 A˚ = 10 m, 12 15 1 pm = 10 m, 1 fm = 10 m 09/03/2021 Jinniu Hu The size of atom
Elements A Density ! (g/cm3) Radius r (nm)
Li 7 0.7
Al 27 2.7
Cu 63 8.9
S 32 2.07
Pb 207 11.34
09/03/2021 Jinniu Hu The size of atom
Elements A Density ! (g/cm3) Radius r (nm)
Li 7 0.7 0.16
Al 27 2.7 0.16
Cu 63 8.9 0.14
S 32 2.07 0.18
Pb 207 11.34 0.19
09/03/2021 Jinniu Hu The size of atom The unit is pm H 1 He 25 Li Be B C N O F 2 Ne 145 105 85 70 65 60 50 Na Mg Al Si P S Cl 3 Ar 180 150 125 110 100 100 100 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br 4 Kr 220 180 160 140 135 140 140 140 135 135 135 135 130 125 115 115 115 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I 5 Xe 235 200 180 155 145 145 135 130 135 140 160 155 155 145 145 140 140 Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po 6 * At Rn 260 215 155 145 135 135 130 135 135 135 150 190 180 160 190 Ra 7 Fr ** Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og 215
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Lanthanides * 195 185 185 185 185 185 185 180 175 175 175 175 175 175 175 Ac Th Pa U Np Pu Am Actinides ** Cm Bk Cf Es Fm Md No Lr 195 180 180 175 175 175 175
09/03/2021 Jinniu Hu 2.4 The Size of Atoms 29 is transferred between adjacent layers x a and x is necessary to maintain a constant mass flux dM/dt of a = = a dx(Fig.2.36). This momentum transfer depends on gas with density ρ through a circular pipe with radius r and + the velocity difference between adjacent layers x a length L. This force is πr 2 ∆p,where∆p is the pressure = · and x a ∆x and on the frictional force between the difference between the two ends of the pipe. The mass flux = + molecules and therefore on the viscosity η.Thefrictional dM/dt ρ dV/dt is related to the viscosity by = · force exerted per area A in the plane x const. can be = written as ρ dV/dt) πr 4∆p/(8ηL) (2.59) · =
Fr η A dvy/dx (2.57a) In summary: = · · This force is due to collisions between the atoms. The momentum transferred per sec through the area A between Measurements of diffusion coefficient D or heat con- adjacent layers x a and x a ∆x of the gas is: duction coefficient λ or viscosity η yield the corre- = = + sponding collision cross sections and therefore the size d p/dt n m A
η n m
→ c ∆x ∆x → X a x = a
Fig. 2.36 Momentum transfer in a gas flow with velocity gradients dvy/dx.Atx 0 is a wall which forces, due to friction, vy 0 Fig. 2.37 Elementary cell of a regular crystal = =
09/03/2021 Jinniu Hu 28 2 The Concept of the Atom
kT For density gradients, diffusion takes place where mass is Λ (2.52c) transported, for temperature gradients, heat conduction occurs = p σ √2 · · where energy is transported and for velocity gradients, the momentum of the molecules is transferred. Measuring the mean free pathlength Λ gives informa- All these transport phenomena are realized on a micro- tion on the collision cross section and therefore on the scopic scale by collisions between atoms or molecules and size of the colliding atoms. therefore the mean free path length Λ (i.e., the mean distance an atom travels between two collisions) plays an important role for the quantitative description of all these phenomena. The above mentioned transport phenomena are directly In a gas at thermal equilibrium with atom number density related to Λ.Theycanthereforebeusedtomeasurethesize n the mean free path length is given by of atoms or molecules.
2 2 Λ (1/n σ ) < v >/
∆x where λ is the coefficient of heat conduction. It is related (b) to the specific Heat cv of the gas at constant volume by (a) B r1 A 1 2cv kTm λ nmcvvΛ . (2.56) = 3 = 3σ π B " r 2 A x Measuring the coefficient λ therefore yields the collision cross section σ and with it the atomic radius.
• Viscosity of a Gas d =r1 +r2
If a velocity gradient dvy/dx existsinagasflowingintothe Collision probability y-direction with the velocity vy,partoftheirmomentum per unit volume σ = π⋅d2 P=n⋅σ⋅∆x
Fig. 2.35 Determination of atomic size from the collision cross section p n m
09/03/2021 Jinniu Hu Element symbol
09/03/2021 Jinniu Hu The discovery of electron
Cathode ray tube
09/03/2021 Jinniu Hu Cathode ray in external field 32 2 The Concept of the Atom
Electric forces are responsible for the interaction between the (a) constituents of an atom. Why the attractive Coulomb force between the positive and negative atomic constituents does not lead to the collapse of atoms has only been later answered by quantum mechanics (see Sect. 3.4.3).
2.5.1 Cathode Rays and Kanalstrahlen
Investigations of gas discharges by J. Plucker (1801–1868), Johann Wilhelm Hittorf (1824–1914), Joseph John Thom- (b) son (1856–1940), Phillip Lenard (1862–1947) (Nobel Prize 1905), and many others have all contributed much to our understanding of the electric structure of atoms. It is worth- while to note that the essential experimental progress was only possible after the improvement of vacuum technology (the invention of the mercury diffusion pump, for example, 6 allowed one to generate vacua down to 10− hPa). In a gas discharge tube at low pressures, Hittorf observed particle rays emitted from the cathode that followed (with- out external fields) straight lines. He could prove this by the Fig. 2.42 a,b. Experimental arrangement of Thomson for the determi- 09/03/2021nation of the ratio e/m of cathodeJinniu rays through Hu their deflection (a)ina shadow that was produced on a fluorescent screen when obsta- magnetic field and (b)inanelectricfield cles were put in the path of the cathode rays. From the fact that these particle rays could be deflected by magnetic fields, Hittorf correctly concluded that they must be charged parti- Sect. 2.6). This was the first example of a cathode ray oscil- cles and from the direction of the deflection it became clear loscope. Thomson could also show that the ratio e/m was that they were negatively charged (Fig. 2.41). The first quanti- independent of the cathode material, but was about 104 times tative, although not very accurate, determination of the mag- larger than that for the “Kanalstrahlen” discovered in 1886 nitude of their charge was obtained in 1895 by J.B. Perrin by Eugen Goldstein (1850–1930) in a discharge tube, which (and with an improved apparatus in 1897 by Thomson,who fly through a hole in the cathode in the opposite direction collimated the particles through a slit in the anode, deflected of the cathode rays (Fig. 2.43). Wilhelm Wien (1864–1928) them after the anode by 90◦ through a magnetic field and measured in 1897 the value of e/m for the particles in the detected them by an electrometer (Fig. 2.42a)). Kanalstrahlen and he proved that they are positively charged With the design of Fig. 2.42b, where the cathode rays are atoms of the gas inside the discharge tube [44]. better collimated by two slits B1 and B2,thusproducinga The negative light particles of the cathode rays were named small spot on the fluorescent screen, Thomson could mea- electrons after a proposal by J. Stoney and G. Fitzgerald in sure the ratio e/m of charge e to mass m of the particles by 1897. The positively charged heavy particles were named ions applying electric and magnetic fields for beam deflection (see according to the existing name for charged atoms or molecules in the electrolysis.
− + Cathode rays = electrons R
Anode Fluorescent screen
Cathode
N Valve S To vacuum pump
Fig. 2.41 Schematic drawing of the experimental setup for observing cathode rays. The deflection of the rays by an external magnet can be Fig. 2.43 Apparatus for demonstrating “channel-rays” (positive observed on the screen charged ions) in a discharge with a hole (channel) in the massive cathode The discovery of electron
1897, J. J. Thomson found electron (corpuscles)
http://web.lemoyne.edu/~giunta/thomson1897.html 1. They travel in straight lines. 2. They are independent of the material composition of the cathode. 3. Applying electric field in the path of cathode ray deflects the ray towards positively charged plate. Hence cathode ray consists of negatively charged particles. 09/03/2021 Jinniu Hu 2.5 The Electric Structure of Atoms 39
2.5 The Electric Structure of Atoms 39
2.5 The Electric Structure of Atoms 39
2.5 The Electric Structure of Atoms 39
Fig. 2.56 Helical path of electrons that enter a homogeneous magnetic field under an angle α 90 against the field lines Fig. 2.54 Experimental device (“Fadenstrahlrohr”) for measuring the != ◦ Fig. 2.56 Helical path of electrons that enter a homogeneous magnetic ratio e/m field under an angle α 90◦ against the field lines Fig. 2.54 Experimental device (“Fadenstrahlrohr”) for measuring the != ratio e/m The path of theThe electrons can path be made of visible, the if the glasselectrons can be made visible, if the glass bulb is filled with a gas at low pressure so that the mean free This shows that the electrons are bent into the x-direction path of the electrons is comparable to the circumference of and acquire a velocity component vx but remain within the the circle. Throughbulb collisions is filled with the electrons, with the a atoms gas at low pressure so that the mean free plane z const. The Lorentz force is always perpendicular This shows that the electrons are= bent into the x-directionare excited and emit light (see Sect. 3.4). This visible circular to their velocity v vx ,vy and therefore does not change ={ } path of the electronspath allows of the the measurement electrons of its radius R is comparable to the circumference of the magnitude of the velocity. The path of the electrons is and acquire a velocity component v but remain within theand of the ratio therefore a circlex with a radius R (Fig. 2.55)definedbythe compensation of centrifugal and Lorentz force the circle.e 2V Through collisions with the electrons, the atoms . (2.72) plane z const. The Lorentz force is always perpendicular m = R2 B2 = mv2 are excited and emit light (see Sect. 3.4). This visible circular evBz. (2.71a) to their velocity v vx ,vy and thereforeR = does not changeIf the electrons enter the homogeneous magnetic field under the angle α against the field direction, the electron velocity Fig. 2.56 Helical path of electrons that enter a homogeneous magnetic ={ } This gives the radius v vx , 0,vpathz can be ofcomposed the of the electrons two components vx allows the measurement of its radius R the magnitude of the velocity. The path of the electrons is={ } field under an angle α 90 against the field lines Fig. 2.54 Experimental deviceand vz (Fig. (“Fadenstrahlrohr”)2.56). The vx component is perpendicular for measuring to the the ◦ mv 1 != R 2Vm/e, (2.71b) and of the ratio therefore a circle with a radius TheR (Fig.ratio discovery=2.55eeB/m= B)definedbythe of electron ! compensation of centrifugal andbecause Lorentz the velocity v of force the electrons is determined by the e 2TheV path of the electrons can be made visible, if the glass acceleration voltage V according to (m/2)v2 eV. 38 Charge-to-Mass2 The Concept= of the Ratio Atom for the Electron bulb is filled. with a gas at low(2.72) pressure so that the mean free This shows that the electrons are bent into the x-directionm = R2 B2 2 path of the electrons is comparable to the circumference of mv and acquire a velocity component vx but remain within the evB . (2.71a) the circle. Through collisions with the electrons, the atoms 2.5.5 The Mass of thez Electronplane z const. The LorentzIf force the is electrons always perpendicular enter the homogeneous magnetic field under Fig. 2.56 HelicalR = path of electrons that enter a homogeneousFig. 2.57 Wien filter magnetic = are excited and emit light (see Sect. 3.4). This visible circular to their velocity v vx ,vy theand angle thereforeα against does not the change field direction, the electron velocity Fig. 2.54 Experimental device (“Fadenstrahlrohr”) for measuring the field under an angle α 90◦ against the field lines All methods for the determination!= of the electron={ mass} use path of the electrons allows the measurement of its radius R ratio e/m This gives the radius the magnitude of the velocity.v The pathvx , 0 of,v thez can electrons be composed is of the two components vx and of the ratio the deflection of electrons in electrictherefore or a magnetic circle with fields, a radius whereR={(Fig. 2.55 )definedbythe} The path of the electrons can be made visible, ifand thevz glass(Fig. 2.56). The vx component is perpendicular to the the Lorentz forcemv 1 compensation of centrifugal and Lorentz force e 2V bulb is filled with a gas at low pressure so that the mean free . (2.72) R Fig.2.55 2CircularVm path/ of ane electron, beam in a homogeneous(2.71b) magnetic 2 2 This shows that the electrons are bent into the x-direction Lorentz Force CircleFig.2.58 Precisionmotion methods for the measurement of e/m with two radio field perpendicular to the initial velocity v0 of the electrons m = R B path of= theeB electrons= B is comparable to the circumferencefrequency2 deflection plates separated of by the distance L and acquire a velocity component vx but remain within the F q(E v B) mv(2.70a) the circle. Through! collisions with the electrons, theevB atomsz. (2.71a) If the electrons enter the homogeneous magnetic field under plane z const. The Lorentz force is always perpendicular = + × R = = because the velocityare excitedv of and the emitAcceleration electrons light (see is determined Sect. voltage3.4). by This V the visible circular the angle α against the field direction, the electron velocity to their velocity v vx ,vy and therefore does not changeacts on a particle with charge q, which moves with a velocity ={ } path of the electrons allowsThis gives the2 measurement the radius of its radius R v vx , 0,vz can be composed of the two components vx the magnitude of the velocity. The path of theacceleration electronsv isacross voltage theV fieldsaccording (Fig. 2.53 to (m)./ Inserting2)v eV Newton’s. equation ={ } and of the ratio = and vz (Fig. 2.56). The vx component is perpendicular to the therefore a circle with a radius R (Fig. 2.55)definedbytheF mr into (2.70a )weobtainthethreecoupleddifferential Charge-to-Mass rationmv 1 R 2Vm/ e, (2.71b) compensation of centrifugal and Lorentz force = ¨ e 2V = eB = B equations 2 2 . ! (2.72) m = R B 2 q mv x E v Bbecausev B the velocity, v of the electrons is determined by the evBz. (2.71a) If the electrons enterx y thez homogeneousz y magnetic field under R = ¨ = m +09/03/2021acceleration− voltage V accordingJinniu to Hu(m/2)v2 eV. the angle αq against! the field direction," the electron velocity = y Ey vz Bx vx Bz , Fig. 2.57 Wien filter This gives the radius v ¨vx=, 0m,vz can+ be composed− of the two components vx Fig. 2.51 Electron impact ion source ={ q !} " and vzz(Fig. 2.56Ez). Thevx Bvyx componentvy Bx . is perpendicular(2.70b) to the mv 1 ¨ = m + − R 2Vm/e, (2.71b) ! " = eB = B These equations show that it is not the mass m directly, but ! Fig. 2.57 Wien filter only the ratio q/m that can be obtained from measuring the because the velocity v of the electrons is determined by thepath of a charged particle in these fields. One therefore needs acceleration voltage V according to (m/2)v2 eV. an additional measurement (for instance the Millikan experi- = ment) in order to determine the mass which yields the charge q seperately. The mass m can then be obtained from one of Fig.2.55 Circular path of an electron beam in a homogeneous magnetic the following experiments, which illustrate (2.70)byseveralFig.2.58 Precision methods for the measurement of e/m with two radio field perpendicular to the initial velocity v of the electrons examples. 0 frequency deflection plates separated by the distance L a) Fadenstrahlrohr Fig. 2.57 Wien filter Fig.2.55 Circular path of an electron beam in a homogeneous magnetic Electrons emitted from a hot cathode in a glass bulb are accel- Fig.2.58 Precision methods for the measurement of e/m with two radio field perpendicular to the initial velocity v of the electrons erated in the y-direction and enter a magnetic field that points 0 frequency deflection plates separated by the distance L into the z-direction (Fig. 2.54). Since here v 0,vy, 0 and ={ } B 0, 0, Bz ,(2.70b)reduceswithq e to ={ } = − e x vy Bz. (2.70c) ¨ = −m
Fig. 2.52Fig.2.55Duo plasmatronCircular path ion of source an electron beam in a homogeneous magnetic Fig.2.58 Precision methods for the measurement of e/m with two radio field perpendicular to the initial velocity v of the electrons 0 frequency deflection plates separated by the distance L pressures is maintained. One example is the duo-plasmatron source (Fig. 2.52)wherealowvoltagegasdischargeisiniti- ated between the heated cathode and the anode. The ions are extracted by a high voltage (several kV) through the small hole in an auxiliary electrode that compresses the plasma and therefore increases its spatial density. A magnetic field keeps the plasma away from the walls and further increases the ion density. Even substances with low vapor pressure can be vaporized (for example by electron- or ion impact) and can then be ionized inside the discharge. Fig. 2.53 Lorentz-force F acting on an electron e− that moves with Amoredetaileddiscussionofdifferenttechniquesforthe velocity v in a homogeneous magnetic field B, pointing perpendicularly production of ions can be found in [46–48]. into the drawing plane Contemporary theories of atom
1. Plum pudding model (Lord Kelvin and J. J. Thomson)
2. Saturnian model of the atom (H. Nagaoka)
09/03/2021 Jinniu Hu Geiger–Marsden experiment
09/03/2021 Jinniu Hu Implications of plum pudding model
Using classical physics, the alpha particle's lateral change in momentum p can be approximated using the impulse of force relationship and the Coulomb force expression:
Q↵Qn 2r p = F t = k 2 r v↵ The maximum deflection angle:
p 2Q↵Qn ✓ Qn: positive charge of gold atom m": mass of alpha particle Q": charge of alpha particle. v": velocity of alpha particle 09/03/2021 Jinniu Hu Implications of plum pudding model Using classical physics, the alpha particle's lateral change in momentum p can be approximated using the impulse of force relationship and the Coulomb force expression: Q↵Qn 2r p = F t = k 2 r v↵ The maximum deflection angle: p 2Q↵Qn ✓ where, r: radius of a gold atom k: Coulomb’s constant Qn: positive charge of gold atom m": mass of alpha particle Q": charge of alpha particle. v": velocity of alpha particle 09/03/2021 Jinniu Hu 2.8 The Structure of Atoms 61 More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where = − of the incident particles A, that is scattered into a defined solid r r1 r2 is the distance between the two particles. The =| − | angle dΩ,andwhichisdeterminedbythedifferentialcross description of the scattering of the two particles by each other section. in this center of mass frame is named “potential scattering,” because it demands, besides the reduced mass µ and the initial conditions (r0, v0) solely the knowledge of the interaction While for the determination of the integral cross section, potential Epot(r). the decrease of the intensity of the incident particles A We will here restrict the discussion to the most simple (that is, the unscattered particles) is measured, the dif- case of spherically symmetric potentials Epot(r),whichis ferential cross section dσ ()/dΩ is a measure for those adequate for many real collision events. In such a potential, particles that have been deflected by a certain angle Θ the angular momentum L of the particle remains constant into the solid angle dΩ. (see Problem 2.20). This implies that the path of the parti- cle is planar. It always stays within the so-called “scattering plane.” Therefore, polar coordinates (r, ϕ) are best suited for We will now derive an expression for the differential cross the description of the particles time-dependent position. The section. deflection angle of our particles, measured in the center of Assume NAincident particles pass per second through the mass frame is named ϑ,whileitisdescribedbyΘ in the ˙ area A in the scattering volume V A ∆x,and∆N(Θ, ∆Ω) laboratory frame (Fig. 2.95). = · ˙ 2.8 The Structure of Atoms is the rate of particles scattered by a deflection angle Θ and The deflection61 of the incident particle A depends on its detected by the detector with an acceptance solid angle ∆Ω. impact parameter b.ThisisthesmallestdistanceofAtothe Then target particle B, if there is no deflection, i.e., if A passes More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where = − along a straight line (Fig. 2.96a). For the potential scattering of the incident particles A, that is scattered into a defined solid r r r is the distance between the two particles. The ∆N nB1 dσ 2 dσ (i.e., the description of the scattering process in the center 2.8 The Structure of Atoms ˙ =| V− ∆Ω| nB∆x ∆Ω (2.133) 61 angle dΩ,andwhichisdeterminedbythedifferentialcrossNAdescription= A dΩ of the= scatteringdΩ of the two particlesof by mass each frame) other the particle B is fixed at the origin of our ˙ section. in this center of mass frame is named “potentialcoordinate scattering,” frame and it can be regarded as a point-like particle More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where is the fractionbecause of incident it demands, particles= besides that− is the scattered reduced into mass the µ withand the infinite initial mass. of the incident particles A, that is scattered into a defined solidsolid angler ∆Ωr1 acceptedr2 is the by the distance detector. between It is determined the two particles. by The =|conditions− | (r0, v0) solely the knowledge of theWhen interaction the initial velocity of particle A is v( ) v0, angle dΩ,andwhichisdeterminedbythedifferentialcrossthe particledescription density n ofof the scatterers scattering B, theof the length two∆ particlesx which by the each other | −∞ |= While for the determination of the integral cross section, potential BE (r). energy conservation demands section. incidentin beam this ofcenter particles ofpot mass A traverse frame is through named the “potential scattering scattering,” the decrease of the intensity of the incident particles A We will here restrict the discussion to the most simple volume becauseV and the it differential demands, besides scattering the cross reduced section mass dσµ/dandΩ the initial 1 2 1 2 case of spherically symmetric potentials E (r),whichisµv Epot(r) µv0 const, (2.134) (that is, the unscattered particles) is measured, the dif-which dependsconditions on the(r0 interaction, v0) solely potential the knowledge between particles of the interactionpot 2 + = 2 = ferentialWhile for cross the sectiondetermination dσ ()/ ofdΩ theis integral a measure cross for section, thoseA and B.potentialadequateEpot for(r) many. real collision events. In such a potential, because Epot(r ) 0. The angular momentum L,with particlesthe decrease that have of the been intensity deflected of the by incident a certain particles angle AΘFor measuringtheWe angular will dσ here/dΩ momentumthe restrict setup the of Fig.L discussionof2.94 thebcanbeused. particle to the most remains simple constant =±∞ = respect to the scattering center at r 0is into(that the is, solid the unscatteredangle dΩ. particles) is measured, the dif-Two beams,case(see collimated of Problem spherically by 2.20). the symmetric apertures This implies S potentials1 and S2 thatcrossE thepot each(r path),whichis of the parti- = ferential cross section dσ ()/dΩ is a measure for thoseother inadequate the scattering for many volume realV collisionA∆x events..TheparticlesA In such a potential, cle is planar. It always= stays within the so-called “scattering dr dϕ particles that have been deflected by a certain angle Θscatteredthe by angular an angle momentumΘ into the solidL of angle the of particle∆Ω are remains mon- constant L µ(r v) µ r er r et plane.” Therefore, polar coordinates2(r, ϕ) are best suited= for × = × dt ˆ + dt ˆ Ω itored by(see the Problem detector with 2.20). sensitive This implies area AD that theR ∆Ω pathin of a the parti- Weinto will the now solid derive angle an d expression. for the differential cross the description of the particles time-dependent= position. The ! " #$ distance R from the scattering volume V where R ∆x. µrϕ r et , (2.135) cle is planar. It always stays within the so-called! “scattering = ˙ × ˆ section. We willdeflection now look angleinto the of relation our particles, between measuredE (r) and in the center of plane.” Therefore, polar coordinates (r, ϕpot) are best suited for % & Assume NAincident particles pass per second throughdσ/ thedΩ inmass more detail. frame is named ϑ,whileitisdescribedbyΘ in the We will˙ now derive an expression for the differential cross the description of the particles time-dependent position. The area A in the scattering volume V A ∆x,and∆N(Θ, ∆Ω) laboratory frame (Fig. 2.95). section. ˙ deflection angle of our particles, measured in the center of = · The deflection of the incident particle A depends on its is theAssume rate of particlesNAincident scattered particles by pass a deflection per second angle throughΘ theand mass frame is named ϑ,whileitisdescribedbyΘ in the ˙ 2.8.2 Basicimpact Concepts parameter ofb Classical.ThisisthesmallestdistanceofAtothe Scattering detectedarea A byin the the scattering detector volume with anV acceptanceA ∆x,and solid∆N angle(Θ, ∆Ω∆Ω) . laboratory frame (Fig. 2.95). Then = · ˙ target particle B, if there is no deflection, i.e., if A passes is the rate of particles scattered by a deflection angle ΘAsand is generallyThe shown deflection in classical of the mechanics, incident particle the movements A depends on its impactalong parameter a straightb.ThisisthesmallestdistanceofAtothe line (Fig. 2.96a). For the potential scattering detected by the detector with an acceptance solid angle ∆Ωof two. particles with masses m1, m2,velocitiesv1, v2 and a Then∆N nB dσ dσ target(i.e., particle the description B, if there of is the no deflection,scattering i.e., process if A in passes the center ˙ V ∆Ω nB∆x ∆Ω (2.133)mutual interaction potential Epot( r1 r2 ) can be represented NA = A dΩ = dΩ alongofClassical mass a straight frame) line scattering the (Fig.| particle2.96− a).| ForB is the fixed potential at the scattering origin of our ˙ in the center of mass coordinate frame as the movement of a 62 2 The Concept of the Atom ∆N nB dσ dσ single particle(i.e.,coordinate the with description reduced frame mass and of the it can scattering be regarded process as a in point-like the center particle ˙ V ∆Ω nB∆x ∆Ω (2.133) In a real experiment the path (r(t), ϕ(t)) of a single particle is the fractionNA = ofA incidentdΩ particles= thatdΩ is scattered into the of massThe frame) reduced the particle mass B is fixed at the origin(a) of our ˙ with infinite mass. cannot be followed. However, the measured deflection angle solid angle ∆Ω accepted by the detector. It is determined by coordinate frame andm it1 canm2 be regarded as a point-like particle ϑ allows to determine the asymptotic values of the path for When theµ initial velocity of particle A is v( ) v0, is the fraction of incident particles that is scattered into the Fig. 2.95 Scattering of a charged particle A in a potentialr .SinceforasphericallysymmetricpotentialthisV (r) 1/r, the particle density n of scatterers B, the length ∆x which the with infinite mass.= m1 m2 | −∞ |= →∞ B energy conservation+ demands where r is the distance between A and B path must be mirror-symmetric∝ to the line OS through the solid angle ∆Ω accepted by the detector. It is determined by point S of closest approach in Fig. 2.96b. (This means that incident beam of particles A traverse through the scattering When the initial velocity of particle A is v( ) v0, Energy conservation demands| −∞ |= the scattering process is invariant against time-reversal.) We the particle density nB of scatterers B, the length ∆x which the energy conservation demands volume V and the differential scattering cross section dσ/dΩ 1 2 1 2 can relate the asymptotic scattering angle ϑ to the polar angle incident beam of particles A traverse through the scattering µv Epot(r) µv0 const, (2.134) ϕmin ϕ(rmin) by which depends on the interaction potential between particles 2 + = 2 = = volume V and the differential scattering cross section dσ/dΩ 1 1 ϑ π 2ϕmin. 2 2 = − A and B. µv Epot(r) µv0 const, (2.134) which depends on the interaction potential between particles v0 2is the+ initial= velocity2 = (b) because Epot(r ) 0. The angular momentum L,with This yields the relation AFor and measuring B. dσ/dΩ the setup of Fig. 2.94bcanbeused. =±∞ = respect to the scattering center at r 0is ϕmin rmin Two beams,For measuring collimated dσ/d byΩ thethe setup apertures of Fig. S12.94andbcanbeused. S2 cross each because Epot(r ) 0. The angular momentum L,with dϕ dt The angular=±∞ momentum= L= ϕmin dϕ dr respect to the scattering center at r 0is = = dt dr otherTwo in beams, the scattering collimated volume by the aperturesV A∆ S1xand.TheparticlesA S2 cross each ϕ'0 r ' = = dr dϕ = =−∞ scatteredother in by the an scattering angle Θ into volume theV solid angleA∆x.TheparticlesA of ∆Ω are mon- rmin L µ(r v) µ r er r et +∞ϕ = 2 dr dt dϕ dt (ϕ/r) dr ˙ dr. itoredscattered by the by detector an angle withΘ into sensitive the solid area angleA ofD ∆ΩRare∆Ω mon-in a L µ(=r v×) µ= r ×e rˆ +e ˆ = ˙ ˙ = r ! "r t #$ r ' rmin' ˙ = 2 = × = × dt ˆ + dt ˆ Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ =−∞ distanceitored byR from the detector the scattering with sensitive volume areaV whereAD RR ∆Ω∆inx. a µrϕ r e!t , " #$ (2.135) = = mAmB/(mA mB) in a potential V (r) with the origin in B. (b)Relation With (2.139a)and(2.139b)thescatteringangleintheCM- !∆ µ= ϕ ˙ ×,ˆ between scattering+ angle ϑ in the center of mass system and the polar distanceWe willR nowfrom look the scattering into the relation volume V betweenwhere RE (rx). and r r et (2.135) frame becomes: !pot = ˙ × ˆ angle ϕmin at the closest approach between A and B (point S) We will now look into the relation between E (r) and % & dσ/dΩ in more detail. pot % & +∞ (L/(µr 2)) dr ϑ(E0, L) π 2 1/2 . dσ/dΩ in more detail. = − 2 L2 because er is parallel to r. The unit vector et points along the E E (r) ˆ ˆ rmin' µ 0 pot 2µr2 tangent to the path of A. For L we obtain: − − ( ! ")(2.140) Jinniu Hu 09/03/2021 2 L L µr ϕ µv0b, (2.136) 1 2 2.8.2 Basic Concepts of Classical Scattering | |= = ˙ = With the total energy E0 µv the amount of the angular = 2 0 2.8.2 Basic Concepts of Classical Scattering momentum because L(x ) µ v0 r sin ϕ µ v0 b.The =! −∞ = · · · = · · 2 2 2 2 2 kinetic energy in the center of mass frame is L µrv sin ϕ µbv0 L µ b v 2µb E0 As is generally shown in classical mechanics, the movements = = ⇒ = 0 = As is generally shown in classical mechanics, the movements (2.141) 1 2 1 2 2 2 of twoof two particles particles with with masses massesmm11,, mm22,velocitiesvv11,,vv2 2andand a a Ekin µv µ r r ϕ = 2 = 2 ˙ + ˙ is uniquely defined by the initial energy E0 and the impact mutualmutual interaction interaction potential potentialEEpot(( rr1 r2 ))cancan be be represented represented 1 L2! " pot 1 2 µr 2 . (2.137) parameter b of the incident particle B. Inserting these relations || −− || 2 in thein the center center of of mass mass coordinate coordinate frame frame as the the movement movement of of a a = 2 ˙ + 2µr into (2.140)weobtain The total energy E T Epot in the center of mass frame +∞ singlesingle particle particle with with reduced reduced mass mass = + dr can then be written as ϑ(E0, b) π 2b 1/2 . (2.142) = − 2 b2 Epot(r) r ' r 1 2 m m 2 min − r − E0 m11m22 1 2 L µ Etotal E0 µr Epot(r) const. (2.138) ( ) µ = = 2 ˙ + 2µr 2 + = ==mm1 mm2 Fig.Fig. 2.95 2.95ScatteringScattering of a of charged a charged particle particle A in a A potential in a potentialV (r) V1(/rr), 1/r, This shows that the deflection angle ϑ is determined by the 1 + 2 ∝ interaction potential Epot(r),theimpactparameterb and the + where r is the distance between A and B Solving (2.138∝)and(2.136)forr and ϕ yields where r is the distance between A and B ˙ ˙ initial energy E0. The lower integration limit rmin is fixed by the condition 2 1/2 2 L r(r ) 0. This gives with (2.139)and(2.141) r E0 Epot(r) (2.139a) min ˙ = µ − − 2µr 2 ˙ = # $ %& b L rmin 1/2 . (2.143) ϕ . (2.139b) = Epot(r ) ˙ = µr 2 1 min − E0 ( ) 2.8 The Structure of Atoms 61 More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where = − of the incident particles A, that is scattered into a defined solid r r1 r2 is the distance between the two particles. The =| − | angle dΩ,andwhichisdeterminedbythedifferentialcross description of the scattering of the two particles by each other section. in this center of mass frame is named “potential scattering,” because it demands, besides the reduced mass µ and the initial conditions (r0, v0) solely the knowledge of the interaction While for the determination of the integral cross section, potential Epot(r). the decrease of the intensity of the incident particles A We will here restrict the discussion to the most simple (that is, the unscattered particles) is measured, the dif- case of spherically symmetric potentials Epot(r),whichis ferential cross section dσ ()/dΩ is a measure for those adequate for many real collision events. In such a potential, particles that have been deflected by a certain angle Θ the angular momentum L of the particle remains constant into the solid angle dΩ. (see Problem 2.20). This implies that the path of the parti- cle is planar. It always stays within the so-called “scattering plane.” Therefore, polar coordinates (r, ϕ) are best suited for We will now derive an expression for the differential cross the description of the particles time-dependent position. The section. deflection angle of our particles, measured in the center of Assume NAincident particles pass per second through the mass frame is named ϑ,whileitisdescribedbyΘ in the ˙ area A in the scattering volume V A ∆x,and∆N(Θ, ∆Ω) laboratory frame (Fig. 2.95). = · ˙ 2.8 The Structure of Atoms is the rate of particles scattered by a deflection angle Θ and The deflection61 of the incident particle A depends on its detected by the detector with an acceptance solid angle ∆Ω. impact parameter b.ThisisthesmallestdistanceofAtothe Then target particle B, if there is no deflection, i.e., if A passes More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where = − along a straight line (Fig. 2.96a). For the potential scattering of the incident particles A, that is scattered into a defined solid r r r is the distance between the two particles. The ∆N nB1 dσ 2 dσ (i.e., the description of the scattering process in the center 2.8 The Structure of Atoms ˙ =| V− ∆Ω| nB∆x ∆Ω (2.133) 61 angle dΩ,andwhichisdeterminedbythedifferentialcrossNAdescription= A dΩ of the= scatteringdΩ of the two particlesof by mass each frame) other the particle B is fixed at the origin of our ˙ section. in this center of mass frame is named “potentialcoordinate scattering,” frame and it can be regarded as a point-like particle More information is obtained by measuring that fraction and relative velocity v v1 v2 in a potential Epot(r),where is the fractionbecause of incident it demands, particles= besides that− is the scattered reduced into mass the µ withand the infinite initial mass. of the incident particles A, that is scattered into a defined solidsolid angler ∆Ωr1 acceptedr2 is the by the distance detector. between It is determined the two particles. by The =|conditions− | (r0, v0) solely the knowledge of theWhen interaction the initial velocity of particle A is v( ) v0, angle dΩ,andwhichisdeterminedbythedifferentialcrossthe particledescription density n ofof the scatterers scattering B, theof the length two∆ particlesx which by the each other | −∞ |= While for the determination of the integral cross section, potential BE (r). energy conservation demands section. incidentin beam this ofcenter particles ofpot mass A traverse frame is through named the “potential scattering scattering,” the decrease of the intensity of the incident particles A We will here restrict the discussion to the most simple volume becauseV and the it differential demands, besides scattering the cross reduced section mass dσµ/dandΩ the initial 1 2 1 2 case of spherically symmetric potentials E (r),whichisµv Epot(r) µv0 const, (2.134) (that is, the unscattered particles) is measured, the dif-which dependsconditions on the(r0 interaction, v0) solely potential the knowledge between particles of the interactionpot 2 + = 2 = ferentialWhile for cross the sectiondetermination dσ ()/ ofdΩ theis integral a measure cross for section, thoseA and B.potentialadequateEpot for(r) many. real collision events. In such a potential, because Epot(r ) 0. The angular momentum L,with particlesthe decrease that have of the been intensity deflected of the by incident a certain particles angle AΘFor measuringtheWe angular will dσ here/dΩ momentumthe restrict setup the of Fig.L discussionof2.94 thebcanbeused. particle to the most remains simple constant =±∞ = respect to the scattering center at r 0is into(that the is, solid the unscatteredangle dΩ. particles) is measured, the dif-Two beams,case(see collimated of Problem spherically by 2.20). the symmetric apertures This implies S potentials1 and S2 thatcrossE thepot each(r path),whichis of the parti- = ferential cross section dσ ()/dΩ is a measure for thoseother inadequate the scattering for many volume realV collisionA∆x events..TheparticlesA In such a potential, 62 cle is planar. It always= stays within the so-called “scattering 2 Thedr Conceptd ofϕ the Atom particles that have been deflected by a certain angle Θscatteredthe by angular an angle momentumΘ into the solidL of angle the of particle∆Ω are remains mon- constant L µ(r v) µ r er r et plane.” Therefore, polar coordinates2(r, ϕ) are best suited= for × = × dt ˆ + dt ˆ Ω itored by(see the Problem detector with 2.20). sensitive This implies area AD that theR ∆Ω pathin of a the parti- Weinto will the now solid derive angle an d expression. for the differential cross the description of the particles time-dependent= In a real position. experiment The the path (r(t!), ϕ(t")) of a single particle#$ (a) distance R from the scattering volume V where R ∆x. µrϕ r et , (2.135) cle is planar. It always stays within the so-called! cannot “scattering be followed.= However,˙ × ˆ the measured deflection angle section. We willdeflection now look angleinto the of relation our particles, between measuredE (r) and in the center of plane.” Therefore, polar coordinates (r, ϕpot) are best suited for % & Assume NAincident particles pass per second throughdσ/ thedΩ inmass more detail. frame is named ϑ,whileitisdescribedbyϑ allows toΘ determinein the the asymptotic values of the path for We will˙ now derive an expression for the differential cross the description of the particles time-dependentr position..Sinceforasphericallysymmetricpotentialthis The area A in the scattering volume V A ∆x,and∆N(Θ, ∆Ω) laboratory frame (Fig. 2.95). →∞ section. = · ˙ deflection angle of our particles, measured inpath the center must be of mirror-symmetric to the line OS through the is the rate of particles scattered by a deflection angle Θ and The deflection of the incident particle A depends on its Assume NAincident particles pass per second through the mass frame is named ϑ,whileitisdescribedbypointΘ Sin of the closest approach in Fig. 2.96b. (This means that ˙ 2.8.2 Basicimpact Concepts parameter ofb Classical.ThisisthesmallestdistanceofAtothe Scattering detectedarea A byin the the scattering detector volume with anV acceptanceA ∆x,and solid∆N angle(Θ, ∆Ω∆Ω) . laboratory frame (Fig. 2.95). the scattering process is invariant against time-reversal.) We Then = · ˙ target particle B, if there is no deflection, i.e., if A passes is the rate of particles scattered by a deflection angle ΘAsand is generallyThe shown deflection in classical of the mechanics, incident particle the movements A dependscan relate on the its asymptotic scattering angle ϑ to the polar angle impactalong parameter a straightb.ThisisthesmallestdistanceofAtothe line (Fig. 2.96a). For the potential scattering detected by the detector with an acceptance solid angle ∆Ωof two. particles with masses m1, m2,velocitiesv1, v2 andϕmin a ϕ(rmin) by Then∆N nB dσ dσ target(i.e., particle the description B, if there of is the no deflection,scattering i.e., process if A= in passes the center ˙ V ∆Ω n ∆x ∆Ω (2.133)mutual interaction potential Epot( r1 r2 ) can be represented ϑ π 2ϕmin. B alongClassical a straight line scattering (Fig.| 2.96− a).| For the potential scattering = − NA = A dΩ = dΩ in the centerof of mass mass frame) coordinate the frame particle as the B movement is fixed at of the a 62 origin of our 2 The Concept of the Atom ˙ ∆N n dσ dσ (i.e., the description of the scattering process in the center ˙ B (b) single particlecoordinate with reduced frame mass and it can be regarded asThis a point-like yields the particle relation V ∆Ω nB∆x ∆Ω (2.133) The reduced mass In a real experiment the path (r(t), ϕ(t)) of a single particle is the fractionNA = ofA incidentdΩ particles= thatdΩ is scattered into the ofwith mass infinite frame) mass. the particle B is fixed at the origin(a) of our ˙ m m cannot be followed. However, the measured deflection angle solid angle ∆Ω accepted by the detector. It is determined by coordinate frame and it1 can2 be regarded as a point-like particle ϕmin r ϑ allows to determine the asymptotic values of the path for When theµ initial velocity of particle A is v( ) v0, min is the fraction of incident particles that is scattered into the Fig. 2.95 Scattering of a charged particledϕ d At in a potentialr .SinceforasphericallysymmetricpotentialthisV (r) 1/r, the particle density n of scatterers B, the length ∆x which the with infinite mass.= m1 m2 | −∞ |= →∞ B energy conservation+ demands where r ϕismin the distanced betweenϕ A and B dr path must be mirror-symmetric∝ to the line OS through the solid angle ∆Ω accepted by the detector. It is determined by = = dt dr point S of closest approach in Fig. 2.96b. (This means that incident beam of particles A traverse through the scattering When the initial velocity of particle A is v( ) v0, Energy conservation demands| −∞ |= ϕ'0 r ' the scattering process is invariant against time-reversal.) We the particle density nB of scatterers B, the length ∆x which the energy conservation demands = =−∞ volume V and the differential scattering cross section dσ/dΩ 1 2 1 2 rmin can relate the asymptotic scattering angle ϑ to the polar angle incident beam of particles A traverse through the scattering +∞ µv Epot(r) µv0 const, (2.134) ϕ ϕmin ϕ(rmin) by which depends on the interaction potential between particles 2 + = 2 = = volume V and the differential scattering cross section dσ/dΩ 1 1 (ϕ/r) dr ˙ dr. ϑ π 2ϕmin. 2 2 r = − A and B. µv Epot(r) µv0 const, (2.134) = ˙ ˙ = which depends on the interaction potential between particles 2 + = 2 = (b) r ' rmin' ˙ becausev0 is Ethe(r initial )velocity0. The angular momentum L,with=−∞ This yields the relation AFor and measuring B. dσ/dΩ the setup of Fig. 2.94bcanbeused.Fig. 2.96 (a)ScatteringofaparticleAwithreducedmasspot µ m m /(m mB) in a potential V (r) with=± the∞ origin= in B. (b)Relation= Two beams, collimated by the apertures S and SA crossB A each becauserespect E to( ther scattering) 0. center The angular at r momentum0isWith (2.139aL,with)and(2.139b)thescatteringangleintheCM- ϕmin rmin For measuring dσ/dΩ the setup of Fig.12.94bcanbeused.between2 scattering+ angle ϑ inpot the center of mass system and the polar dϕ dt The angular=±∞ momentum= L= ϕmin dϕ dr respect to the scattering center at r 0is frame becomes: = = dt dr otherTwo in beams, the scattering collimated volume by the aperturesV A∆ S1xand.TheparticlesA Sangle2 crossϕmin eachat the closest approach between A and B (point S) ϕ'0 r ' = = dr dϕ = =−∞ scatteredother in by the an scattering angle Θ into volume theV solid angleA∆x.TheparticlesA of ∆Ω are mon- +∞ 2 rmin L µ(r v) µ r er r et (L/(µr )) dr +∞ϕ = 2 dr dϕ (ϕ/r) dr ˙ dr. scattered by an angle Θ into the solid angle of ∆Ω are mon- = × = × dt ˆ + ϑd(tE0ˆ, L) π 2 . = ˙ ˙ = r itored by the detector with sensitive area AD R ∆Ω in a L µ(r v) µ r e r e 2 1/2 ! "r t #$= − 2 L r ' rmin' ˙ =because2 e is parallel to r. The unit vector e points along the (a)ScatteringofaparticleAwithreducedmassµ =−∞ itored by the detector with sensitive area A R ∆Ω inr a = × = t × dt ˆ + dt ˆ Fig. 2.96 rmin' E0 Epot(r) 2 distance R from the scattering volume V whereD R ∆ˆx. µrϕ r ˆe!t , " #$m m /(m (2.135)m ) V (r)µ = 2µr = A B A B in a potential with the origin− in B. (b)Relation− With (2.139a)and(2.139b)thescatteringangleintheCM- tangent!∆ to the path of A. For µL=weϕ obtain:˙ ×,ˆ between scattering+ angle ϑ in the center of mass system and the polar distanceWe willR nowfrom look the scattering into the relation volume V betweenwhere RE (rx). and r r et (2.135) ( ! frame") becomes:(2.140) !pot = ˙ × ˆ angle ϕmin at the closest approach between A and B (point S) We will now look into the relation between E (r) and % & dσ/dΩ in more detail. pot 2 +∞ (L/(µr 2)) dr % & ϑ(E , L) π 2 . L L µr ϕ µv0b, (2.136) 1 2 0 1/2 dσ/dΩ in more detail. = − 2 L2 | |= = ˙ = Withbecause the totaler is energy parallel to rE. The0 unit vectorµv ethet points amount along the of the angular E E (r) ˆ 2 0 ˆ rmin' µ 0 pot 2µr2 tangent to the path of A. For L=we obtain: − − momentum ( ! ")(2.140) because L(x ) µ v0 r sin ϕ µ v0 b.TheJinniu Hu 09/03/2021 2 =! −∞ = · · · = · · L L µr ϕ µv0b, (2.136) 1 2 2.8.2 Basic Concepts of Classical Scatteringkinetic energy in the center of mass frame is | |= = ˙ = 2 2 2 2 With the2 total energy E0 µv the amount of the angular 2.8.2 Basic Concepts of Classical Scattering L µrv sin ϕ µbv0 L µ b v0 2µb E0 = 2 0 = = ⇒ = = momentum because L(x ) µ v0 r sin ϕ µ v0 b.The (2.141) =! −∞ = · · · = · · 2 2 2 2 2 1 1 kinetic energy in the center of mass frame is L µrv sin ϕ µbv0 L µ b v 2µb E0 As is generally shown in classical mechanics, the movements 2 2 2 2 = = ⇒ = 0 = As is generally shown in classical mechanics, the movements Ekin µv µ r r ϕ (2.141) = 2 = 2 ˙ + ˙ 1 2 1 2 2 2 of two particles with masses m1, m2,velocitiesv1, v2 and a is uniquely definedEkin byµv theµ initialr r energyϕ E0 and the impact of two particles with masses m1, m2,velocitiesv1, v2 and a 2! " = 2 = 2 ˙ + ˙ 1 L is uniquely defined by the initial energy E0 and the impact mutualmutual interaction interaction potential potentialEEpot(( rr1 r2 ))cancan be be represented represented 2 parameter b of the incident1 particleL2! B." Inserting these relations pot 1 2 µr . (2.137) µr 2 . (2.137) parameter b of the incident particle B. Inserting these relations || −− || 2 2 in thein the center center of of mass mass coordinate coordinate frame frame as the the movement movement of of a a = 2 ˙ + 2µr into (2.140)weobtain= 2 ˙ + 2µr into (2.140)weobtain singlesingle particle particle with with reduced reduced mass mass The total energy E T Epot in the center of mass frame +∞ dr The total energy E T Epot in the center of mass frame = ++∞ ϑ(E , b) π 2b . (2.142) can then be written as dr 0 1/2 = + = − 2 b2 Epot(r) ϑ(E , b) π 2b . (2.142) rmin' r 1 2 m m can then be written as 0 2 1/2 − r − E0 m11m22 = −1 2 L 2 Epot(r) Etotal E0 µr 2 Epot(r) b const. (2.138) ( ) µµ r ' 2r 1 2 = m m Fig. 2.95 Scattering of a charged particle A in a potential V (r) 1=/r, = 2 ˙ +min2µr + =r E0 This shows that the deflection angle ϑ is determined by the = m11 m2 1Fig. 2.95 LScattering2 of a charged particle A in a potential V (r) 1/r, − − + where2r is the distance between A and B ∝ interaction potential Epot(r),theimpactparameterb and the + Etotal E0 whereµr r is the distanceEpot(r between) const A and. (2.138) B Solving (2.138∝)and(2.136)forr and( ϕ yields ) = = 2 ˙ + 2µr 2 + = This shows that the deflection˙ angle˙ ϑ is determinedinitial energy by theE0. The lower integration limit rmin is fixed by the condition 2 1/2 2 L r(r ) 0. This gives with (2.139)and(2.141) interactionr potentialE0 EEpotpot(r()r),theimpactparameter(2.139a) minb and the ˙ = µ − − 2µr 2 ˙ = Solving (2.138)and(2.136)forr and ϕ yields # $ %& b ˙ ˙ initial energy EL 0. rmin 1/2 . (2.143) ϕ . (2.139b) = Epot(rmin) The lower˙ = µ integrationr 2 limit rmin is fixed by the condition 1 2 1/2 − E0 2 L r(r ) 0. This gives with (2.139)and(2.141) ( ) r E0 Epot(r) (2.139a) min ˙ = µ − − 2µr 2 ˙ = # $ %& b L rmin 1/2 . (2.143) ϕ . (2.139b) = Epot(r ) ˙ = µr 2 1 min − E0 ( ) 62 2 The Concept of the Atom 62 In a real experiment the2 path The( Conceptr(t), ϕ( oft)) theof Atom a single particle (a) cannot be followed. However, the measured deflection angle In a real experiment the path (r(t), ϕ(t)) of a single particle (a) ϑ allows to determine the asymptotic values of the path for cannot ber followed..Sinceforasphericallysymmetricpotentialthis However, the measured deflection angle 62 ϑ allows to→∞ determine the asymptotic2 The values Concept of of the the path Atom for path must be mirror-symmetric to the line OS through the r .Sinceforasphericallysymmetricpotentialthis In a real→∞ experimentpoint S the of closest path (r( approacht), ϕ(t)) of in a Fig. single2.96 particleb. (This means that (a) path must be mirror-symmetric to the line OS through the cannotpoint be S followed. ofthe closest scattering However, approach process the in measuredFig. is invariant2.96b. deflection (This against means angle time-reversal.) that We ϑ allowsthe scattering tocan determine relate process the the is asymptotic asymptotic invariant against values scattering time-reversal.) of the angle pathϑ forto We the polar angle r .Sinceforasphericallysymmetricpotentialthisϕ ϕ(r ) by →∞can relate themin asymptoticmin scattering angle ϑ to the polar angle path must be mirror-symmetric= to the line OS through the ϕmin ϕ(rmin) by ϑ π 2ϕmin. point S= of closest approach in Fig. 2.96=b. (This− means that ϑ π 2ϕmin. (b) the scattering process is invariant= − against time-reversal.) We can relate theThis asymptotic yields the scattering relation angle ϑ to the polar angle (b) This yields the relation ϕmin ϕ(rmin) by = ϕmin rmin ϑ ϕminπ 2ϕminrmin. dϕ dt =ϕmin− dϕdϕ dt dr ϕmin dϕ = = dr dt dr (b) This yields the relation= = ϕ'0 dt drr ' ϕ'0 r =' =−∞ = =−∞rmin r +∞ ϕmin min rmin ϕ dϕ (dϕ+t/∞rϕ) dr ˙ dr. ϕmin dϕ (ϕ=/r) dr ˙ d˙r˙ dr. = r = = = ˙ ˙r 'd=t dr r rmin' ˙ Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ ϕ'0r ' r ' =−∞ rmin' ˙ Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ = = =−∞ =−∞ mAmB/(mA mB) in a potential V (r) with the origin in B.= (b)Relation mAmB/(mA mB) in a potential V (r) with the origin in B. (b)Relation With (2.139armin )and(2.139b)thescatteringangleintheCM- between scattering+ angle ϑ in the center of mass system and theWith polar (2.139a)and(2.139b)thescatteringangleintheCM-+∞ between scattering+ angle ϑ in the center of mass system and the polar ϕ angle ϕmin at the closest approach between A and B (point S) frame becomes:(ϕ/r) dr ˙ dr. angle ϕmin at the closest approach between A and B (point S) frame becomes:= ˙ ˙ = r r ' +∞rmin' ˙ 2 Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ =−∞+∞ (L/(µr 2)) d(rL/(µr )) dr = ϑ(E , L)ϑ(Eπ0, L) π 2 . . mAmB/(mA mB) in a potential V (r) with the origin in B. (b)Relation 0 2 1/2 1/2 because+ e is parallel to r. The unit vector e points alongWith ( the2.139a)and(= 2.139b− =)thescatteringangleintheCM-2− 2 L2 L2 betweenbecause scatteringer is parallelangler ϑ in to ther. The center unit of mass vector systemet points and the alongt polar the ' Er0min' Epot(rE)0 Epot(r) 2 ˆ ˆ ˆ ˆ rmin µ µ −2µr2 − 2µr angletangentϕmintangentat to the the closest to path the approach of path A. For of between A.L we For A obtain: andL we B (point obtain: S) frame becomes: − − ( ! ( ! ")(2.140) ")(2.140) +∞ (L/(µr 2)) dr 2 2 L LL µrL ϕ µµvr ϕ0b, µv b, (2.136) (2.136)ϑ(E0, L) π 2 1/2 . 0 = − 2 1 2 1 2L2 because er is parallel| to|=r. The| =|= unit vector=˙ = et˙ points= along the With the totalWith energy the totalE0 energyE2 µv0EEthe0 ( amountr)µv the of the amount angular of the angular ˆ ˆ rmin' µ= 0 pot= 2 02µr2 tangent to the path of A. For L we obtain: momentummomentum − − because L(x ) µ v0 r sin ϕ µ v0 b.The ( ! ")(2.140) because=!L(−∞x = )· ·µ · v0 r =sin·ϕ · µ v0 b.The kinetic energy in the=! center−∞ of mass= · frame· is· = · · 2 2 2 2 2 kinetic energy in the center2 of mass frame is L µrv sin ϕ µbv0 L µ b v02 2µ2b 2E02 2 L L µr ϕ µv0b, (2.136) = L µ=rv sin ϕ⇒ µb=v0 L = µ b v0 2µb E0 = 1 =2 ⇒ = (2.141)= | |= = ˙ = With the total energy E0 2 µv0 the amount of the angular (2.141) 1 2 1 2 2 2 = Classical scatteringEkin µv 1 µ2 r 1 r ϕ2 2 2 momentum because L(x =)E2kinµ v=0 µv2r sin˙ ϕ+µ µr˙ v0r bϕ.The =! −∞ = · · · = · · is uniquely defined by the initial energy E0 and the impact = 2 2!= 2 ˙ "+ ˙ is uniquely defined2 by the2 2 initial2 energy2 E and the impact kinetic energy in the center1 of massL frame is L µrv sin ϕ µbv0 L µ b v 2µb E0 0 The kinetic energy in centerµr 2 1 of mass. L2! frame (2.137) " parameter= b of the= incident⇒ particle= B. Inserting0 = these relations 2 2µ2r 2 parameter b of the incident particle B. Inserting(2.141) these relations = ˙ +µr 2 . (2.137)into (2.140)weobtain 1 2 = 12 ˙ 2+ 2µ2 r2 into (2.140)weobtain Ekin µv µ r r ϕ = 2 = 2 ˙ + ˙ The total energy E T Epot in the center of mass frameis uniquely defined by the+∞ initial energydrE0 and the impact The total energy1 = E+ LT2! Epot in" the center of mass frame +∞ can then be written as 2 parameterϑ(E0,bbof) theπ incident2b particle B. Inserting these1/d2 relationsr. (2.142) µr = +. (2.137) = − 2 Epot(r) can then be written2 as2µr 2 ϑ(E0, b) πr 2 21b b . (2.142) = ˙ + into (2.140)weobtain rmin' r2 E 2 1/2 The total energy 2 = − − −2 0 b Epot(r) 1 L r ' r 1 2 E E µr 2 E (r) const. (2.138) ( min )r E0 The totaltotal energy0 E T 1 E 2in theL2pot center of mass frame − − E= = E2 ˙ +µ2rpotµ2r + E= (r) const. (2.138)This shows that the+ deflection∞ angledr ϑ is( determined by the) total =0 + 2 pot ϑ(E , b) π 2b . (2.142) can then be written= as = 2 ˙ + 2µr + = interaction0 This potential showsE that(r the),theimpactparameter deflection angle1/2 ϑ bisand determined the by the = − pot2 b2 Epot(r) Solving (2.138)and(2.136)forr and ϕ yields r ' r 1 2 The derivatives of radii and2 angle˙ are˙ initial energyinteractionE0. min potential− rEpot−(r)E,theimpactparameter0 b and the Solving1 (2.1382 )and(L 2.136)forr and ϕ yields Etotal E0 µr Epot(r) const1/2 . (2.138) The lowerinitial integration energy E limit(0. rmin is fixed) by the condition = = 22 ˙ + 2µr 2 + L2 =˙ ˙ This shows that the deflection angle ϑ is determined by the ( ) r(rmin) 0.The This lower gives with integration (2.139)and( limit2.141r )is fixed by the condition r E0 Epot r 2 1/(2.139a)2 ˙ = min ˙ = µ 2− − 2µr L2 interaction potential Epot(r),theimpactparameterb and the Solving (2.138#)and(r $ 2.136E)forrEand(ϕr)yields%& (2.139a) r(rmin) 0. This givesb with (2.139)and(2.141) L 0 ˙ pot ˙ 2 initial energy˙ E0. r= . (2.143) ϕ ˙ =. µ − − 2µr (2.139b) min 1/2 2 # $ %& The lower integration= limit rminEpotis(rmin fixed) by the condition ˙ = µr 2 1/2 1 b 2 L L rmin E0 . (2.143) r(rmin) 0. This gives with (2.139− )and(2.141) 1/2 r ϕ E0 Epot. (r) (2.139a) (2.139b) = Epot(rmin) Since for a ˙ spherically= µ ˙ = −µr 2symmetric− 2µr 2 potential this path˙ = ( 1 ) # $ %& b − E0 L must be mirror-symmetric to the line OS rmin (1/2 . (2.143)) ϕ . (2.139b) = Epot(r ) ˙ = µr 2 1 min 09/03/2021 Jinniu Hu − E0 ( ) 62 2 The Concept of the Atom 62 2 The Concept of the Atom 62 2 The Concept of the Atom In a real experiment the path (r(t), ϕ(t)) of a single particle (a) In a real experiment the path (r(t), ϕ(t)) of a single particle (a) cannotIn a real be followed.experiment However, the path the(r measured(t), ϕ(t)) deflectionof a single angle particle (a) cannot be followed. However, the measured deflection angle ϑcannotallowsϑ allows be to followed. determineto determine However, the the asymptotic asymptotic the measured values values of deflection the the path path for for angle r .Sinceforasphericallysymmetricpotentialthis ϑ→∞rallows to.Sinceforasphericallysymmetricpotentialthis determine the asymptotic values of the path for path must→∞ be mirror-symmetric to the line OS through the r path must.Sinceforasphericallysymmetricpotentialthis be mirror-symmetric to the line OS through the pointpoint→∞ S of S ofclosest closest approach approach in in Fig. Fig.2.962.96b.b. (This (This means means that that path must be mirror-symmetric to the line OS through the Classicalthe the scatteringscattering scattering process process is is invariant invariant against against time-reversal.) time-reversal.) We We canpointcan relate relateS of the closestthe asymptotic asymptotic approach scattering scattering in Fig. angle angle2.96ϑϑ totob. the (This polar polar means angle angle that the scattering process is invariant against time-reversal.) We ϕminϕmin ϕ(rϕmin(rmin) by) by The relation= = the asymptotic scattering angle to the can relate the asymptoticϑ scatteringπ 2ϕ angle. ϑ to the polar angle ϑ π 2ϕminmin. polar angleϕmin byϕ(r min ) by == −− = (b) (b) This yields the relation This yields the relation ϑ π 2ϕmin. = − This yields the relationϕ min rmin (b) This yields the relationϕmin rmin dϕ dt ϕmin dϕ dϕ dt dr ϕmin = dϕ = dt dr dr ϕ'0 r ' dt dr = ϕmin = =−∞rmin ϕ'0= r ' rmin = =−∞ +d∞ϕ dt ϕmin rmin dϕ ϕ dr (ϕ/r) dr +∞dt ˙ddrr. = = ϕr =ϕ'0 (ϕ/˙r)˙ drr '= ˙ dr. r =' =−∞rmin' ˙ Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ = =−∞ ˙ ˙ = r rmin = r ' rmin' +˙∞ Fig. 2.96mA(am)ScatteringofaparticleAwithreducedmassB/(mA mB) in a potential V (r) with the origin in B. (bµ)Relation With (2.139a)and(=2.139b−∞ )thescatteringangleintheCM-ϕ between scattering+ angle ϑ in the center of mass system and the polar m m /(m mB) in a potential V (r) with the origin in B. (b)RelationThe= scattering angle in the( ϕCM-frame/r) dr becomes˙ dr. A B angleA ϕ at the closest approach between A and B (point S) Withframe (2.139a becomes:)and(2.139b= )thescatteringangleintheCM-˙ ˙ = r + min ϑ between scattering angle in the center of mass system and the polar r ' rmin' ˙ +∞ 2 Fig.angle 2.96ϕmin(aat)ScatteringofaparticleAwithreducedmass the closest approach between A and B (point S) µ frame becomes: =−∞ (L/(µr )) dr = ϑ(E , L) π 2 . mAmB/(mA mB) in a potential V (r) with the origin in B. (b)Relation 0 1/2 because+ e is parallel to r. The unit vector e points along theWith (2.139a)and(= − 2.139b+∞ 2)thescatteringangleintheCM-(L/(µr 2)) drL2 between scattering angler ϑ in the center of mass systemt and the polar rmin' E0 Epot(r) 2 ˆ ˆ ϑ(E0, L) π 2 µ − − 2µr . tangent to the path of A. For L we obtain: frame becomes: 2 1/2 angle ϕmin at the closest approach between A and B (point S) = − 2 L (2.140) because er is parallel to r. The unit vector et points along the ' ( !E0 Epot(r) ") ˆ ˆ rmin µ − − 2µr2 tangent to the path of A. For L we obtain:2 09/03/2021 +∞Jinniu Hu (L/(µr 2)) dr L L µr ϕ µv0b, (2.136) ( !1 2 ")(2.140) | |= = ˙ = ϑWith(E0 the, L total) energyπ 2 E0 µv the amount of the angular . 2 0 2 1/2 because e is parallel to r. The unit2 vector e points along the = − = 2 L r L L µr ϕ µv b,t (2.136) momentum r ' E0 Epot(r) 2 ˆ because L(x ) µ v0 r0 ˆsin ϕ µ v0 b.The min 1µ 2 2µr tangent to the path| of|= A.=! For−∞= L =we˙ obtain:=· · · = · · With the total energy E0 µv the− amount of− the angular kinetic energy in the center of mass frame is = (2 !02 2 2 2 2 ")(2.140) momentumL µrv sin ϕ µbv0 L µ b v0 2µb E0 because L(x ) µ v0 r sin ϕ µ v0 b.The = = ⇒ = = =! −∞ = · 2 · · = · · (2.141) kinetic energy inL the centerL µ of1r massϕ2 frame1µv0b2, is 2 2 (2.136) 1 2 2 2 2 2 2 Ekin µv µ r r ϕ WithL theµr totalv sin energyϕ µbEv0 Lµv µtheb amountv0 2µ ofb theE0 angular | |= = ˙ = = = 0 ⇒ 2 =0 = = 2 = 2 ˙ + ˙ is uniquely defined by the= initial energy E0 and the impact 2! " momentum (2.141) because L(x )1 µ2 1v012 r 2sinL ϕ2 2 µ v0 b.The parameter b of the incident particle B. Inserting these relations Ekin µv µr µ r r. ϕ (2.137) =! −∞2= = ·2 2˙ · +·2µr 2 = · · into (2.140)weobtain 2 2 2 2 2 kinetic energy in the= center of= mass frame˙ + is˙ is uniquelyL µ definedrv sin ϕ by theµbv initialL energyµEb0 andv the2µ impactb E 2! " 0 0 0 1 2 L parameter= b of the incident= particle⇒ B. Inserting= these= relations The total energyµEr T E. in the center of(2.137) mass frame +∞ (2.141) 1 1 2 pot dr = 2 ˙2 =+ 2µ+r 2 2 2 into (ϑ2.140( ,)weobtain) π . canE thenkin be writtenµv as µ r r ϕ E0 b 2b 1/2 (2.142) = − 2 Epot(r) = 2 = 2 ˙ + ˙ r 2 1 b is uniquely defined byrmin' the initialr2 energyE0 E0 and the impact The total energy E T1 Epot 2inL!2 the center" of mass frame +∞ − − 1 2 2 L parameter b of the incident particled B.r Inserting these relations Etotal E0=µr µ+r . Epot(r) const(2.137). (2.138) ϑ(E , b) π 2b ( ) . (2.142) can then be written= as= 2 ˙ + 2µ2r 2 + = This0 shows that the deflection angle ϑ is determined1/2 by the = 2 ˙ + 2µr into (2.140=)weobtain− 2 b2 Epot(r) r ' r 1 2 2 interaction potentialminEpot(r),theimpactparameter− r − E0 b and the Solving (12.1382)and(L2.136)forr and ϕ yields TheE total energyE E µr T E inE the˙( centerr) ˙ const of mass. (2.138) frame initial energy E0. ( ) total 0 2 2µrpot2 pot +∞ dr = = =˙ ++ + = This showsThe lower that integration the deflection limit anglermin isϑ fixedis determined by the condition by the can then be written as 2 1/2 ϑ(E0, b) π 2b . (2.142) 2 L 2 1/2 r E E (r) (2.139a)interactionr(rmin) potential=0. This− givesEpot( withr),theimpactparameter2 (2.139)and(b 2.141Epot(r)) b and the 0 pot 2 ˙ = r ' r 1 2 Solving (2.138˙)and(= µ2.136)for−2 r and−ϕ2yieldsµr min r E0 1 # $ L ˙ ˙ %& initial energy E0. b− − E E µr 2L E (r) const. (2.138) ( ) total 0 2 pot The lower integrationrmin limit rmin is fixed1/2 . by the condition(2.143) ϕ2 . 2µr 2 1/2 (2.139b) = Epot(rmin) = =2˙ = ˙µr 2+ + L = This shows that the deflection1 angle ϑ is determined by the ( ) r(rmin) 0. This gives with (2.139E0)and(2.141) r E0 Epot r 2 (2.139a) ˙ = − ˙ = µ − − 2µr interaction potential Epot( (r),theimpactparameter) b and the Solving (2.138# )and($ 2.136)forr and ϕ%&yields b L ˙ ˙ initial energy Er0. . (2.143) ϕ . (2.139b) min 1/2 2 The lower integration= limitEpotr(rminmin) is fixed by the condition ˙ = µr 2 1/2 1 2 L r(r ) 0. This gives with− (2.139E0 )and(2.141) r E0 Epot(r) (2.139a) min ˙ = µ − − 2µr 2 ˙ = ( ) # $ %& b L rmin 1/2 . (2.143) ϕ . (2.139b) = Epot(r ) ˙ = µr 2 1 min − E0 ( ) 62 62 2 The Concept of the2 Atom The Concept of the Atom In a real experimentIn the a real path experiment(r(t), ϕ(t)) theof path a single(r(t), particleϕ(t)) of a single particle (a) (a) cannot be followed.cannot However, be followed. the measured However, deflection the measured angle deflection angle ϑ allows to determineϑ allows the asymptotic to determine values the of asymptotic the path for values of the path for 62 2 The Concept of the Atom r .Sinceforasphericallysymmetricpotentialthisr .Sinceforasphericallysymmetricpotentialthis →∞ →∞ 62 path must be mirror-symmetricpath must be2 to mirror-symmetric The the Concept line OS of the through Atom to the the line OS through the In apoint real Sexperiment of closest approach thepoint path S of( inr( closest Fig.t), ϕ2.96(t approach))b.of (This a single in means Fig. particle2.96 thatb. (This means that (a) In a real experiment the path (r(t), ϕ(t)) of a single particle (a) cannotthe scattering be followed. process However,the is scattering invariant the measured process against istime-reversal.) invariant deflection against angle We time-reversal.) We cannot be followed. However, the measured deflection angle ϑ allowscan relate to determine the asymptoticcan the relate scattering asymptotic the asymptotic angle valuesϑ to scattering ofthe the polar path angle angle forϑ to the polar angle ϑ allows to determine the asymptotic values of the path for r ϕmin .Sinceforasphericallysymmetricpotentialthisϕ(rmin) by ϕ ϕ(r ) by r = .Sinceforasphericallysymmetricpotentialthismin min →∞→∞ = pathpath must must be be mirror-symmetric mirror-symmetricϑ π to2 to theϕmin the line. line OSϑ OS throughπ through2ϕ themin. the = − = − pointpoint S S of of closest closest approach approach in in Fig. Fig.2.962.96b. (Thisb. (This means means that that (b) (b) thetheThis scattering scattering yields processthe process relationThis is is invariant invariant yields againstthe against relation time-reversal.) time-reversal.) We We cancan relate relate the the asymptotic asymptotic scattering scattering angle angleϑ toϑ theto polar the polar angle angle ϕ r ϕ ϕ(r ) by min min ϕmin rmin ϕminmin ϕ(rminmin) by dϕ dt dϕ dt == ϕ dϕ dr min ϑ π 2ϕminϕmin. dϕ dr =ϑ = π−= 2ϕmin.=dt dr = dt dr ϕ'=0 −r ' ϕ'0 r ' (b) = =−∞ = =−∞ This yields the relation r (b) min rmin This yields the relation +∞ϕ +∞ϕ ϕ (ϕ/rr) dr ˙ dr(.ϕ/r) dr ˙ dr. =min ˙ ˙min == r ˙ ˙ = r ' dϕ drt' ϕmin ϕrmindϕ rmin minrdr'˙ rmin' ˙ Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassFig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ µ = =−∞= dt ddϕr =d−∞t = = ' r ' mAmB/(mA mB) in am potentialAmB/(mVA(r)mwithB) in the a origin potential in B.V ( (rb))Relationwith the origin in B. (b)Relationϕmin ϕ 0 dϕ dr + + With (2.139a)and(=With2.139b (2.139a)thescatteringangleintheCM-=−∞)and(2.139b)thescatteringangleintheCM- between scattering anglebetweenϑ in the scattering center of angle massϑ systemin the and center the of polar mass system and the polar= rmin = dt dr ' ' +∞ angle ϕ at the closestangle approachϕ betweenat the closest A and approach B (point between S) A and Bframe (point becomes: S) ϕ frame0 becomes:r ϕ min min = (ϕ/r) d=r−∞ ˙ dr. = rmin ˙ ˙ = r +∞ (L/(µ+∞r+2∞)) dr 2 r ' rmin' ˙ ϕ (L/(µr )) dr ϑ(E , L) π =2−∞ϑ(E , L) π 2 . . Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ 0 (ϕ0 /r) dr ˙ dr. 1/2 1/2 = = − 2 = − 2 L2 L2 becausemAmBe/(r misA parallelmB) inbecause ato potentialr. TheerV unitis(r) parallelwith vector the origin toet rpoints. Thein B. ( unitb along)Relation vector the et points along the= ' ˙ ˙ E0 =Epot(rr) E E (r) ˆ + ˆ ˆ ˆWith (2.139a)and(2.139brmin )thescatteringangleintheCM-µ rmin' µ2µr2 0 pot 2µr2 tangentbetween to the scattering path ofangletangent A.ϑ Forin thetoL the centerwe path obtain: of mass of A. system For L andwe the obtain: polar r ' − rmin' ˙ − − − Fig. 2.96 (a)ScatteringofaparticleAwithreducedmassµ frame becomes: =−∞ ( ! ")(2.140) (2.140) angle ϕmin at the closest approach between A and B (point S) = ( ! ") mAmB/(mA mB) in a potential V (r) with the origin in B. (b)Relation + 2 2 With (2.139a)and(2.139b+∞ )thescatteringangleintheCM-(L/(µr 2)) dr between scattering angle ϑ inL the centerL µ ofr massϕ µv systemL 0b,L and theµr polarϕ(2.136)µv0b, ϑ(E , L)(2.136)π 2 1 2 1 2. | |= = ˙ = | |= = ˙ = With the0 total energyWithE0 the totalµv energythe amountE0 ofµv the1/2 angularthe amount of the angular angle ϕmin at the closest approach between A and B (point S) frame becomes:= − 2 2 0 L2 2 0 because er is parallel to r. The unit vector et points along the ' = E0 Epot(r) = ˆ ˆ momentum rminmomentumµ − − 2µr2 becausetangentL( tox the pathbecause of) A.µ ForLv(L0xwer obtain:sin ϕ) µµ vv00 br.Thesin ϕ µ v0 b.The +∞ 2 =! −∞ = · · · = · · ( ! (L/(µr )) dr")(2.140) =! −∞ = · · · = · · ϑ(E , L) π 2 2 2 2 2 22 .2 2 2 2 kinetic energy in thekinetic center energy of mass in frame the center is of mass frame is 0L µrv sin ϕ µbLv µrLv sin ϕµ bµvbv0 2µLb E1/µ2 b v 2µb E0 2 = − 20 0 L2 0 0 because er is parallel to r. TheL unitL vectorµr ϕ etµvpoints0b, along(2.136) the = = =⇒ = = =⇒ = = rmin' µ1 E2 0 Epot(r) 2 (2.141) (2.141) ˆ | |= = ˙ =ˆ With the total energy E0 2 µv0 the amount of the2µ angularr tangent to the path of A. For1L we2 obtain:1 2 12 22 1 2 2 2 = − − Ekin µv µEkinr r µvϕ µ r r momentumϕ ( ! ")(2.140) because L(x = 2 ) =µ 2v0 r˙ =+sin2ϕ ˙ µ= v20 b.The˙ + ˙ =! −∞ = · · · = · · is uniquely definedis by uniquely the initial defined energy byE the0 and initial the energy impact E0 and the impact kinetic energy in the center2 of mass2! frame is " 2! " 2 2 2 2 2 L L 1µr 2ϕ Lµv b, 1 2 (2.136)ClassicalL parameter scatteringL µrbv sinof theϕ incidentparameterµbv0 particleLb ofµ the B.b Inserting incidentv0 2µ particle theseb E0 relations B. Inserting these relations µr 0. µr (2.137). With the= total(2.137) energy= E ⇒1 µv=2 the amount= of the angular | |= ==2 ˙ ˙+=2µr 2 = 2 ˙ + 2µr 2 into (2.140)weobtaininto0 (2.1402 )weobtain0 (2.141) 1 2 1 2 2 2 = Ekin µv µ r r ϕ The momentumamount of the angular momentum because L(x ) µ= 2v0 r= sin2 ϕ˙ +µ ˙v0 b.The The total energy E TheT totalE energyin theE centerT ofE masspot in frame the centeris uniquely of mass defined frame by the initial energy E0 and+∞ the impact =! −∞ = · · pot· 2! = · " · +∞ 2 dr2 2 2 2 dr kinetic energy in the center= of1 mass+ 2 frameL is = + parameter b of the incident particle B. Inserting these relations can then be written ascan thenµr be written. as (2.137) Lϑ(E0µ,rbv) sinπϕ 2µbϑb(vE00, b)L π µ2bb v0 2µ. b(2.142)E0 . (2.142) 2 1/2 2 E (r) 1/2 = 2 ˙ + 2µr into= (2.140)weobtain= −= ⇒2 = =b2− Epot(r)=2 b pot r 1 r ' r 1 (2.141)2 rmin' r2 minE0 r E0 1 1 2 2 Therefore − − − − 1 2 2 L 2 12 2 2 L E TheEkin totalE energyµvµEr EtotalTµ ErEpot0 inEr theϕµ(rr center) const of mass. (2.138)E framepot(r) const. (2.138) +∞ ( ) ( ) total =0 2 ==2 + 2˙ + pot˙ 2 dr = = 2 ˙ + 2µ=r += 2 ˙ =+ 2µr + is= uniquelyThisϑ(E shows, b) defined thatπ the2b byThis deflection the shows initial angle that energy theϑ is deflectionE determined0 and. (2.142) the angle byimpactϑ theis determined by the can then be written as 2 0 1/2 1 L ! " = − 2 b2 Epot(r) 2 parameterinteractionb of potential the incidentrinteraction'Epotr particle(r1),theimpactparameter potential2 B. InsertingEpot(r these),theimpactparameterb relationsand the b and the µr .2 (2.137) ϕ min r E0 Solving (2.138)and(1Solving2.136)for2 (L2.138r and)and(ϕ yields2.136)forr and yields initial energy− E− . E =E2 ˙ µ+r 22µr ˙ E ˙(r) const. ˙(2.138)˙ intoinitial (2.140 energy)weobtainE0. ( 0 ) total 0 2 2µr 2 pot The lower integration limit rmin is fixed by the condition = = ˙ + + = 1This/2 The shows lower that integration the deflectionThe limit lower anglermin integrationϑ isis fixed determined by limit the byr conditionmin theis fixed by the condition 2 2 1/2 L2 The total energy E 2 T Epot in the centerL of mass frame interactionr(r ) potential0. This givesErpot+(r∞min(r with),theimpactparameter) (2.1390. This)and( gives2.141 withb) (and2.139 the)and(2.141) r E0 Epotr(r) E0ϕ Epot(r) (2.139a) min (2.139a) dr Solving (2.138= )and(+ 2.136˙ =)forµr and2 −yields − 2 µr 2 ϑinitial(˙E , b energy)= πE . 2b ˙ = . (2.142) can then be written˙ = asµ − −# 2˙ µ$r ˙ %& 0 0 1/2 # $ %& = − b2 Epot(r) b L which givesThe lower integration limit2 rmin isb fixedr by the condition . (2.143) L 2 1/2 rminrmin' r 1 2 min E . (2.143)1/2 ϕ .2 2 ϕ .L (2.139b) (2.139b) r 1/=20 Epot(rmin) 1r 2 EL E ˙ =(r)µr 2 (2.139a) r(rmin) 0. This gives with= (2.139−Epot)and((rmin−)2.141) 1 ˙ = µr2 0 pot 2 ˙ = 1 E0 Etotal E0 ˙µ=r µ − Epot−(r2)µr const. (2.138) (− E0 )− = = 2 ˙# +$2µr 2 + = %& This shows that the deflection angleb ϑ is determined( by) the L rmin ( ). (2.143) ϕ . The(2.139b) angular for Coulomb potentialE (r ) 1 /2is 2 interaction potential E=pot(r),theimpactparameterpot min b and the qQ ˙ = µr 1 E πε Solving (2.138)and(2.136)forr and ϕ yields − −1 04 0 2 Epot = initial energy E0. ϑ = 2 cot μv b 4πε r ˙ ˙ ( ( qQ) 0 ) 0 The lower integration limit rmin is fixed by the condition 2 1/2 2 L 09/03/2021r(r ) 0. This gives withJinniu (2.139 Hu)and(2.141) r E0 Epot(r) (2.139a) min ˙ = µ − − 2µr 2 ˙ = # $ %& b L rmin 1/2 . (2.143) ϕ . (2.139b) = Epot(r ) ˙ = µr 2 1 min − E0 ( ) 2.8 The Structure of Atoms 63 Note We should keep in mind that the only quantity obtained from a scattering experiment is the differential or integral •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- = Whether the integral remains finite depends on the expo- not be directly measured! The measured scattering cross nent n in the power dependence of the interaction potential section yields, however, the wanted information on the deflec- n (Epot(r )). tion curve ϑ(b) from which the interaction potential can be •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. = = ⇒ = = central collisions with B. They are reflected back into the Let us assume a parallel beam of incident particles A with incident direction. particle flux density NA nAvA that passes through a layer ˙ = •Ifϑmin is the smallest still detectable deflection angle then of particles B in rest with density nB.AllparticlesApassing all particles with ϑ < ϑmin are regarded as not scattered. through an annular ring with radius b and width db around an These are all particles with b > bmax(ϑmin).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- 2 ± gral scattering cross section is then σint πb .This cally symmetric interaction potential (Fig. 2.98). Through this = max shows that with such a definition the cross section, which annular ring dNA NA dA nAvA2πb db particles A pass ˙ = ˙ = should be solely dependent on the particle characteristics, per second. One particle B therefore scatters the fraction becomes dependent on the design of the apparatus. This 1 contradiction is removed by the quantum-mechanical treat- dNA ϑ dϑ db ˙ ± 2 2πb db 2πb dϑ (2.144) ment of collisions. N = = dϑ ! ˙A " •FormonotonicpotentialsEpot(r) (for example pure repul- 1 sive potentials Epot r )thereis,foragivenenergyE0, of all particles A, incident per second and unit area onto the ∝ − awell-defineduniquedeflectionangleϑ for each value target, into the range of deflection angles ϑ dϑ/2. The detec- 2 2 ± b of the impact parameter (Fig. 2.97a). This is no longer tor with area AD R dΩ R sin ϑ dϑ dφ in a distance R = = true for non-monotonic potentials (Fig. 2.97b), where, for from the scattering center B, receives the fraction example, two different impact parameters b1 and b2 may lead to the same deflection angle ϑ.Plottingthecurves dNA(ϑ, φ) dφ db ˙ b dϑ dφ, (2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 2π = dϑ ˙A such as those shown in Fig. 2.97.Theirformdependson E0 and Epot(r). which passes through the segment b db dφ of the annular ring in Fig. 2.98. Classical scattering The fraction of all incident particles A, scattered by all atoms B with density nB in the volume V A∆x is then: = (b) (a) dNA(ϑ, dΩ) db ˙ nB A∆xb dϑ dφ. (2.146) N = dϑ ˙A 09/03/2021Fig. 2.97 a,b. Qualitative relationJinniu interaction Hu potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ The Geiger-Marsden experiment Trajectories in Rutherford scattering Trajectories in Scattering Trajectories depend on the impact parameter. Trajectories are strongly dependent on the impact parameter 09/03/2021 Jinniu Hu NUCS 342 (Lecture 22) March 16, 2011 9 / 29 The impact parameter/scattering angle relationship Rutherford Scattering Formula The impact parameter/scattering angle relationship TheThe key key concept in in Rutherford Rutherford scattering scattering is the relationship is the relationship between betweenthe impact the parameter impactb parameterand the scattering b and angle the ✓scattering. angle . Trajectories are hyperbolic 09/03/2021 Jinniu Hu NUCS 342 (Lecture 22) March 16, 2011 11 / 29 Rutherford Scattering Formula 1. Basic knowledge The Coulomb force; The Newton’s laws; The conservation of linear momentum; The conservation of angular momentum. 2. Assumptions Single scattering Only consideration Coulomb force The effect of electrons in nuclei is neglected The target is static 09/03/2021 Jinniu Hu The impactThe impact parameter/scattering parameter/scattering angle angle relationship relationship RutherfordThe impact parameter/scatteringMomentum angle relationship Scattering change Formula in Rutherford scattering The impactMomentum parameter/scattering angle change relationship in Rutherford scattering The key concept in Rutherford scattering is the relationship between Momentumthe impact parameter b and change the scattering angle in✓ . Rutherford scattering p The momentum ∆ f p pf Thechange momentum is ∆p θ change is θ p~ = p~ = p pi | i | | f | p~ = ~pp~= =pp pi i | |f | | | | 1 pf NUCS 342 (Lecture 22) ~p = p 2March 16,p 2011 11 / 29 p ∆p Elastic scatteringsin(✓/2) = = | | 1 p 2p pθ/2 p p f ∆p θ sin( ✓/p2)=2 =p 2sin(✓/=2) θ/2 p~i = p~f = p p 2p θ/2 | | | | pi θ Momentum changep =2p sin(✓/2) θ/2 p p =2p sin(✓/2) i NUCS 342 (Lecture 22) March 16, 2011 13 / 29 09/03/2021 Jinniu Hu NUCS 342 (Lecture 22) March 16, 2011 13 / 29 Rutherford Scattering Formula From the Newton’s second law dp~ F~ = = p~ = Fdt~ dt ) Z The force is the Coulomb force 2 ~ 1 Z1Z2e ~r F = 2 4⇡"0 r r Before we start integrating let us note that the trajectories are symmetric with respect to the line defined by the distance of the closest approach 09/03/2021 Jinniu Hu Rutherford Scattering Formula Trajectories are symmetric with respect to angle # 09/03/2021 Jinniu Hu Rutherford Scattering Formula The symmetry with respect to the line at # = 0 implies p~ = Fdt~ = p = F cos dt ) So, Z Z Z Z e2 1 p = 1 2 cos dt 4⇡" r2 0 Z This integral can be carried over with a help of conservation of angular momentum. The angular momentum is d~r d ~ d ~ L~ = ~r p~ = m~r ~v = m~r + r = mr~r ⇥ ⇥ ⇥ dt dt ! ⇥ dt 09/03/2021 Jinniu Hu Rutherford Scattering Formula The magnitude of angular momentum d ~ L = L~ = mr2 | | dt From the initial condition L = mv0b Since the angular momentum is conserved d mr2 = mv b dt 0 dt d 2 = r v0b 09/03/2021 Jinniu Hu Rutherford Scattering Formula Thus the change of momentum Z Z e2 dt Z Z e2 d p = 1 2 cos = 1 2 cos 4⇡" r2 4⇡" v b 0 Z 0 Z 0 Z Z e2 1 > = 1 2 d cos 4⇡" v b 0 0 Z < The limits for integration are 1 = (⇡ ✓) > 2 1 = (⇡ ✓) < 2 09/03/2021 Jinniu Hu Rutherford Scattering Formula The integral is Z Z e2 1 > Z Z e2 1 p = 1 2 d cos = 1 2 (sin sin ) 4⇡" v b 4⇡" v b > < 0 0 Z < 0 0 Z Z e2 2 ✓ = 1 2 cos 4⇡"0 v0b 2 Since p =2p sin(✓/2) The impact parameter is expressed as Z Z e2 1 1 b = 1 2 4⇡"0 pv0 tan(✓/2) Z Z e2 1 1 = 1 2 4⇡"0 2E tan(✓/2) with E being the initial kinetic energy for the projectile 09/03/2021 Jinniu Hu 2.8 The Structure of Atoms 63 Note We should keep in mind that the only quantity obtained from a scattering experiment is the differential or integral •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- = Whether the integral remains finite depends on the expo- not be directly measured! The measured scattering cross nent n in the power dependence of the interaction potential section yields, however, the wanted information on the deflec- n (Epot(r )). tion curve ϑ(b) from which the interaction potential can be •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. = = ⇒ = = central collisions with B. They are reflected back into the Let us assume a parallel beam of incident particles A with incident direction. particle flux density NA nAvA that passes through a layer ˙ = •Ifϑmin is the smallest still detectable deflection angle then of particles B in rest with density nB.AllparticlesApassing all particles with ϑ < ϑmin are regarded as not scattered. through an annular ring with radius b and width db around an These are all particles with b > bmax(ϑmin).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- 2 ± gral scattering cross section is then σint πb .This cally symmetric interaction potential (Fig. 2.98). Through this = max shows that with such a definition the cross section, which annular ring dNA NA dA nAvA2πb db particles A pass ˙ = ˙ = should be solely dependent on the particle characteristics, per second. One particle B therefore scatters the fraction becomes dependent on the design of the apparatus. This 1 contradiction is removed by the quantum-mechanical treat- dNA ϑ dϑ db ˙ ± 2 2πb db 2πb dϑ (2.144) ment of collisions. N = = dϑ ! ˙A " •FormonotonicpotentialsEpot(r) (for example pure repul- 1 sive potentials Epot r )thereis,foragivenenergyE0, of all particles A, incident per second and unit area onto the ∝ − awell-defineduniquedeflectionangleϑ for each value target, into the range of deflection angles ϑ dϑ/2. The detec- 2 2 ± b of the impact parameter (Fig. 2.97a). This is no longer tor with area AD R dΩ R sin ϑ dϑ dφ in a distance R = = true for non-monotonic potentials (Fig. 2.97b), where, for from the scattering center B, receives the fraction example, two different impact parameters b1 and b2 may lead to the same deflection angle ϑ.Plottingthecurves dNA(ϑ, φ) dφ db ˙ b dϑ dφ, (2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 2π = dϑ ˙A such as those shown in Fig. 2.97.Theirformdependson E0 and Epot(r). which passes through the segment b db dφ of the annular ring in Fig. 2.98. The fraction of all incident particles A, scattered by all atoms B with density nB in the volume V A∆x is then: = (b) (a) dNA(ϑ, dΩ) db ˙ nB A∆xb dϑ dφ. (2.146) The cross sectionN = dϑ ˙A Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum09/03/2021angle ϑ and differential crossJinniu section Hu dσ/dΩ 2.8 The Structure of Atoms 63 Note We should keep in mind that the only quantity obtained 2.8 The Structure of Atoms from a scattering experiment is the differential or integral 63 •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- = Whether the integral remains finite depends on the expo- not be directly measured! The measured scattering cross Note nent n in the power dependence of the interaction potential Wesection should yields, however, keep in the mind wanted that information the only on the quantity deflec- obtained n The cross section (Epot(r )). fromtion a curve scatteringϑ(b) from experiment which the interaction is the potential differential can be or integral •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. •Forr rmin the integrand= in= (2.140⇒ =)becomesinfinite.= scattering cross section. The impact parameter b itself can- = central collisions with B. They are reflected backLet into theus assumeLet us assume a parallel a parallel beam beam of of incident incident particles particles A with A with Whether the integralincident remains direction. finite depends on the expo- notparticle be directly flux density measured!NA nAvA Thethat passes measured through a scattering layer cross particle flux density ˙ = that passes through a layer nent n in the power•If dependenceϑmin is the smallest of the still interaction detectable deflection potential angle thensectionof particles yields, B in however, rest with density the wantednB.AllparticlesApassing information on the deflec- all particles with ϑ < ϑ are regarded as not scattered.of particlesthrough B an in annular rest ring with with radiusdensityb and n widthB. db around an (E (r n)). min tion curve ϑ(b) from which the interaction potential can be pot These are all particles with b > bmax(ϑmin).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- 2 ± •Forb 0isL 0gral scatteringϑ π. cross Particles section with is thenbσint 0sufferπb .Thisderived.cally symmetric This can interaction be seen potential as follows. (Fig. 2.98). Through this = maxAll particles A passing through an annular ring with radius = = shows⇒ that= with such a definition the cross= section, which annular ring dNA NA dA nAvA2πb db particles A pass central collisions with B. They are reflected back into the Let us assume˙ = a parallel˙ = beam of incident particles A with should be solely dependent on the particle characteristics,b and widthper second. db One around particle Ban therefore atom scattersB are the deflected fraction by the incident direction. particle flux density N n v that passes through a layer becomes dependent on the design of the apparatus. This ˙A A A angle $ and$ ±d$/2,1 assuming= a spherically symmetric •Ifϑmin is the smallestcontradiction still detectable is removed bydeflection the quantum-mechanical angle then treat-of particlesdNA Bϑ in restdϑ with density ndBb .AllparticlesApassing ˙ ± 2 2πb db 2πb dϑ (2.144) ment of collisions. N = = dϑ all particles with ϑ < ϑmin are regarded as not scattered.interactionthrough potential. an annular! ˙A " ring with radius b and width db around an •FormonotonicpotentialsEpot(r) (for example pure repul- These are all particles with b > b 1 (ϑ ).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- sive potentials Epot r max)thereis,foragivenenergymin E0, of all particles A, incident per second and unit area onto the ∝ − 2 The angular ring ± gral scattering crossawell-defineduniquedeflectionangle section is then σint πbmaxϑ for.This each valuecallytarget, symmetric into the range interaction of deflection potential angles ϑ d (Fig.ϑ/2. The2.98 detec-). Through this 2 2 ± b of the impact parameter (Fig.=2.97a). This is no longer tor with area AD R dΩ R sin ϑ dϑ dφ in a distance R shows that with such a definition the cross section, which annular ring dNA= NA=dA nAvA2πb db particles A pass true for non-monotonic potentials (Fig. 2.97b), where, for from the scattering˙ center= ˙ B, receives= the fraction should be solely dependent on the particle characteristics,particlesper second. A pass One per particle second B therefore scatters the fraction example, two different impact parameters b1 and b2 may becomes dependentlead on to the the same design deflection of the angle apparatus.ϑ.Plottingthecurves This dNA(ϑ, φ) dφ db ˙ b dϑ dφ, (2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 1 2π = dϑ contradiction is removed by the quantum-mechanical treat-09/03/2021dNA ϑ ˙A dϑ Jinniu Hu db such as those shown in Fig. 2.97.Theirformdependson ˙ ± 2 2πb db 2πb dϑ (2.144) ment of collisions.E and E (r). which passesN through the segment= b db dφ=of the annulardϑ ring 0 pot ! ˙A " •FormonotonicpotentialsEpot(r) (for example pure repul- in Fig. 2.98. 1 The fraction of all incident particles A, scattered by all sive potentials Epot r − )thereis,foragivenenergyE0, of all particles A, incident per second and unit area onto the ∝ atoms B with density nB in the volume V A∆x is then: awell-defineduniquedeflectionangleϑ for each value target, into the range of deflection angles= ϑ dϑ/2. The detec- (a) (b) 2 2 ± b of the impact parameter (Fig. 2.97a). This is no longer tor with areadNA(Aϑ,DdΩ) R dΩ dRb sin ϑ dϑ dφ in a distance R ˙ nB A∆xb dϑ dφ. (2.146) N = = =dϑ true for non-monotonic potentials (Fig. 2.97b), where, for from the scattering˙A center B, receives the fraction example, two different impact parameters b1 and b2 may lead to the same deflection angle ϑ.Plottingthecurves dNA(ϑ, φ) dφ db ˙ b dϑ dφ, (2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 2π = dϑ ˙A such as those shown in Fig. 2.97.Theirformdependson E0 and Epot(r). which passes through the segment b db dφ of the annular ring in Fig. 2.98. The fraction of all incident particles A, scattered by all atoms B with density nB in the volume V A∆x is then: = (b) (a) dNA(ϑ, dΩ) db ˙ nB A∆xb dϑ dφ. (2.146) N = dϑ ˙A Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ 2.8 The Structure of Atoms 63 Note We should keep in mind that the only quantity obtained 2.8 The Structure of Atoms from a scattering experiment is the differential or integral63 2.8 The Structure of Atoms 63 •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- = Note We should keep in mind that the only quantity obtained Whether the integral remainsNote finite depends on the expo- not be directlyWe should measured! keep in mind The that measured the only quantity scattering obtained cross from a scattering experiment is the differential or integral nent n in the power dependence of the interaction potential section yields,from however, a scattering the experiment wanted is information the differential on or the integral deflec- •Forr r the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- n •Formin r rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- (Epot(r )). = = tion curve ϑ(b) from which the interaction potential can be Whether theWhether integral the remains integral finite remains depends finite depends on the on expo- the expo-notnot be directlybe directly measured! measured! The The measured measured scattering scattering cross cross •Forb 0isLnent n0 in theϑnent powernπin. dependence Particles the power dependence with of theb interaction of0suffer the interaction potentialderived. potentialsection Thissection yields, can yields, be however, seen however, as the follows. the wanted wanted information information on on the the deflec- deflec- = = ⇒n = n = central collisions(Epot with(r )). B.(E Theypot(r )) are. reflected back into the Lettion us assumetion curve curveϑ a(bϑ parallel)(bfrom) from which beam which the the of interaction interaction incident particlespotential potential can A can be with be •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. incident direction.•Forb 0isL =0 ϑ = π⇒. Particles= with b 0sufferparticle= derived. flux density This canNA be seennA asvA follows.that passes through a layer = central= collisions⇒ = with B. They are reflected= back into the Let us assume˙ a= parallel beam of incident particles A with •Ifϑmin is the smallestcentral collisions still detectable with B. They deflection are reflected angle back then intoof the particlesLet B us in assume rest with a parallel density beamnB of.AllparticlesApassing incident particles A with incident direction. particle flux density NA nAvA that passes through a layer incident direction. particle flux density NA˙ =nAvA that passes through a layer all particles with ϑ <•Ifϑminϑminareis the regarded smallest still as not detectable scattered. deflectionthrough angle then anof annular particles ring B in rest with˙ with= radius densitybnandB.AllparticlesApassing width db around an •Ifϑmin is the smallest still detectable deflection angle then of particles B in rest with density nB.AllparticlesApassing These are all particles withall particlesb > withbmaxϑ <(ϑϑminmin )are.Theinte- regarded as notatom scattered. B arethrough deflected an annular by the ring angle withϑ radiusdϑb/and2, assumingwidth db around a spheri- an all particlesThese with ϑ are< allϑmin particlesare regarded with b 2> asbmax not(ϑ scattered.min).Theinte-throughatom an B are annular deflected ring by with the angle radius±ϑ bdϑand/2, assuming width db aaround spheri- an gral scattering cross section is then σint πbmax.This cally2 symmetric interaction potential (Fig.± 2.98). Through this These are allgral particles scattering with crossb section>= bmax is then(ϑminσ)int.Theinte-πbmax.Thisatomcally B are symmetric deflected interaction by the angle potentialϑ (Fig.dϑ/2,2.98 assuming). Through a spheri-this =The cross section 2.8 The Structure of Atoms ± 63 shows that with such a definitionshows that with the such cross a definition section, the which cross2 section,annular which ringannular dNA ring dNNA dANA dAnAvnAAv2Aπ2πbbddbb particlesparticles A passA pass gral scattering cross section is then σint πb .This cally symmetric˙ interaction˙˙ Note˙ potential (Fig. 2.98We should keep). in Throughmind that the only quantity obtained this max = = = = from a scattering experiment is the differential or integral = •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- should be solely dependent on the particle characteristics,per second.per One second. particle One particle B therefore= B therefore scatters scatters the the fraction fraction should be solely dependent on the particle characteristics, Whether the integral remains finite depends on the expo- not be directly measured! The measured scattering cross shows that with such a definition the cross section,One which particleannular B ringtherefore dNA NscattersAnentdn inA the power dependence ntheA of thev interactionA 2fraction potentialπb sectiondb yields,particles however, the wanted information A on the pass deflec- n becomes dependent on the design of the apparatus. This ˙ = ˙ (Epot(r )). = tion curve ϑ(b) from which the interaction potential can be •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. = = ⇒ = = becomes dependentshould be on solely the design dependent of onthe the apparatus. particle characteristics, This per second. One particle1 central B collisions therefore with B. They are reflected back scatters into the Let us assume the a parallel fraction beam of incident particles A with incident direction. particle flux density NA nAvA that passes through a layer contradiction is removed by the quantum-mechanical treat- dNA ϑ dϑ db ˙ = ˙1 2 •Ifϑmin is the smallest still detectable deflection angle then of particles B in rest with density nB.AllparticlesApassing contradictionbecomes is removed dependent by the quantum-mechanical on the design of the apparatus. treat- This dNA ϑ dϑ ± all particles2 withπϑb< ϑdmin bare regarded as2 notπ scattered.bdbthroughd anϑ annular ring with radius(2.144)b and width db around an These are all particles with b > bmax(ϑmin).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- ment of collisions. 2 ± gral= scattering cross section is then=σ πb2 .Thisdϑcally symmetric interaction potential (Fig. 2.98). Through this ˙ N int max A = ± ! 1 2"πshowsb thatd withb such a definition2 the crossπ section,b which annulard ringϑ dN N dA n v 2π(2.144)b db particles A pass A A A A dN ϑ ˙ dϑ ˙ = ˙ = ment of collisions.contradiction•Formonotonicpotentials is removed by the quantum-mechanicalEpot(r) (for example treat- pure repul- A 2 should be solely dependent on the particle characteristics, dper second.b One particle B therefore scatters the fraction N˙ = becomes dependent on= the design of the apparatus.d Thisϑ A ± contradiction2π is removedb d by theb quantum-mechanical2π treat-b dN dϑ ϑ1 dϑ (2.144)db A 1 ! " ˙ ± 2 2πb db 2πb dϑ (2.144) ˙ ment of collisions. of all particles A, incident per second and unitNA area= onto= dϑ the ment of collisions.sive potentials Epot r − )thereis,foragivenenergyof all particlesE0, A, incident per second and unit! ˙ " area onto •FormonotonicpotentialsE (r) (for example pure repul- N •Formonotonicpotentials= Epot(r) (for= example pure repul- dϑ pot A 1 ∝ ! " sive potentials Epot r − )thereis,foragivenenergyE0, of all particles A, incident per second and unit area onto the ˙ ∝ awell-defineduniquedeflectionangleϑ for each value target, into the range ofawell-defineduniquedeflectionangle deflection anglesϑ for each valueϑ target,d intoϑ the/ range2. of The deflection angles detec-ϑ dϑ/2. The detec- •FormonotonicpotentialsE (r) (for example pure repul- 2 2 ± 1 b of the impact parameter (Fig. 2.97a). This is no longer tor with area AD R dΩ R sin ϑ dϑ dφ in a distance R pot = = sive potentials Epot r − )thereis,foragivenenergyE0, theof alltarget, particles into A,the incident range of per2true fordeflection non-monotonic second potentials2 (Fig. 2.97 andb), where, angles for± unitfrom the scattering area center B,±d receives onto the fraction/2 . the b of the impact1 parameter (Fig. 2.97a). This is no longer tor with area AD R example,dΩ two differentR impact parameterssin ϑb1 anddb2ϑmay dφ in a distance R of all particles A, incidentlead to the per same deflection second angle ϑ.Plottingthecurves and unitdNA(ϑ, φ) areadφ db onto the sive potentials∝ Epot r − )thereis,foragivenenergyE0, ˙ b dϑ dφ, (2.145) = ϑ(b) at a given= initial energy E0 yields deflection curves N 2π = dϑ ˙A awell-defineduniquedeflectionangletrue for non-monotonic∝ ϑ potentialsfor each (Fig. value2.97b),Thetarget, where, detector for intofrom the with range the scatteringarea of deflection centersuch as those B, shown receives angles in Fig. 2.97.Theirformdependson theϑ fractiondϑ/2. The detec- which passes through the segment b db dφ of the annular ring awell-defineduniquedeflectionangleϑ for each value target, into the range of deflectionE0 and Epot(r). angles ϑ dϑ/2. The detec- 2 2 ±in Fig. 2.98. example, two different impact parameters b and b may The fraction of all incident particles A, scattered by all 1 tor with2 area A R dΩ2 R sin2 ϑ dϑ datomsφ± Bin with density an in distance the volume V A∆x is then: R b of the impact parameter (Fig. 2.97a). This is no longer D B b of the impact parameter (Fig. 2.97a). This is no longer tor with area A R dΩ R sin ϑ dϑ dφ in a distance= R D (a) (b) =dN (ϑ, φ)=dφ db dNA(ϑ, dΩ) db lead to the same deflection angle ϑ.Plottingthecurves A ˙ n A∆xb dϑ dφ. (2.146) B ˙ = = NA = dϑ true for non-monotonictrue for non-monotonic potentials (Fig. potentials2.97 (Fig.b), where,2.97b), for where,from for thefrom scattering the scattering center center B, receives B, receivesb d theϑ d theφ fraction, fraction˙ (2.145) ϑ(b) at a given initial energy E0 yields deflection curves NA 2π = dϑ receives the fraction ˙ example, twoexample, different two impactsuch different as those parameters impact shown in parameters Fig.b1 and2.97.Theirformdependsonb2b1mayand b2 may which passesdN through(ϑ, φ) d theφ segmentdbb db dφ of the annular ring lead to the samelead deflectionto the sameE0 and angle deflectionEpot(rϑ). .Plottingthecurves angle ϑ.Plottingthecurves dNA(ϑ,˙Aφ) dφ db in Fig.˙ 2.98. b b dϑddϑφd,φ, (2.145)(2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 2π = dϑ ϑ(b) at a given initial energy E0 yields deflection curves NA ˙A2π = dϑ such as those shown in Fig. 2.97.Theirformdependson The˙ fraction of all incident particles A, scattered by all such as those shown in Fig. 2.97.Theirformdependson atoms B with density nB in the volume V A∆x is then: ( ) which passes through theFig. 2.97 segmenta,b. Qualitative relation interactionb potentiald andb deflectiondφ=Fig.of 2.98 Relative the relation annular between impact parameter b, scatteringring E0 and Epot r . angle ϑ and differential cross section dσ/dΩ E0 and Epot(r). which passes through the segmentfunction ϑ(b).(a)Monotonicpotential.(b db)Potentialwithaminimumb dφ of the annular ring (a) (b) in Fig. 2.98. Jinniu Hu in09/03/2021 Fig. 2.98. dNA(ϑ, dΩ) db ˙ nB A∆xb dϑ dφ. (2.146) The fractionN of all incident= particlesdϑ A, scattered by all The fraction of all˙A incident particles A, scattered by all atoms B with density nB in the volume V A∆x is then: atoms B with density nB in the volume V =A∆x is then: (b) = (a) dNA(ϑ, dΩ) db (a) (b) ˙ nB A∆xb dϑ dφ. (2.146) dNA(ϑ, dΩNA) = dbdϑ ˙ ˙ nB A∆xb dϑ dφ. (2.146) N = dϑ ˙A Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ 2.8 The Structure of Atoms 63 Note We should keep in mind that the only quantity obtained from a scattering experiment is the differential or integral •Forr rmin the integrand in (2.140)becomesinfinite. scattering cross section. The impact parameter b itself can- = Whether the integral remains finite depends on the expo- not be directly measured! The measured scattering cross nent n in the power dependence of the interaction potential section yields, however, the wanted information on the deflec- n (Epot(r )). tion curve ϑ(b) from which the interaction potential can be •Forb 0isL 0 ϑ π. Particles with b 0suffer derived. This can be seen as follows. = = ⇒ = = central collisions with B. They are reflected back into the Let us assume a parallel beam of incident particles A with incident direction. particle flux density NA nAvA that passes through a layer ˙ = •Ifϑmin is the smallest still detectable deflection angle then of particles B in rest with density nB.AllparticlesApassing all particles with ϑ < ϑmin are regarded as not scattered. through an annular ring with radius b and width db around an These are all particles with b > bmax(ϑmin).Theinte- atom B are deflected by the angle ϑ dϑ/2, assuming a spheri- 2 ± gral scattering cross section is then σint πb .This cally symmetric interaction potential (Fig. 2.98). Through this = max shows that with such a definition the cross section, which annular ring dNA NA dA nAvA2πb db particles A pass ˙ = ˙ = should be solely dependent on the particle characteristics, per second. One particle B therefore scatters the fraction becomes dependent on the design of the apparatus. This 1 contradiction is removed by the quantum-mechanical treat- dNA ϑ dϑ db ˙ ± 2 2πb db 2πb dϑ (2.144) ment of collisions. N = = dϑ ! ˙A " •FormonotonicpotentialsEpot(r) (for example pure repul- 1 sive potentials Epot r )thereis,foragivenenergyE0, of all particles A, incident per second and unit area onto the ∝ − awell-defineduniquedeflectionangleϑ for each value target, into the range of deflection angles ϑ dϑ/2. The detec- 2 2 ± b of the impact parameter (Fig. 2.97a). This is no longer tor with area AD R dΩ R sin ϑ dϑ dφ in a distance R = = true for non-monotonic potentials (Fig. 2.97b), where, for from the scattering center B, receives the fraction example, two different impact parameters b1 and b2 may lead to the same deflection angle ϑ.Plottingthecurves dNA(ϑ, φ) dφ db ˙ b dϑ dφ, (2.145) ϑ(b) at a given initial energy E0 yields deflection curves N 2π = dϑ ˙A such as those shown in Fig. 2.97.Theirformdependson which passes through the segment b db dφ of the annular ring E0 and Epot(r). The cross section in Fig. 2.98. The fraction ofThe all fraction incident of allparticles incident particlesA, scattered A, scattered by all by all atoms B with density nB in the volume V A∆x is then: atoms B with density nB in the volume V = A =x is then: (b) (a) dNA(ϑ, dΩ) db ˙ nB A∆xb dϑ dφ. (2.146) N = dϑ ˙A We define a differential cross section to be the ratio of scattered particles with per target and per unit solid angle to the number of incoming particles per unit area, Therefore, · dσ dNA(ϑ, dΩ) = · dΩ NAnBAΔxdΩ 09/03/2021 Jinniu Hu Fig. 2.97 a,b. Qualitative relation interaction potential and deflection Fig. 2.98 Relative relation between impact parameter b, scattering function ϑ(b).(a)Monotonicpotential.(b)Potentialwithaminimum angle ϑ and differential cross section dσ/dΩ 20.4 The scattering cross-section 193 with the appealing result that All connected, amputated Feynman 4 (4) 64 i (2π) δ pf pi = diagrams with incoming momentum pi . 2 The Concept of the Atom M − and outgoing momentum p !" " # " f (20.20) We now need to design an experiment which measures and that The crossThe comparisonsectionturns out to be a with scattering (2.133 experiment.)gives,withdM Ω sin ϑ dϑ dφ, = the differential scattering cross section Therefore 20.4 The scattering cross-section (a) (a) When firing one particle at another, we want to think about the ampli- tude of scattering of our incomingdσ particled inb terms1 of how big the other particle appears to be in cross-sectional area terms. Imagine driving at night, where your car headlights give youb a beam of particles. with which (2.147) you probe the inky blackness.dΩ A sheep= showsdϑ upsin moreϑ in your headlights (b) than a field mouse, simply because it’s bigger and scatters more photons 64 [Fig. 20.7(a)]. Similarly a cow will scatter more light if seen from the 2 The Concept of the Atom with side than from the front, because it offers a larger cross-sectional area1 Fromin the(2.142 former)weobtaind case [Fig. 20.7(b)]. Howeverb/dϑ it’s not( justdϑ about/db area)− – as a function white sheep are more visibleΩ than black sheepϑ –=ϑ but weφ fold this all in The comparison withof ( the2.133 interactionto our)gives,withd definition of potential the scattering cross-sectionEsinpot(rd)σ.Thisshows:ind which, we measure the rate R of a scattering process= occurring as R = σL, where L is the Fig. 20.7 Scattering cross-sections. the differential scattering crossluminosity section5 of our incoming beam of particles. (a) σsheep > σfield mouse. (b) (a) σ > σ . Scattering can occur in all directions, and so in each bit of solid angle cow, side cow, front dΩ there will be a bit of scattering dσ. A detector usually only covers a 5Luminosity is a quantity with dimen- range of solid angle, so it is usual to think in terms of the differential sions Time 1 Area 1 to be evaluated The relationship between b and for the Rutherford− × − dTheσ cross-section differentialdb 61defined cross by dσ/d sectionΩ. Fermi’s$gives golden rule information allows us to for the on case underthe study. relateb the scattering cross-section. to the modulus(2.147) squared of a matrix 6The relationship between σ and dinteractionΩ =element.dϑ After sin potential takingϑ properE account(r of) allbetween the normalization the factors collidingdσ/dΩ is given par- by scattering yields 7 pot one can show, for example, that for a scattering process involving two 2π 1 dσ dφ d(cos θ) = σ. equal-mass particles scattering off each other then 0 1 dΩ ticles A and B at a distance r. Z Z− 1 dσ 2 7This result is discussed in Peskin and From (2.142)weobtaindb/dϑ (dϑ/db)=− |M|as a, function (20.21) dΩ 64π2E2 Schroeder, Chapter 4. = 2CM 2 of the interaction potential Epotd (r).Thisshows:Z Z e 1 1 2 1 2 where ECM is the total energy in the centre of mass frame. With dΩ sin=ϑdϑdφ we can write (2.1464)as d⌦ 4⇡"0 4E sin (✓/2) = ✓ ◆ The differential cross sectionExample 20.4 gives information on the (b) (c) We’re finallydN goingA( tod workΩ) out some scattering amplitudesdσ and cross-sections for ψ†ψφ theory.˙ We start with the scatteringnB ofA two∆ distinguishablex d psionΩ. particles. In (2.148) interaction09/03/2021 potential Epotthis( caser) onlybetween the t-channel diagram the colliding [Fig.Jinniu 20.6(a)] contributes.Hu par- Let’s further simplify by considering onlyNA a non-relativistic= case. This meansd thatΩ we can take the four- momentum p =(E˙p , p) to be p (m, p). We have ticles A and B at a distance r. ≈ p (m, p),p (m, k), 1i ≈ 2i ≈ (20.22) p (m, p ),p (m, k"), Measuring the relative1f ≈ fraction" 2f ≈ dN /N yields the differen- ˙A ˙A tial cross section and with it the interaction potential. With dΩ sin ϑdϑdφ we can write (2.146)as = The integral scattering cross section is obtained by integra- tion over dΩ, where the integration limits are(b)ϑ(b 0) π (c) dNA(dΩ) dσ = = ˙ and ϑn(bB A∆)x ϑdΩ:. (2.148) N = max dΩ min ˙A = ϑmin 2π Measuring the relative fraction ddNσA/NA yields the differen-dσ Fig. 2.99 a–c. Collision of hard spheres with diameter D.(a)Scattering σint ˙ dΩ˙ sin ϑ dϑ dφ, (2.149a) tial cross section and with= it thed interactionΩ = potential.dΩ angle for impact parameters b < D.(b)PotentialV (r).(c)Deflection The integral scattering crossΩ! section is obtainedϑ !π φ! by0 integra- function ϑ(b) = = tion over dΩ, where the integration limits are ϑ(b 0) π = = and ϑ(bmax) ϑminwhere: ϑmin is the smallest detectable deflection angle. The = integration over φ gives 2π.With(2.147)weget: ϑ 2π The impact parameters for b D are therefore min ≤ dσ dσ ϑmin Fig. 2.99 a–c. Collision of hard spheres with diameter D.(a)Scattering σint dΩ sin ϑ dϑ dφ, (2.149a) = dΩ = dΩ b db angle for impact parameters b < D.(b)Potentialb DV sin(r).(ϕc)DeflectionD cos(ϑ/2). ! ! ! σint 2π sin ϑ dϑ m Ω ϑ π φ 0 = sin ϑ dϑ function ϑ(b) = = = = ! " " ϑ π " " where ϑ is the smallest detectable deflection= angle." The" Then the derivative db/dϑ becomes min bmax " " integration over φ gives 2π.With(2.147)weget: 2 2π b db πbmax. The impact(2.149b) parameters for b D are thereforedb D = = ≤ sin ϑ/2 ϑmin b!0 dϑ = 2 = " " b db b D sin ϕ D cos(ϑ/2"). " σint 2π sin ϑ dϑ m " " = sin ϑ dϑ = and the differential= scattering" " cross section is: ϑ !π " " = " " Then the derivative db/dϑ becomes Exampleb " " max " " dσ b db D cos(ϑ/2)D sin(ϑ/2) D2 Collisions of hard2 spheres A and B with equal diameters D. 2π b db πbmax. (2.149b) db D =The potential= energy in this case is: dΩ =sinsinϑ/ϑ2 dϑ = 2sinϑ = 4 b!0 dϑ = 2 2 = " " dσ D 2 " " σint dΩ 4π π D . for r D " " ⇒ = dΩ = 4 = Epot(r) ∞ ≤ and. the differential scattering" " cross section! is: = 0forr > D # $ Example The deflection function ϑ(b) for2 hard spheres (Fig. 2.99c) is Collisions of hard spheres A and B with equal diameters D. dσ b db D cos(ϑ/2)D sin(ϑ/2) D From Fig. 2.99aitisseenthatattheclosestapproachsinϕm The potential energy in this case is: dΩ = sin ϑ=dϑ = ϑ(2sinb) ϑπ 2ϕ = π4 2arcsin(b/2D). b/D,whichimpliesthatacollisioncanonlytakeplacefor 2 m dσ D = −2 = − b D. For the scattering angle ϑ we find ϑ/2 πσ/int2 ϕm. dΩ 4π π D . ≤ for r D =⇒ −= dΩ = 4 = Epot(r) ∞ ≤ . ! = 0forr > D # $ The deflection function ϑ(b) for hard spheres (Fig. 2.99c) is From Fig. 2.99aitisseenthatattheclosestapproachsinϕm = ϑ(b) π 2ϕm π 2arcsin(b/2D). b/D,whichimpliesthatacollisioncanonlytakeplacefor = − = − b D. For the scattering angle ϑ we find ϑ/2 π/2 ϕm. ≤ = − The cross section If N incident particles strike a foil of thickness t containing n scattering centers per unit volume, the average number dN of particles % scattered into the solid angle d% around % is given by dσ dN = Nnt dΩ dΩ Therefore, 2 dN Z Z e2 1 = nt 1 2 dΩ 4 N ( 16πε0E ) sin (θ/2) This is the Rutherford result explaining the Geiger- Marsden experiment 09/03/2021 Jinniu Hu The Geiger-Marsden experiment Number of particles scattered at a given angle in Rutherford scattering is calculable and well understood, since it is defined by the well understood electromagnetic 70 2 The Concept of the Atom force. 70 2 The Concept of the Atom (a) (a) (b) (c) (b) (c) Fig. 2.106 Comparison of experimental results (points)withthepre- Fig.dictions 2.106 Comparison by Thomson of (dashed experimental curve)andRutherford( results (points)withthepre-solid curve) dictions by Thomson (dashed curve)andRutherford(solid curve) Fig. 2.107 (a) Paths of particles with different initial energies, all scat- Fig. 2.107tered by(aϑ) Paths60 of.(b particles) Deviation with from different Coulomb initial potential energies, for all paths scat- with = ◦ teredE bykinϑ> 2560 MeV◦.(b) (i.e., Deviationb < bc from)forϑ Coulomb60 .( potentialc)DeviationfromCoulomb for paths with Jinniu Hu = = ◦ 09/03/2021 Ekin scattering> 25 MeV at (i.e., fixedb < initialbc)for energyϑ 60Ekin◦.(c)DeviationfromCoulomb10 MeV for scattering angle = = longer valid. The attractive nuclear force is much strongerscatteringϑ > 100 at fixed initial energy Ekin 10 MeV for scattering angle ◦ = longerthan valid. the repulsive The attractive Coulomb nuclear force force and changes is much the stronger deflectionϑ > 100◦ thanof the the repulsiveα particles. Coulomb force and changes the deflection of the αThusparticles. the impact parameter at which the measured distri- The volume of the nucleus therefore only accounts for Thus the impact parameter at which the measured distri- The volume of the nucleus3 therefore13 only accounts15 for bution deviates from the predicted one gives a measure for the fraction (RN/RA) 10 10 of the atomic 3 13 − 15 − bution deviates from the predicted one gives a measure for the fraction (RN/RA) 10≈ 10− of the atomic the size of the atomic nucleus. One obtains values of volume VA! ≈ − − − the size of the atomic nucleus. One obtains values of volume VA! R r A1/2, N 10/2 RN r0≈A , ≈ While the atomic volume of the gold atom is about VA 15 While29 the3 atomic volume of the gold atom is about42 3V = where A is the atomic mass in AMU and r0 1.3 10− m. 10 m that of its nucleus is only VN 10 m .A 15 29 − 3 42 − 3 = where A is the atomic mass in AMU and r0 1=.3 10×− m. 10 m that of its nucleus is only VN 10≈ m . = × − ≈ − The key points Impact parameter: b 2 2 d Z1Z2e 1 1 = 4 d⌦ 4⇡"0 4E sin (✓/2) Scattering angle: $ ✓ ◆ Differential cross-section: the ratio of the number of particles scattered into an element of solid angle d% in the direction $ per unit area (unit 1 barn=10-28 m2) The important quantities in Rutherford formula: 1. impact parameter 3.charges 2. Scattering angle. 4. Initial kinetic energy 09/03/2021 Jinniu Hu 4.2 Passage of Alpha Particles Through Matter (Rutherford Scattering) 47 For very large impact parameters (small deflection angles), the Rutherford formula is likewise no longer exactly valid. The Coulomb potential of the nucleus is perturbed by the atomic electrons. These effects occur for b 10 -10 cm (deflection angles of a few seconds of arc) and are very difficult to detect experimentally. Completely analogous scattering formulae and scattering problems occur in the scattering of protons by atomic nuclei. The angular dependence of the scattering is related to the scattering potential; the latter can thus be determined from experiment. Scattering pro- cesses play an important role in nuclear and elementary particle physics, in the investi- gation of the internal structure of nuclei and of certain elementary particles. For example, Hofstadterwas granted the Nobel Prize in 1961 for his scattering experiments using fast electrons (10 9 eV) on protons and neutrons. From the angular dependence of the scattering intensity, he was able to obtain information about the inner structure of the proton and of the neutron. 4.2.5 What is Meant by Nuclear Radius? We can summarise our considerations in the above sections as follows: an alpha par- What is Meaning by Nuclearticle, whichRadius? approaches a nucleus from outside the atom, is acted on at first only by the repulsive Coulomb potential. If it approaches the nucleus sufficiently closely, it will also be acted upon by the attractive nuclear force. The nuclear radius is defined as the The nuclear radius is defineddistance as atthe which distance the effect ofat the which nuclear potentialthe is comparable to that of the Coulomb potential (Fig. 4.9). For such investigations, alpha particles of high kinetic effect of the nuclear potentialenergies is are comparable used, so that they tocan thatapproach of the nucleus sufficiently closely. the Coulomb potential I 2 \ Z1Z2e 1 \ E = \ \ Coulomb potential 4⇡"0 r \/ "- "- "- 2 Z1Z2e 1 R rmin = 4⇡"0 E Empirical nuclear radius Nuclear force potential Fig. 4.9. Nuclear force and Coulomb potentials, R =1.2A1/3 fm used for defining the nuclear radius R Jinniu Hu 09/03/2021 The empirical result of such measurements on nuclei with the mass number A is found to be R = (1. 3 ± 0.1) A 113 • 10 - 15 m . Numerical examples for A = 12 and A = 208 are: = 2.7 . 10- 15 m, = 7.1 .10- 15 m. Homework The Physics of Atoms and Quanta 2.2, 2.4, 4.3, 4.4, 4.5, 4.6, 4.8 09/03/2021 Jinniu Hu Exercise 1 The main constituents of air are: 78% N2, 21% O2 and 1% Ar. Using these numbers calculate the mass density of air under normal conditions. 09/03/2021 Jinniu Hu Exercise 2 The density of gold (197Au) is 19.3 g/cm3. How many gold atoms are present in a piece of gold whose volume is 3.50 cm3 ? 09/03/2021 Jinniu Hu Exercise 3 Calculate the impact parameter for scattering a 7.7 MeV particle from gold at an angle of (a) 1° and (b) 90°. 09/03/2021 Jinniu Hu Exercise 4 Rutherford found deviations from his scattering formula at backward angles when he scattered 7.7 MeV particles (Z1=2) on aluminum (Z2=13). He suspected this was because the particle might be affected by approaching the nucleus so closely. Estimate the size of the nucleus based on these data. 09/03/2021 Jinniu Hu