Chronological Table for the Development of Atomic and Molecular Physics

≈ 440 Empedocles assumes, that the whole world is constant proportions”: Each chemical element BC composed of 4 basic elements: Fire, Water, Air consists of equal atoms which form with sim- and Soil. ple number ratios molecules as building blocks ≈ 400 Leucippos and his disciple Democritus claim, of chemical compounds. BC that the world consists of small indivisible 1811 Amedeo Avogadro derives from the laws particles, called atoms, which are stable and of Gay-Lussac (Δp/p = ΔT/T ) and Boyle- nondestructable. Marriot (p · V = constant for T = constant) ≈ 360 Plato attributes four regular regular geometric the conclusion that all gases contain under BC structures composed of triangles and squares equal conditions (pressure and temperature) (Platonic bodies) to the four elements and the same number of particles per volume. postulates that these structures and their inter- 1820 John Herapath publishes his conception of a change represent the real building blocks of the Kinetic theory of gases. world. 1857 Rudolf J.E. Clausius develops further the ki- ≈ 340 Aristotle contradicts the atomic theory and as- netic gas theory, founded by J. Herapath and BC sumes that the mater is continuous and does D. Bernoulli. not consist of particles. 1860 Gustav Robert Kirchhoff and Robert Bunsen 300 Epicurus revives the atomic model and as- create the foundations of spectral analysis of BC sumes that the atoms have weight and spatial the chemical elements. extension. 1865 Joseph Lohschmidt calculates the absolute 200 BC no real progress in atomic physics. number of molecules contained in 1 cm3 of Ð1600 AC a gas under normal conditions (p = 1atm, 1661 Robert Boyle fights in his book: “The Sceptical T = 300 K). Chemist” for the atomic model, which states, 1869 Lothar Meyer and D.I. Mendelejew estab- that all matter consists of atoms which differ lish (independent of each other) the Periodic in size and form for the different elements. System of the Elements. He defines the terms “chemical element” and 1869 Johann Wilhelm Hittorf discovers the cathode “chemical compound”. rays in gas discharges. 1738 Daniel Bernoulli assumes that heat can be ex- 1870 James Clark Maxwell gives the mathematical plained as the movement of small particles. He foundations to the kinetic gas theory. He de- may be regarded as the father of the kinetic gas fines the atoms as “absolute and unchangeable theory. building blocks of matter”. 1808 John Dalton supports in his book “ANew 1884 Ludwig Boltzmann develops from statistical System of Chemical Philosophy” the atomic grounds the distribution function for the en- hypothesis by describing his experiments on ergy of a system of free atoms in a constant careful weighing the masses of reactants and volume. Together with the Austrian physicist reaction products of a chemical reaction. The Josef Stefan he derives the Stefan–Boltzmann results of these experiments lead to the “law of radiation law.

W. Demtröder, Atoms, Molecules and Photons, 2nd ed., Graduate Texts in Physics, DOI 10.1007/978-3-642-10298-1, c Springer-Verlag Berlin Heidelberg 2010 520 Chronological Table

1885 Johann Jakob Balmer finds the Balmer- shows that both the characteristic and the con- formula for the spectral lines of the hydrogen tinuum radiation could be polarized. atom. 1913 Niels Bohr (Nobelprize 1922) develops his 1886 Eugen Goldstein discovers the “Kanal- new atomic model, based on the Rutherford Strahlen” (anode rays). model and the quantum hypothesis of Planck. 1886 Heinrich Hertz detects experimentally the 1913 Henry Moseley finds periodic regularities for electromagnetic waves predicted by Maxwell’s the absorption frequencies of X-rays by differ- theory and discovers 1887 the photo-electric ent atoms and is able to determine the nuclear effect and performs first experiments on the charge number Z of the atoms from his mea- absorption of cathode rays. surements of absorption edges. 1888 Phillip Lenard further investigates the absorp- James Franck and Gustav L. Hertz investigate tion of the cathode rays. the inelastic collisions of electrons with atoms 1895 Wilhelm Conrad Röntgen discovers, while (FranckÐHertz experiment). Nobel prize 1925. working on the properties of cathode rays 1919 Arnold Sommerfeld comprises all known facts a new kind of radiation which he called X-rays and models of atoms in his famous textbook: (first Nobel Prize in Physics 1901). “Atombau und Spektrallinien” and refines the 1896 Henry Becquerel first discovers radioactivity atomic model of Bohr. (Nobel prize 1903). 1921 Otto Stern and Walter Gerlach investigate the 1898 Marie Curie separates different radioactive el- deflection of atoms in an inhomogeneous mag- ements (Polonium and Radium) from minerals netic field and demonstrate the quantization of (Nobel prizes for Physics 1903 and Chemistry the component of the atomic angular momen- 1911). tum. 1900 Max Planck presents his new theory of black 1923 Arthur Holly Compton (Nobel prize 1927) ex- body radiation, introducing the energy quanta plains the inelastic scattering of X-rays by h · ν of the radiation field. This is nowadays electrons (Compton effect) using the model of regarded as the birth year of quantum physics light quanta. (Nobel prize 1918). 1924 Louis de Broglie (Nobel prize 1929) introduces 1905 Albert Einstein develops his theory of Brow- the concept of matter waves. nian motion. He explains the photoelectric 1925 S.A. Goudsmit and G.E. Uhlenbeck explain the effect using Planck’s light quantum hypothesis anomalous Zeeman effect by introducing the (Nobel prize 1921). electron spin, postulated theoretical already 1906 Charles Glover Barkla discovers the charac- 1924 by W. Pauli. teristic X-rays of the elements (Nobel prize 1925 W. Pauli (Nobel prize 1945) introduces the ex- 1917). clusion principle (Pauli-principle) which states 1909 Robert Millikan measures the elementary that every existing atomic state occupied by charge e with his oil-droplet experiment more than on electron must be described by (Nobel prize 1923). a wavefunction (product of spatial part and 1911 Ernest Rutherford and his coworkers investi- spin function) which is antisymmetric with gate the scattering of α-particles by gold nuclei respect to the exchange of two electrons. and postulates his atomic model. This can be 1925 Erwin Schrödinger (Nobel prize 1933) ex- regarded as the foundation of modern atomic tends the ideas of deBroglie about matter physics (Nobel prize for Chemistry 1908). waves to a general wave-mechanics which is 1912 Max von Laue (Nobelprize 1914) and his based on a special wave equation, called the coworkers demonstrate, that X-rays repre- Schrödinger equation. sent electro-magnetic waves by observing the 1927 Wolfgang Pauli gives a mathematical descrip- diffraction of X-rays by crystals. tion of the electron spin in form of quadrat- Shortly later William Henry Bragg (Nobelprize ic “spin-matrices” with two rows and two 1915) confirms this result and furthermore columns. Werner Heisenberg (Nobel prize Chronological Table 521

1932) develops together with Max Born 1954 N.G. Basow, A.M. Prochorov and Ch. Townes (Nobel prize 1954) and Pascual Jordan the (Nobel prize 1964) develop the theoretical mathematical concept of quantum mechan- foundations of the maser principle, based on ics, represented by matrices. He derives the Kastler‘s idea of optical pumping. First exper- uncertainty relations. imental verification of the NH3-maser by J.P. 1928 J.C. Davisson (Nobel prize 1937) and L.H. Gordon, H.J. Zeiger and Ch. Townes. Germer prove experimentally the wave na- 1957 Explanation of supra-conductivity by John ture of electrons by observing the diffraction Bardeen, Leon Cooper and J. Robert Schrieffer pattern of electrons passing through thin crys- (BCS theory) Nobel prize 1972. talline foils. 1958 Rudolf Mößbauer: Recoil-free emission and Paul Dirac (Nobel prize 1933) develops a rel- absorption of γ -quants by atomic nuclei ativistic theory of Quantum Mechanics. (Mößbauer effect) (Nobel prize 1972). Chandrasekhara Venkata Raman (Nobel prize 1959 Arthur Schawlow (Nobel prize 1995) and 1930) discovers the inelastic scattering of light Charles Townes give detailed description for by molecules (Raman-effect). the extension of the maser principle to the 1932 E. Ruska (Nobel prize 1986) constructs the optical range. first electron microscope. 1960 First experimental realization of an optical 1936 I. Rabi (Nobel prize 1944) demonstrates a new maser (ruby laser) by Th. Maiman. techniques of radiofrequency spectroscopy in 1961 The first He-Ne-laser is constructed by W.R. molecular beams for the precise measurement Bennet and A. Javan, based on detailed inves- of magnet moments. tigations of atomic collision processes in gas 1944 G.Th. Seaborg (Nobelprize for Chemistry discharges. 1951) identifies the first tran-uranium ele- 1966 The dyelaser is developed indepently by F. P. ments. Schäfer and P.A. Sorokin. 1947 Polykarp Kusch (Nobel prize 1955): Mea- 1971 G. Herzberg receives the Nobel prize in surement of the magnetic moment of the Chemistry for his centennial work on molec- electron. ular spectroscopy. Willis Lamb (Nobel prize 1955): Measurement 1980 First proposals for optical cooling of atoms by of the energy difference (Lamb-shift) between photon recoil by Th.W. Hänsch, A. Schawlow the 2S1/2 and 2P1/2 levels in the hydrogen and V. Letokhov. atom. 1982 Development of tunnel microscopy by G. 1947 John Bardeen develops together with W.H. Binning and H. Rohrer, where single atoms Brattain and W. Shockley the transistor (Nobel on surfaces can be observed (Nobel prize prize 1956). 1986). 1948 Felix Bloch and Edward Mills Purcell (Nobel 1986 Discovery of high temperature supra conduc- prize 1952) demonstrate the nuclear magnetic tivity by J. Bednarz and K.A. Müller (Nobel resonance technique NMR. prize 1987). 1948 J. Schwinger, R.P. Feynman and S. Tomonaga 1988 Nobel prize to H. Michel, J. Deisenhofer and (Nobel prize 1965) Theoretical formulation R. Huber for the elucidation of the primary of quantum field theory (quantum Electro- process in the photosynthesis of green plants dynamics). using femtosecond laser spectroscopy. 1950 A. Kastler (Nobel prize 1966) and J. Brossel 1989 Nobel prize to Norman Ramsey, H. Dehmelt demonstrate the experimental technique of op- and Wolfgang Paul for the experimental stor- tical pumping using incoherent light sources age and trapping of neutrons, ions and elec- before the invention of the laser. trons in electromagnetic traps. 1953 F. H . C r i ck and J.D. Watson prove experimen- 1991 Pulse-Fourier-transform NMR spectroscopy: tally by X-ray diffraction the double helix Nobel prize to Richard Ernst. structure of DNA (Nobel prize for medicine 1992 Manipulation of single atoms on surfaces us- 1963). ing the atomic force microscope. 522 Chronological Table

1994 Optical cooling of free atoms in the gas phase; 1998 First demonstration of a continuous coherent observation of optical molasses. beam of cold atoms from a BEC (atom laser) 1995 Realization of magneto-optical traps; cooling by Th.W. Hänsch and coworkers. of atoms below 1 μK by Sysiphos cooling, de- 2000 Development of optical frequency comb. veloped by C. Cohen-Tannoudji and cowork- 2001 Production of very cold molecules by recom- ers. bination of atoms in a BEC. First realization of BoseÐEinstein Condensa- 2005 First observation of Bose Einstein Condensa- tion (BEC) by C. Wieman, E. Cornell and tion of molecules. indepently by W. Ketterle, using the combina- 2006 Nobel prize to J. Hall and Th. Hänsch for the tion of optical cooling and evaporation cooling development of optical precision spectroscopy to reach temperatures below 100 nK (Nobel based on the optical frequency comb, to R. prize 2001). Glauber for his contributions to the quantum 1997 Nobel prize to St. Chu C. Cohen-Tannoudji theory of optical coherence. and W. Phillips for their developments of experimental techniques to cool and trap atoms with laser light. Solutions to the Exercises

Chapter 2 10−3 b) 1 cm3 He =ˆ V 22.4 M 1. a) The mean distance is × 23 × −3 ⇒ = 6 10 10 = × 19 1 N 2.7 10 . d = 3 m 22.4 2.6 × 1025 − = 3 × 10 9 m ≈ 15 atom diameters . 6 × 1023 c) 1 kg N =ˆ × 103 molecules 2 28 b) The ﬁlling factor η is: ⇒ N = 4.3 × 1025 atoms . 4 4 − η = π R3n = π 10 30 × 2.6 × 1025 3 6 3 5 3 3 d) 10 dm H2 at 10 Pa =ˆ 100 dm at 10 Pa = −4 = 1.1 × 10 = 0.01% . 1atm c) The mean free path length is 100 ⇒ ν = ≈ 4.5 moles 1 22.4 Λ = √ 23 24 σ ⇒ N =4.5 × 6 × 10 =2.7 × 10 molecules= n 2 24 − 5.4 × 10 H-atoms . σ = π(2r)2 = 4π R2 = 1.3 × 10 19 m2 − n = 2.6 × 1025 m 3 4. p = nkT; n = 1cm−3 = 106 m−3 1 − 6 −23 ⇒ Λ = √ m = 2.2 × 10 7 m ⇒ p = 10 × 1.38 × 10 × 10 [Pa] 2 × 3.3 × 106 − − = 1.38 × 10 16 Pa ≈ 1.38 × 10 18 mbar . = 220 nm . Such low pressures cannot be obtained in laborato- 2. The mass density is: ries. Because of outgassing of the walls of a vacuum chamber and backstreaming of gas through the = × + × + × m (0.78 28 0.21 32 0.01 40) vacuum pump the lowest achievable pressure in the × nAMU lab is around 10−10 Pa ≈ 10−12 mbar. − ◦ ◦ 1AMU= 1.66 × 10 27 kg, n = 2.6 × 1025/m3 5. 100 N =ˆ 273 K ⇒ 1 N =ˆ 2.73 K. The mean energy per atom and degree of freedom for a ﬁxed ⇒ m = (21.8 + 6.72 + 0.4) − temperature must be independent on the chosen × 2.6 × 1025 × 1.66 × 10 27 kg/m3 temperature scale. The new Boltzmann constant kN = 1.25 kg/m3 is then obtained from: 1 1 1 ⇒ kBTK = kNTN 3. a) 1 g12C = mol 2 2 12 1 − ◦ ⇒ k = k = 5.1 × 10 24 J/ N. ⇒ N = 6 × 1023/12 = 5 × 1022 . N 2.73 B 524 Solutions to the Exercises

The boiling point of water, measured in the new The vertical distribution is temperature scale is: −m∗gz/kT n(z) = n0e 100 ◦ ◦ n(h1) −( ∗ / ) − TS = 100 + N = 136.6 N. ⇒ = e m g kT (h1 h2) 2.73 n(h2) m∗gΔh ⇒ k = 6. The sound velocity vPh in a gas at pressure p and T ln(n1/n2) density is 7.7 × 10−18 × 9.81 × 6 × 10−5 = / v = κp/ with κ = C /C 290 ln(49 14) Ph p V − = 1.25 × 10 23 J/K. f + 2 = , f The best value accepted today is = × −23 / where C p is the molar speciﬁc heat at constant k 1.38 10 J K. pressure, CV at constant volume and f is the 8.3 N = R/k = number of degrees of freedom. A 1.25 × 10−23 mol 23 ⇒ v2 = κ / ≈ 6.02 × 10 /mol Ph p . 23 −14 M = NAm = 6.02 × 10 × 4.76 × 10 g/mol From the general gas equation for a mole vol- = 3 × 1010 g/mol . ume VM pV p M When a colloid molecule has an average mass M 4 pVM = RT ⇒ R = = , number of 10 AMU the nanoparticle consists of T T about 3 × 106 molecules. (M = mole mass) 8. a) If the ﬁrst diffraction order is at an angle β1 = ◦ α we obtain: 87 , the incidence angle can be obtained from the grating equation (see Fig. S.1): M · v2 v2 = κ RT/M ⇒ R = Ph α − β = λ ⇒ Ph κ · d(sin sin 1) T λ sin α = + sin β For radial acoustic resonances the acoustic wave- d 1 length is 5 × 10−10 = + 0.99863 × −6 nλ = r0 ⇒ vPh = νnλ = (νn/n)r0 0.83 10 = 0.99923 ◦ The general gas constant R is then obtained as ⇒ α = 87.75 . v2 M ν2r 2 M R = Ph = n 0 The second diffraction order appears at: κT n2κT 2λ ◦ sin β = sin α − = 0.99803 ⇒ β = 86.40 . For argon κ = ( f + 2)/f = 5/3, M = 40 g/mole. 2 d 2 Measuring the frequencies ν for different values ◦ ◦ n The difference is Δβ = 0.6 .Forα = 88.94 we of the integers n (n = 1, 2, 3, ...) yields the gas ◦ would obtain: Δβ = 0.75 . constant R, because M, r0, κ, T are known. 7. The effective mass of the collodial particle is:

∗ 4 4 m = m − πr 3 = πr 3( − ) 3 liquid 3 part liquid − = 7.74 × 10 18 kg, −17 m = 4.0 × 10 kg . Fig. S.1. Chapter 2 525

The radius r0 of the spheres is (according to Sect. 2.4.3) √ 1 − r = 2a = 2.33 × 10 10 m 0 4 4 − ⇒ V = πr 3 = 5.3 × 10 29 m3 . sph 3 0 The ﬁlling factor is √ 4 × 4 πr 3 16π × 2 2 η = 3 0 = ≈ 0.78 . a3 3.64 9. The van der Waals equation for 1 mole is: a Fig. S.2. p + V − b = RT 2 ( M ) VM V = b) The Bragg condition is ( M mole volume). a ab α = λ ⇒ pV − pb + − = RT 2d sin M 2 . VM V λ 2 × 10−10 M ⇒ d = = α × m b a/p ab/p 2sin 2 0.358 ⇒ pVM 1 − + − = RT . −10 V 2 3 = 2.79 × 10 m. M VM VM This is half the side length a of the elementary cell This can be written as of the cubic crystal (Fig. S.2). pVM (1 − x) = RT with (x 1) . ⇒ = a 0.56 nm . With NaCl has a face-centered cubic (fcc) crystal 1 ≈ 1 + x structure. Each elementary cell is occupied by 1 − x four NaCl molecules. The molecular mass of NaCl b a/p ab/p is (23 + 35) = 58 AMU. The number of molecules ⇒ pV = RT 1 + − + . M V 2 3 per m3 is: M VM VM Comparison with the virial equation (vires = = 4 3 = × 28 −3 N − m 2.5 10 m , forces) 5.63 × 10 30 The mass of one molecule NaCl is: B(T ) C(T ) pV = RT + + 3 M 1 2 2.1 × 10 − VM V m = = kg = 9.2 × 10 26 kg . M NaCl × 28 N 2.28 10 gives the coefﬁcients: In 1 mole of NaCl (= 58 g) are NA molecules. B(T ) = b = 4 times the eigenvolume of all × −2 molecules in VM, and ⇒ = 5.8 10 −1 = × 23 −1 NA − mole 6.3 10 mole . a 9.2 × 10 26 C(T ) =− , p c) From the Bragg condition = / 2 2d sin ϑ = mλ The ratio pi a VM is called the “internal pressure” = pressure caused by the mutual attraction of we obtain for m = 1 the side length a = 2d of the / 2 the molecules. The term C(T ) VM gives the ratio elementary cell of the fcc crystal as pi/p of “internal pressure” to external pressure. 2 λ − 10. a) If a parallel beam of atoms A per m per s a = = 6.6 × 10 10 m. sin ϑ hits atoms B at rest, the scattering cross section is 526 Solutions to the Exercises

Fig. S.3. With an acceleration voltage U the velocity vz is from m 2 eU = v ⇒ vz = 2eU/m .(2) 2 z

Insertingin(1)gives

2πm 4π 2m2 2eU L = v ⇒ L2 = eB z e2 B2 m σ = π + 2 = (Fig. S.3) (r1 r2) . For equal atoms (A e 8π 2 U B) r = r ⇒ σ = 4πr 2 = π D2 with D = 2r. = , 1 2 m L2 B2 The number of particles scattered out of the beam δ / δ δ δ along the path dx through a gas of atoms B with ⇒ (e m) ≤ 2 L + 2 B + U / density nB is e m L B U − − − = 4 × 10 3 + 2 × 10 4 + 1 × 10 4 dN =−NnBσ dx −3 −nBσ x = × ⇒ N(x) = N0e . 4.3 10 . The mean free path length for which N(x) = N0/e is In order to set L = 4 f one has to determine 1 the focal length f . This is achieved by shifting Λ = . n σ the aperture A2 until maximum transmission is B reached. b) In a gas with thermal equilibrium the particles Assume the maximum deviation of the electrons have an isotropic Maxwellian velocity distribution. from the axis is a = 5mm; The mean time between two collisions is: 1 τ = with L = 100 mm we obtain σ|v | n r 5 ⇒ sin α ≈ = 0.2 rad where the relative velocity is 25 vr = v1 − v2 (Fig. S.4). If the position of the focus is shift- ⇒ v2 = v2 + v2 − v · v r 1 2 2 1 2 ed by ΔL from its optimum value, the radius ⇒ v2 = v2 + v2 v · v = of the convergent electron beam is enlarged by r 1 2 , because 1 2 0 ΔL tan α from r0 to r0 + ΔL tan α. If the current I for A = Bis v2 = v2 = v2 of electrons flows through the aparture with radius 1 2 r0 = 0.5 mm (can be measured within ΔI/I = ⇒ v2 = v2 − r 2 . 10 3) the shift can be seen, if the area of the 1 ⇒ τ = √ , 2 nσ v2 Λ = τ v2 1 = √ . 2 nσ 11. a) For a longitudinal magnetic field B with a length L = 4 f , the time of flight between the two apera- tures in Fig. 2.67 is, according to (2.99b), 2πm T = with T = L/v .(1) eB z Fig. S.4. Chapter 2 527 π 2 + −3 12. a) The vertical force is gravitation. If the flight aparture has increased to r0 1 10 . direction is the x-direction the flight time is: − π (r + Δr)2 − r 2 ≤ 10 3πr 2 0 0 0 L 2 −3 tf = = s = 6.7 × 10 s. −3 −4 v 300 ⇒ Δr 10 r0 = 2.5 × 10 mm , Δr = ΔL tan α The vertical deflection is − 1 1 − ⇒ ΔL ≤ 2.5 × 10 4 mm/0.2 Δ z = gt2 = · 9.81 · 6.72 · 10 6 m 1 2 2 = × −3 − 1.25 10 mm , = 2.2 × 10 4 m = 0.22 mm . ⇒ Δ / = × −5 L L 1.25 10 . The divergence of the beam is: Δ −6 The geometrical uncertainty of L from the mea- b + d 4 × 10 − Δϑ = 1 0 = = 2 × 10 5 . surement of L is therefore much larger than that 2 · L 2 of the uncertainty of measuring the transmitted The width of the beam at a distance d = 200 cm current. The maximum relative error is therefore 2 downstream of S is: not affected by the uncertainty of optimising the 1 −3 transmitted current but rather by the mechanical Δ2z = d2 · Δϑ = 4 × 10 m. accuracy of length measurement. The deflection by gravity therefore changes the b) The maximum deflection from the straight path beam intensity, transmitted through a slit S2 by δ < −3 ∝ − − − is x 10 b, because the current I b can the fraction 2.2 × 10 4/4 × 10 3 = 5.5 × 10 2 = Δ / = −3 be measured within I I 10 . The deviation 5.5%. from the z-axis is b) The deflection of atoms with mass M and charge 1 2 1 1 Δq is: x = at with a = eEx − evBy = Fx . 2 m m 1 E · Δq With t = L/v and v2 = 2eU/m we obtain Δz = t2 2 M 2 1 L 1 6 2 −6 x = Fx = · 5 × 10 · Δq · 6.7 × 10 m 2m 2eU 2 Δ ∂x ∂x ∂x 2 q ⇒ δ x = δ F + δ L + δU = 1.1 × 10 · [m] . ∂ F x ∂ L ∂U M x δ δ δ δ Assume M = 4AMU= 6.7×10−27 kg (He-atoms) ⇒ x = Fx + L + U 2 . and a sensitivity of Δzmin = 10 μm than the min- x Fx L U imum value of Δq is Δq = 6 × 10−35 C = δ / = × −5 δ / =δ / + min With L L 1.25 10 , Fx Fx Ex Ex 3.8 × 10−16 e. δ / = × −4 δ / = −4 By By 2 10 , U U 10 we get c) The change Δz of the deflection is 2Δz = δ x − × 2 Δ / = 3.3 × 10 4 2.2 10 q M [m]. The relative change of the x intensity, transmitted through S2 is, according to a) −5 ⇒ δ x = 3.3 × 10 mm ΔI 2.2 × 102Δq/M = . for x = b = 0.1 mm. The uncertainty δ x < 10−3b I 4 × 10−3 = × −4 1 10 mm due to the uncertainty of measur- If a relative change of 10−4 can be still measured, ing the current I is larger here than in problem a) than the minimum charge difference because of the uncertainties of measuring E, B − − / = 2/ 2 10 4 × 4 × 10 3 and U. The ratio e m E (2UB ) can then be Δq = · M measured within 2.2 × 102 δ / δ δ δ − (e m) = E + U + B can be measured. Inserting M = 6.7 × 10 27 kg / 2 2 e m E U B gives δ − x −3 Δ = × 35 ≤ < 10 . q 1.2 10 C. x 528 Solutions to the Exercises

v2/ = v = / ϕ The optimum voltage is then: 13. From m R e B and f0 R sin one + obtains R1 R2 m 2 R2 √ U = φ2 − φ1 = v ln mv mv 1 2R e 0 R B = = = 2meU . 0 1 eR ef0 sin ϕ ef0 sin ϕ mv2 R = 0 ln 2 3 −16 With eU =10 eV=1.6 × 10 J, m =40 AMU√ e R1 = × × −27 ϕ = ◦ = 1 2eV 40 1.66 10 kg, sin sin 60 2 3, = 0 / −2 ln(R2 R1). f0 = 0.8 m ⇒ B = 4.2 × 10 Tesla. e 14. According to (2.100) the focal length f is: b) Assume an ion enters the cylindrical field at √ √ ϕ = 0 and r = R0 = (R1 + R2)/2 with the veloc- 4 φ0 2 φ0/a f = = ity v0. Assume it deviates at time t from its opti- z0 z0 √2a dz dz mum path r = R0 by δr. The equation of motion 2 φ0+az (φ2/a)+z2 = · 0 √ z=0 0 F m a than becomes for the radial motion: 2 2 φ0/a v = mδr¨ = m · + e · E(r + δr). (3) z0 r + φ2/ + 2 ln z ( 0 a) z Expansion of E into a Taylor series yields: √ 0 dE 2 φ0/a + δ = + δ + ... = . E(R0 r) E(R0) r .(4) 2 2 dr z0+ φ /a+z R0 ln 0 0 φ2/ From a) we obtain: 0 a dE U 1 15. a) The potential of the cylindric condenser can = · .(5) dr ln(R /R ) r 2 be obtained from the Laplace equation Δφ = 0, 2 1 Inserting this into (4) and (3)gives: which is written in cylindric coordinates (r, ϕ, z) as: v2 v2 δr δr¨ − 0 R2 + 0 1 − = 0 1 ∂ ∂φ r 3 0 R R r · = 0, 0 0 r ∂r ∂r 1 1 1 3 = ≈ − δr + ... r 3 R3(1 + δr/R )3 R3 R4 with the solution 0 0 0 0 2 φ = + v δr δr c1 ln r c2 . ⇒ δr¨ − 0 1 − 3 − 1 + = 0 R R R = = 0 0 0 The cylinder surfaces r R1 and r R2 are at the v2 φ φ 2 0 fixed potentials 1 and 2. This gives: ⇒ δr¨ + δr = 2 0. R0 c2 = φ1 − c1 ln R1 With ω0 = v0/R0 this becomes: φ2 − φ1 c = 2 1 / δr¨ + 2ω δr = 0 ln(R2 R1) 0 √ φ − φ 2 1 ⇒ δr = R0 sin 2ω0 · t . ⇒ φ(r) = φ1 + · ln(r/R1). ln(R2/R1) The electric field E(r)is: ∂φ φ − φ 1 E(r) =− = 1 2 · . ∂r ln(R2/R1) r The optimum path of the ions through the center R0 = (R1 + R2)/2 of the cylindric sector field is obtained from: v2 φ − φ m 0 2e 1 2 = e · E(R0) = . R0 R1 + R2 ln(R2/R1) Fig. S.5. Chapter 2 529 √ √ Inserting the numerical values for m1 =110 AMU, After the time t = π/ 2ω0 ⇒ ϕ = π/ 2 = m2 = 100 AMU gives: 127◦. The deviation δr becomes zero. The cylin- ◦ 8 −14 drical condenser with an angle ϕ = 127 therefore ΔT2 = 1.49 × 10 × 2 × 10 s acts as focussing device. ≈ 3 μs. 16. An ion produced at the location x travels a distance Two masses can be separated if their flight time = 1 2 = / = s 2 at1 with a eE m in the electric field E difference ΔT2 is at least ΔT1 or larger. The mass U/d to the grid 2 (Fig. S.5). resolution of our example is only Δm/m ≈ 10. It can be greatly increased by the McLaren 2ms ⇒ t1 = arrangement with two different accelerating elec- eE tric fields. 2eEs v = (eE/m) t = . b) The increase of mass resolution in the reflec- 1 1 m tron can be seen as follows: Assume the lengths of = = The drift time in the field-free region between the two arms of the reflectron to be L1 L2 L. grid 2 and 3, where the ion moves with constant Two ions, generated at two different locations have v v v = · / 1/2 velocity, is: velocities 1 and 2 with i (2eE si m) . Their flight time is: Ti = 2L/vi without penetrat- m ing into the reflection field Er. Here they penetrate t2 = L/v1 = L . 2eEs a distance dr determined by the energy balance: m The total flight time is: v2 = e · E · d ⇒ d = mv2/ (2eE ) . 2 i r r r i r m 2s + L T = t + t = √ . The deceleration time is obtained from 1 2 eE 2s 1 E d = e r t2 . The time difference ΔT for ions of equal mass, r 2 m s = d + b / s = d − b / produced at 1 ( ) 2 and 2 ( ) 2 Their time for penetration and reflection is there- (i.e., at the opposite edges of the ionization vol- fore ume) is 1/2 t = 2 · (2dr · m/eEr) . m 2s + L 2s + L ΔT = √1 − √2 1 Inserting si = (d ± b)/2 we obtain for the total eE 2s1 2s2 flight time of an ion through the reflectron: m d + b + L d − b + L = √ − √ . v eE + − 2L m i d b d b Ti = + 2 . vi eEr For m =100 AMU= 1.66 × 10−25 kg, b = 2mm, Inserting s = (d ± b)/2 ⇒ v = [(2eE/m)(d ± d = 30 mm, L = 1 m one obtains: i i b)/2]1/2. −5 ΔT1 = 1.018 × 10 (5.769 − 6.143)s We can calculate the maximum time difference for =−3.811 μs. ions generated at si = (d ± b)/2as: 2L 1 1 The ion with s = (d + b)/2 has a shorter flight ΔT = √ √ − √ / − + time than an ion starting from s = (d − b)/2, be- eE m b d b d cause it is accelerated longer and has a larger √ √ + 2m eE − − + velocity. d b d b . eEr m Two ions, with masses m and m , both starting 1 2 Δ / = from the middle of the ionization volume (s = With d T db 0 one obtains the optimum re- d/2) have the flight time difference: tarding field Er as / + √ √ E d − b − (d2 − b2)1 2 d L E = . ΔT2 = √ m1 − m2 . r 2 − 2 1/2 − − 2 2 − 2 −1/2 edE L (d b ) (d b) (d b ) 530 Solutions to the Exercises

c) The width of the ion beam at the exit of the The torque is the time derivative of the angular sector field is (see Fig. 2.74a) momentum L = r × p = m · r × v 2m dL b2 = b1 + Δv = D . eB dt mv2 mv = qvB ⇒ R = Since D = 0 ⇒ L = const. R qB 18. a) According to (2.163a) the impact parameter is m v2 = ⇒ = 1 2Um qZe qU R b = cot(ϑ/2) , with 2 B q πε μv2 4 0 0 R2 B2q = = ⇒ m = . q 2e , Z 79 , 2U −12 ε0 = 8.85 × 10 As/Vm, In order to seperate two masses m1 and m2,the ◦ μ − cot 45 = 1, v2 = 5MeV= 8 × 10 13 J condition 2 0 ⇒ = × −14 = 2(R1 − R2) ≥ b2 b 2.27 10 m 22.7 Fermi . has to be met. The mass resolution is then b) For the backwards-scattered particles (ϑ = 180◦) we obtain the minimum distance r at the m R2 B2q 2U min = turning point from the energy balance Δ 2 2 − 2 m 2U B q R1 R2 μ qZe R2 v2 = = 0 πε r 2 − 2 2 4 0 min R1 R2 qZe −14 Rm ⇒ r = = × ≈ min 2 4.54 10 m. 2πε0μv 2(R1 − R2) 0 ◦ c) For ϑ = 90 the impact parameter is b0 = with −14 2.27 × 10 m. All particles with b < b0 are scat- 1 tered into the angular range 90◦ <ϑ≤ 180◦. Rm = (R1 + R2). 2 In order to estimate the maximum value of b (i.e., ϑ = Since 2(R1 − R2) > b2 the smallest deflection angle ) we assume bmax d/2, where d is the average distance between two m R ⇒ ≤ m gold atoms in the scattering gold foil. In this case Δ . m b2 the α-particle passes between two gold atoms and Assuming Rm = 0.3 m, b2 = 1mm experiences a net deflection force of zero. Δ The number density of gold atoms is ⇒ m ≤ 300 . m NA 22 3 nV = = 6 × 10 /cm This is better than our simple TOF spectrometer, M but worse than the reflectron. with the mass density = 19.3 g/cm3;the 17. The torque D on a particle with mass m is: 23 Avogadro number NA = 6 × 10 /mole and the D = r × F , molar mass M = 197 g/mole. 2 The number density nF per cm of gold atoms where F is the force acting on the particle at in the foil with thickness t = 5 × 10−6 mis a distance r from the center. n = n t = 3 × 1019/cm2. With b = 1 d = F √ V max 2 For centro-symmetric force fields is 1 / = × −11 = × −13 2 nF 9.1 10 cm 9.1 10 mthe Fˆ = f (r) ·ˆr scattering cross section is ⇒ D = f (r) · r ×ˆr = 0. σ = π 2 ≈ × −20 2 bmax 2.6 10 cm . Chapter 3 531

The fraction of atoms scattered into the angular ϑ2 ϑ ≥ ◦ − ϑ/ϑ 2 range 90 is N(ϑ)Δϑ ∝ sin ϑe ( ) dϑ ϑ 2 1 N(ϑ ≥ 90◦) πb2 2.27 × 10−14 = 0 = ≈ ϑ −(ϑ/ϑ)2 ϑ ϑ ≤ ◦ π 2 × −13 e d N( 180 ) bmax 9.1 10 −4 ϑ = 6 × 10 . 2 2 ϑ − ϑ/ϑ 2 = e ( ) ◦ ◦ 2 ϑ d) b(ϑ = 45 ) = a cot 22.5 = 2.71 a with a = 1 ◦ ◦ − − qZe/(4πε μv2) ⇒ a = 2.27 × 10−14 m. N(1 ± 0.5 ) e 0.17 − e 1.56 0 0 ⇒ = ≈ 7.5 × 105 . N ◦ ± ◦ −14 − −21 (5 0.5 ) e e ◦ ≤ ϑ ≤ ◦ π (2.41)2a2 − a2 N(45 90 ) = This shows that the scattering rate decreases much ϑ ≤ ◦ π 2 ϑ N( 180 ) bmax stronger with increasing than for the Rutherford 4.8a2 4.8 × 2.272 × 10−28 scattering. = = 20. a) b2 8 × 10−21 max 2 2 −7 μv Ze = 3.1 × 10 . 0 = , 2 4πε0rmin −15 19. The rate N(ϑ) of particles scattered into the angu- rmin = 5 × 10 m, Z = 29 , lar range ϑ1 ≤ ϑ ≤ ϑ2 is for Rutherford scattering: 1 × 63 μ = = 0.98 AMU 64 ϑ − 2 μ 29 × 1.62 × 10 38 sin ϑ dϑ ⇒ v2 = J N(ϑ)Δϑ ∝ 2 0 4π × 8.85 × 10−12 × 5 × 10−15 4 ϑ/ sin ( 2) = × −12 ϑ1 1.33 10 J ϑ m 2 −12 2 ϑ/ ⇒ v = 1.36 × 10 J = 8.5 MeV . 2 cos( 2) 2 0 = dϑ ◦ sin3(ϑ/2) b) For ϑ<180 is ϑ 1 −1/2 ϑ E (r ) 2 2 r = b − pot min = − min 1 μ 2 . 2 . v sin (ϑ/2) ϑ 2 0 1 −15 −15 With rmin =5 × 10 m⇒ b = 1.775 × 10 m This allows the calculation of the ratio Ze2 ⇒ cot(ϑ/2) = b/a with a = πε μv2 ◦ ◦ 4 0 0 N(1 ± 0.5 ) 46,689 ◦ ◦ ◦ = = 218 . ⇒ ϑ ≥ 113.4 . N(5 ± 0.5 ) 214.4

For the Thomson model we obtain for a medium Chapter 3 scattering angle ϑ = 2 × 10−4 rad and an average number m of scattering events in the gold 1. With the de Broglie relation foil, according to the numerical value given in Problem 12.15c: λ = h dB p 19 −16 4 − m = nFσ = 3 × 10 × 3 × 10 ≈ 10 h 6.63 × 10 34 m √ − ◦ ⇒ v = = ⇒ ϑ = m ϑ = 2 × 10 4 × 102 rad ≈ 1.2 mλ 10−10 × 1.67 × 10−27 s − ◦ = 2 × 10 2 rad ≈ 1.2 , = 3.97 × 103 m/s. 532 Solutions to the Exercises

Thermal neutrons have at T = 300 K a mean ve- This equation can be only solved numerically. The locity v = 2.2 × 103 m/s and a kinetic energy of solution is: Ethermal = 40 meV. Our neutron is slightly super kin = thermal. Its kinetic energy is xm 2.8215 − ⇒ ν = 2.8215 kT/h = 5.873 × 1010 s 1 · T [K]. m 2 −20 m Ekin = v = 1.31 × 10 J = 82 meV . 2 With λ = c/ν ⇒ dλ =−(c/ν2)dν we obtain in- 2. The average energy per mode is ∗ stead of (3.16)forSλ the expression = · ν E n h , 2 ∗ 2πhc 1 Sλ = . where n is the average number of photons in this λ5 ehc/(λkT) − 1 mode. If P is the probability, that a mode con- n ∗/ λ = tains n photons, it follows for thermal equilibrium With dSλ d 0 we obtain in a similar way ∞ −3 −n·hν/kBT 2.88 × 10 [m] = e ⇒ = λ = . Pn − ν/ Pn 1. m e nh kBT T [K] n n=0 λ/ ν Pn+1 − ν/ Note, that d d decreases with increasing fre- ⇒ = e h kBT = x . ν ∗ P quency . The distribution Sλ, which gives the n λ ∞ radiation ﬂux per constant interval d , therefore n < ∗ The geometrical series n=0 x has for x 1the differs from Sν which is given for constant interval value dν. The maximum of S∗ at ν is not at ν = c/λ ! ν m m m 1 4. a) Energy conservation demands xn = . 1 − x n ν = Δ el h Ekin⎡ ⎤ (1) With the relation: 1 1 ∞ = m c2⎣ − ⎦ n d n x 0 n · x = x · x = − v2/c2 − v2/c2 dx (1 − x)2 1 2 1 1 n=0 we obtain The conservation of momentum requires: ∞ m v m v xn x h¯ k = 0 2 − 0 1 (2) n = n · P(n) = n · = xn 1 − x 1 − v2/c2 1 − v2/c2 n=0 2 1 1 2ν2 = 2 2 h ν/ . h¯ k = eh kBT − 1 c2 The mean energy per mode is then m2v2 m2v2 = 0 1 + 0 2 (3) ν 1 − v2/c2 1 − v2/c2 = h 1 2 E ν/ . 2 h kBT − 2m v · v e 1 − 0 1 2 ∗ . 3. Differentiation of Sν in (3.16)gives: − v2/ 2 − v2/ 2 1 1 c 1 2 c ∗ ν/ ∂ ν2 ν3 h kBT · / Sν = 6h 1 − 2h e h kBT ν/ ν/ Taking the square of (1)gives: ∂ν c2 eh kBT − 1 c2 (eh kBT − 1)2 ⎡ = 0 2ν2 = 2 4⎣ 1 + 1 hν −1 h m0c (4) hν/kBT hν/kBT − v2/ 2 − v2/ 2 ⇒ 3 − e e − 1 = 0. 1 1 c 1 2 c kBT ⎤ = ν/ With x h kBT this gives 2 − ⎦ . x − 3 = ⇒ x = 3 1 − e x . 1 − v2/c2 1 − v2/c2 1 − e−x 1 2 Chapter 3 533

A comparison of (3) and (4) gives, after rearrang- which should be larger than the slit width b.This ing the terms: gives the condition 2 1/2 2 − v2 2 − v2 = 2 − v · v 2Dh c 1 c 2 c 1 2 b < √ . 2mEkin ⇒ (v1 − v2) = 0 ⇒ v1 = v2 For D = 1 m and E = 1 keV = 1.6 × 10−16 J ⇒ ν = 0 ! kin we obtain: 1/2 This means that photoabsorption by a free electron 2 × 6.6 × 10−34 is not possible. The absorption can only take place bmax = √ m × × −31 × × −16 in the presence of an atom, which can compensate 2 9.11 10 1.6 10 − the photon recoil. In the Compton effect the scat- = 8.81 × 10 6 m = 8.81 μm. = tered photon has the momentum h¯ ks h¯ k and the 6. The radii of the Bohr orbitals are energy hν < hν. s n2 b) The momentum of the photon is: r = a . n Z 0 hν = n = Z = ⇒ r = a = × pphot . a) For 1, 1 1 0 5.29 c 10−11 m. −20 −13 For hν = 0.1 eV = 1.6 × 10 J(λ = 12 μm) b) For n = 1, Z = 79 ⇒ r1 = 6.70 × 10 m. −20 The velocities of the electron are 1.6 × 10 Js − ⇒ p = = 5.3 × 10 29 Ns h Zh¯ phot × 8 v = = 3 10 m 2 . 2πmern mea0n For hν = 2eV(λ = 600 nm) a) Z = 1, n = 1: ⇒ = × −27 pphot 1.07 10 Ns 6 −3 c ⇒ v1 = 2.19 × 10 m/s = 7.3 × 10 c = . For hν = 2MeV 137 b) Z = 79, n = 1: ⇒ = × −21 pphot 1.07 10 Ns 8 v1 = 1.73 × 10 m/s = 0.577 c . The recoil velocity of a hydrogen atom with the In case b) the relativistic effects become very large above momenta would be and have to be taken into account. We can calculate p − v = = 3.2 × 10 2 m/sforhν = 0.1 eV , the relativistic velocity: 1 m p v = = × −1 / ν = 2 2 1 2 6.4 10 m sforh 2eV, Ekin = (m − m0)c = m0c − 1 m 1 − v2/c2 p 5 ∗ v3 = = 6.4 × 10 m/sforhν = 2MeV. Ry m =−E = Z 2 n 2 . In the ﬁrst case the atom would not be pushed n = out of resonance for the Lyman α-line, for the last For n 1 we obtain: case it would be completely Doppler-shifted out of 2 2 m0c resonance. v = c 1 − , m c2 + E 5. The ﬁrst diffraction minimum appears at the 0 1 2 diffraction angle α with with m0c = 0.5 MeV we obtain: λ h h 792 × 13.5 sin α = = = √ . E = eV = 0.084 MeV b bp b 2mEkin 1 The full width between the two minima at both 0.5 2 ⇒ v = c 1 − sides of the central maximum is 0.584 2Dh B = 2D sin α = √ > b , = 0.517 c . b 2mEkin 534 Solutions to the Exercises

The relative error of the nonrelativistic calculation where mN is the mass of the nucleus. is 3 1 ⇒ Ry H = Ry∞ Δv 0.06 1 + m /m = c = 0.116 c = 11.6% . e N v 0.517 1 ≈ Ry∞ + 1 c) The relativistic mass increase is 1 3·1836 = 0.999818Ry∞ 1 − Δm = m − m0 = m0 − 1 = × 7 1 − v2/ 2 1.0971738 10 m . 1 c The wavelength of Lyman α n = 2 → n = 1is = √ 1 − m0 1 then: 1 − 0.5172 = 4 − 0.17 m0 . λ = = 1.215 × 10 7 m = 121.5 nm . 3Ry The relativistic energy correction is (see Sect. 5.4) b) For positronium (e+e−) ΔE n = Z = = × −4 r( 1, 1) 9 10 eV . − + 1 μ = me/2 ⇒ Ry(e e ) = Ry∞ For Z = 79 it is 2 ⇒ λ = 243.0 nm . ΔE (n = 1, Z = 79) = 5.6 eV . r 9. At room temperature (T = 300 K) only the ground 7. After the mean life time τ the number of neutrons state is populated. Therefore all absorbing transi- have decayed to 1/e of the initial value and after tions start from the ground state with n = 1. The τ / photon energies are then the time ln2to12 of the initial value. During this time they travel a distance x = vτ ln 2. 1 hν = a − The velocity of the neutrons is n 1 2 n h hτ ln 2 1 v = ⇒ x = hνn+1 = a 1 − mλ mλ (n + 1)2 − 6.62 × 10 34 × 900 × 0.69 λ = /ν = = 2.4 × 105 m. with c we obtain 1.67 × 10−27 × 10−9 λ ν − / + 2 1 = n+1 = 1 1 (n 1) The decay time of the neutrons could be measured 2 . λ2 νn 1 − 1/n by trapping them in a magnetic quadrupole trap λ = λ = ⇒ λ /λ = with the geometry of a circle. With a radius r = With 1 97.5 nm, 2 102.8 nm 1 2 = ⇒ λ /λ = = ⇒ 1m,theytravel(2.4× 105/2π) = 4 × 104 times 0.948. For n 2 1 2 0.843, for n 3 λ /λ = around the circle before they decay, if no other 1 2 0.948. = = losses are present. The two lines therefore belong to n 3 and n 4. The constant a can be determined from 8. The wavelength of the Lyman α-line can be obtained from the relation c a 1 νn = = 1 − λ h n2 hc ∗ 1 n hν = = Ry 1 − λ 4 with λ3 = 102.8 nm we obtain 4 hc 1 ⇒ λ = = ∗/ a = with Ry Ry hc . 2 3 Ry λ3 1 − 1/3 hc 9 − ∗ a) = · = 2.177 × 10 18 J = Ry . λ 8 μ 3 3 memN Ry H = Ry∞ · with μ = , The lines therefore belong to transitions in the me me + mN hydrogen atom with Z = 1, n = 3 and n = 4. Chapter 4 535

10. Since the resolving power of the spectrograph is The total energy is then assumed to be h¯ 2 2e2 λ ν E ≥ − . = = × 5 2ma2 4πεa Δλ Δν 5 10 From the condition dE/da = 0 for the minimum the difference Δν of two adjacent lines in the energy we obtain Balmer spectrum has to be Δν ≥ ν/(5 × 105). The 2πε h¯ 2 a frequencies of the Balmer series are a = 0 = 0 min me2 2 Ry∗ 1 1 νn = − n ≥ 3. 4e2 h 22 n2 ⇒ E (a ) =− pot min 4πε a The ratio ν/Δν is then 0 0 =−4Epot(H, n = 1) ν ν 1 − 1 = n = 4 n2 ≤ × 5 =−108 eV , 1 1 5 10 Δν νn+1 − νn − n2 (n+1)2 =−1 =+ n2 − 4 Ekin Epot 54 eV . ⇒ ≤ × 5 2 2 5 10 − n 4 4 n+1 ⇒ n ≤ 158 . Chapter 4 Another way of solving this problem is as follows: 1. Inserting the ansatz ψ (r, t) = g(t) · f (r) into the Ry∗ 1 1 ν(n) = − . time-dependent Schrödinger equation (4.7b) one h 4 n2 obtains, after division by f (r) · g(t), ν For large n we can regard (n) as a continuous 1 ∂g(t) h¯ 2 1 function of n and obtain by differentiating: ih¯ =− Δf (r) = C . g(t) ∂t 2m f (r) dν 2Ry∗ 1 = Since the left side of this equation depends solely dn h n3 2Ry∗ 1 on t, the right side solely on r, both sides have to ⇒ Δν ≈ Δn be constant, which we name C. The right side gives h n3 the time-independent Schrödinger equation (4.6) with Δn = 1weget: = − for C E Epot. Then the left side becomes ν 1 1 1 3 = − n . ∂g(t) E − Epot Δν 2 4 n2 = g(t) ∂t ih¯ 2 Since n 4: −iEkin/h¯·t ⇒ g(t) = g0e . 3 n 5 3 6 ≤ 5 × 10 ⇒ n ≤ 4 × 10 For a free particle Epot = 0 ⇒ Ekin = E.The 8 function g(t) then represents the phase factor ⇒ n ≤ 158 . − / − ω g(t) = g e i(E h¯)t = g e i t 11. For the uncertainty Δr = a the kinetic energy of 0 0 theelectronis with E = h¯ω. h¯ 2 2. The reflectivity R = 1 − T can be derived E ≥ . kin 2ma2 from (4.26a) when we insert: Its potential energy at a distance a from the nucleus E = 0.4 = is 0.8 , E0 0.5 2 =− 2e 1 Epot . α = 2m(E0 − E) 4πε0a h¯ 536 Solutions to the Exercises √ × −27 × × × −22 4. With = 2 1.67 10 0.1 1.6 10 −1 − m ik1x −ik1x 1.05 × 10 34 ψ1 = Ae + Be , 9 −1 ik2x −ik2x = 2.2 × 10 m ψ2 = Ce + De , ψ = A eik1x = × −9 ⇒ α · = 3 with a 1 10 m a 2.2. 1/2 2 k1 = 2mE/h¯ , 0.2 ⇒ T = = 0.126 , 1/2 + × 2 2 0.2 0.3125 sinh (2.20) k2 = 2m(E − E0)/h¯ = iα i.e., 12.6% of all particles are transmitted, 87.4% we obtain from the boundary condition (4.25)the are reflected. relations: 3. For the negative potential step the same derivation A + B = C + D , can be used as for the positive step. The reflection ik a −ik a ik a coefficient is Ce 2 + De 2 = A e 1 , k1(A − B) = k2(C − D), | |2 − 2 B k k − − R = = . k Ceik2a − De ik2a = k A e ik1a . |A|2 k + k 2 1 This set of equations yields the results: We abbreviate: 2 + 2 √ k1 k2 ik a 1 1 A = cos k a − i sin k a e 1 A , k = 2mE ; k = 2m(E − E ) 2 2k k 2 h¯ h¯ 0 1 2 k2 − k2 2 1 ik1a with E0 < 0 and obtain B = i sin k2ae A . √ 2k1k2 E − E0/2 − E(E − E0) The reflection coefficient with cos2 x = 1 − sin2 x R = √ . E − E0/2 + E(E − E0) 2 2 2 2 2 |B| k − k sin k2a R = = 1 2 For E = 0 ⇒ R = 0, for E →−∞⇒R = 1. |A|2 2 2 2 2 2 2 0 0 4k k + k − k sin k2a For E =−E the kintetic energy of the particle 1 2 1 2 0 and the transmission coefficient is: becomes for x > 02Ekin(x < 0). The reflection |A|2 4k2k2 coefficient is then T = = 1 2 . √ |A|2 2 2 + 2 − 2 2 2 − 4k1k2 k1 k2 sin k2a = 3 2√2 = R 0.029 . + = 2 = 3 + 2 2 One can readily prove that R T 1. With k1 2 2 2 (2mE/h¯ ) and k = 2m(E − E0)/h¯ the transmis- =−1 2 For E 2 E0 we obtain: sion becomes √ 4E(E − E0) 2 − 3 T = √ . R = √ = 0.072 . 4E(E − E ) + E2 sin2 a 2m(E − E ) 2 + 3 0 0 h¯ 0 (1) This shows that with increasing step heights R increases. This is completely analogous to the re- Dividing the nominator and denominator by = flection of optical waves at a boundary between (4EE0) and using the relation sin(ix) isinhx we > > two media with refractive indices n and n , where obtain for E E0 the result (4.26a). For E E0 the 1 2 = transmission becomes T 1for − 2 Δ 2 a n1 n2 n 2m(E − E ) = nπ R = = h¯ 0 n + n (2n)2 1 2 h 2a ⇒ λ = √ = , n = 1, 2, 3 . = 1 + 2m(E − E0) n with n 2 (n1 n2). Chapter 4 537

For a potential well with depth E0 the potential energy is Epot < 0 if we choose Epot = 0 outside the well. In (4.26a) one has to change the sign of E0. With the numerical data of Problem 4.2 (E = 0.4 eV, E0 =−0.5 meV, a = 1 nm) we then obtain from (4.26a) 1 + 0.8 T = √ 1 + 0.8 + 0.31 sin2 a 2m × 0.9 meV/h¯ 1.8 = = 0.994 , 1.8 + 0.31 sin2(6.46) wherewehaveusedsinix = isinhx. 5. For a potential well with inﬁnitely high walls the − bound energy levels for a well with depth ( E0) Fig. S.6. are h¯ 2 π 2 E = n2 ≤ E where the phase ϕ determines the amplitude of ψ n 2 0 . II 2m a for x = 0 and x = a. Inserting the numerical values a = 0.7 nm, E = 0 ψ = −αx ≥ 10 eV gives: III A3e for x a . 2 From the boundary conditions −49 n −18 En = 1.1 × 10 J ≤ 1.6 × 10 J. m ψIII(0) = ψII(0) ; ψII(a) = ψIII(a)(1) − a) Electrons with mass m = 9.1 × 10 31 kg: we obtain: −19 2 En = 1.2 × 10 n J A1 = A2 sin ϕ ,(2) −α 1.6 × 10−18 A = A sin(ka + ϕ)e a .(3) ⇒ n2 ≤ = 12.9 ⇒ n ≤ 3. 3 2 × −19 1.2 10 From the continuity of the derivatives one obtains: There are only three bound levels in the well. − ψ = ψ ⇒ α A = kA ϕ b) Protons with mass m = 1.67 × 10 27 kg: I(0) II(0) 1 2 cos . −23 2 This gives with (2): ⇒ En = 5.59 × 10 J × n α = ϕ ⇒ ϕ = /α ⇒ n2 ≤ 2.4 × 104 k cot arctan(k ). (4) ⇒ n ≤ 155 . From the condition ψ ψ c) With the exact solution for the potential well d(ln II) d(ln III) = with ﬁnite wall heights the wave functions are no dx x=a dx x=a = = longer zero for x 0 and x a, but they penetrate one obtains: a little bit into the wall regions (Fig. S.6). As was discussed in Sects. 4.2.2 and 4.2.4 we now have the ϕ =−ka − arccot(k/α) + nπ .(5) wave functions The comparison of (4) and (5)givesfork the αx ψI = A1e for x ≤ 0 condition: with ka = nπ − 2arccot(k/α) 1 and the energy levels: α = 2m(E0 − E); h¯ 2 2 2 h¯ kn h¯ 2 ψII = A2 sin(kx + ϕ)for0≤ x ≤ a , En = = [nπ − 2 arccotg(k/α)] . 2m 2ma2 538 Solutions to the Exercises

For E =∞ ⇒ α =∞and the arccotg becomes ∂ϑ / 2 0 zy r zero. This gives the results = ∂y x2 + y2 2π 2 ∂ϕ ∂ϕ h¯ 2 y x kn = nπ/a ⇒ En = n , ϕ = arctan ⇒ = , = 0. 2ma2 x ∂y x2 + y2 ∂z which was derived from the well with inﬁnitely Inserting these relations into (1)gives: high walls. 6. At the lowest energy (zero-point energy) the par- ∂ ∂ ˆ =− − y ticle is restricted to the interval Δx = x − x Lx ih¯ 0 2 1 ∂r x2 + y2 ∂ϑ between two points zx ∂ =± / 1/2 − x1,2 (2Epot D) , 2 + 2 ∂ϕ x y which are the intersections of the energy level ∂ ∂ =+ih¯ sin ϕ + cotϑ cos ϕ . 1 ∂ϑ ∂ϕ E(v = 0) = h¯ D/m . 2 ˆ ˆ The components L y and Lz can be obtained in an With the parabolic potential analogous way. In order to obtain Lˆ 2 we use the 1 relation: E = Dx2 pot 2 Lˆ 2 = Lˆ 2 + Lˆ 2 + Lˆ 2 we obtain: x y z 1/2 ˆ 2 ˆ 2 ˆ 2 Δx = 2 h¯ D/m D and we have to calculate L x , L y and Lz : √ 1/2 ∂ ∂ = 2 h¯ D · m . Lˆ 2 =−h¯ 2 ϕ + ϑ ϕ x sin ∂ϑ cot cos ∂ϕ ˆ ∂ ∂ 7. The x-component of L is × ϕ + ϑ ϕ sin ∂ϑ cot cos ∂ϕ . ˆ ∂ ∂ L x =−ih¯ y − z ∂z ∂y The differential operators ∂/∂ϑ and ∂/∂ϕ act on all ∂ ∂ ∂ ∂ϑ ∂ ∂ϕ ∂ = r + + functions, after multiplication of the two brackets, ∂ ∂ ∂ ∂ ∂ϑ ∂ ∂ϕ stand behind these operators, z z r z z ∂r ∂r ∂ This yields the four terms: ⇒ Lˆ =−h¯ y − z x i ∂ ∂ ∂ z y r ∂ ∂ ∂2 ∂ϑ ∂ϑ ∂ sin ϕ sin ϕ = sin2 ϕ ; + y − z ∂ϑ ∂ϑ ∂ϑ2 ∂z ∂y ∂ϑ ∂ ∂ ϕ ϑ ϕ ∂ϕ ∂ϕ ∂ sin ∂ϑ cot cos ∂ϕ + y − z .(1) ∂z ∂y ∂ϕ 1 ∂ ∂ ∂ = sin ϕ cos ϕ − + cot ϑ ; 2 ∂ϕ ∂ϑ ∂ϕ with sin ϑ ∂ ∂ = 2 + 2 + 2 ϑ ϕ ϕ r x y z cot cos ∂ϕ sin ∂ϑ ∂r z ∂r y ∂ ∂ ∂ ⇒ = , = . = ϑ 2 ϕ + ϕ ϕ ∂z r ∂y r cot cos ∂ϑ cos sin ∂ϑ ∂ϕ ; z ∂ ∂ ϑ = arccos cot ϑ cos ϕ cot ϑ cos ϕ x2 + y2 + z2 ∂ϕ ∂ϕ ∂ϑ z2/r 2 − 1 ∂ ∂2 ⇒ = = cot2ϑ − cos ϕ sin ϕ + cos2 ϕ . ∂ϕ ∂ϕ2 ∂z x2 + y2 Chapter 5 539

ˆ 2 ˆ 2 is the de Broglie wavelength of the particle within Similar terms are obtained for L y and Lz .The addition finally gives (4.111) when the relation the range of the barrier. b) E = 0.8 eV, E0 = 1eV ⇒ E/E0 = 0.8, m = ∂ ∂2 1 ∂ ∂ × −31 cot ϑ + = sin ϑ 9.1 10 kg. ∂ϑ ∂ϑ2 sin ϑ ∂ϑ ∂ϑ α = 2m(E − E0)/h¯ is used. × × −31 × × −19 8. The wave function within the range x <0 and x >a = 2 9.1 10 (0.2 1.6 10 ) −1 − m is for a penetration depth δx 1.06 × 10 34 √ 9 −1 / − δ = 2.28 × 10 m ψ δ = (i h¯) 2m(E0 E) x ( x) Ce . − a = 10 9 m ⇒ sinh2(αa) = 24.4 The probability of finding a particle in this range is −0.2 proportional to |ψ(δ x)|2. It decreases to 1/efor ⇒ T = ≈ 0.03 . 0.2 − 0.28 × 24.4 h¯ δ x = √ . For E = 1.2 eV − 2m(E0 E) −0.2 − ⇒ T = = 0.625 . Example: m = 9.1 × 10 31 kg (electron mass), −0.2 − 0.208 sinh2 2.28 = 1 = = × −19 E 2 E0, E0 1eV 1.6 10 J 10. The energy levels in the two-dimensional quadratic 1.06 × 10−34 potential well are, according to (4.66): ⇒ δ x = √ m − − ¯ 2π 2 1.82 × 10 30 × 0.8 × 10 19 = h 2 + 2 ≤ E nx , ny nx ny Emax .(2) = 0.28 nm . 2ma2 1 Inserting the numerical values 9. For E = E0 we obtain from (4.26a): 2 = × −31 0.5 m 9.1 10 kg , T = = −8 0.5 + 0.5 sinh2 2π a 10 m, − = = 1.4 × 10 5 . Emax 1eV gives the conditions With the approximation (4.26b) we obtain for E = 0.5 E0: 2 + 2 ≤ × 2 nx ny 2.66 10 = −4π = × −5 T 4e 1.395 10 , ⇒ ≤ 2 + 2 ≤ nx , ny 16 and nx ny 266 . which is practical identically to the value of the All possible levels can be visualized as points correct calculation. 1 in a two-dimensional space with the axis nx For E = E0: 3 and ny. They fill a quarter of a circular 2/3 π/ 2 + 2 ≥ T = √ area ( 4) nx ny because nx , ny 0. There 2/3 + 3/4sinh2 2π 2 are therefore approximately (π/4) × 266 = 208 = × −8 energy levels that obey these conditions. Some 5.1 10 . of them are degenerate. These are levels with the 2 + 2 = = Maximum transmission T = 1 is reached for E > same value of nx ny. Examples are: nx ny 2 E0 if sin (iaα) = 0 5 and nx = 1, ny = 7 and nx = 7, ny = 1. n ⇒ a 2m(E − E ) = nπh¯ = h 0 2 n Chapter 5 ⇒ a = λ , 2 where 1. The expectation value of r is defined as

h ∗ λ = √ r = ψ rψ dτ 2m(E − E0) 540 Solutions to the Exercises

with dτ = r 2 sin ϑ dr dϑ dϕ.Inthe1s state of the where IP = 13.6 eV is the ionization potential. Hatom Therefore all lines appear in the emission that start from levels n ≤ 6. These are for n = 6 1 − / ψ = r a0 √ / e π 3 2 s → p p p p a0 6 5 ,4 ,3 ,2 , ∞ 6p → 5s ,4s ,3s ,2s ,1s , 1 − / ⇒ r = 4π e 2r a0 r 3 dr 6d → 5p ,4p ,3p ,2p , π 3 a0 0 6 f → 5d ,4d ,3d , 4 3! 3 6g → 5 f ,4f , = = a0 . a3 (2/a )4 2 0 0 and similar expressions for n = 5, 4, 3, 2. The expectation value of r is therefore larger than Since all terms with equal quantum number j = the Bohr radius a0! The expectation value of 1/r is l + s are degenerate, many of these lines are coin- ∞ cident in energy and are not separated. 1 1 − / 3. a) For the ground state 1s of the H atom r = a = 4π e 2r a0 r dr 0 r π 3 according to Bohr’s model. (Note, however, that a0 0 = 3 r 2 a0, see Problem 5.1.) 4 a2 1 The excitation energy of 12.09 eV reaches levels = 0 = . 3 a with energies a0 4 0 ∗ For the 2s state the wave function is Ry E = IP− = n 2 12.09 eV 1 r − / n ψ s = √ − r 2a0 (2 ) 3/2 2 e 2 13.599 4 2πa a0 ⇒ n = = 9 ⇒ n = 3. 0 13.599 − 12.09 ∞ 2 4π r − / ∝ 2 = = ⇒ r = 2 − e r a0 r 3 dr Since r n the Bohr radius becomes r(n 3) × π 3 a 16 2 a0 0 9a0. 0 = ∞ b) For the excitation energy Ee 13.387 eV we 4 1 − / 4r − / obtain in a similar way = 4r 3e r a0 − e r a0 3 a 8a0 0 13.599 0 n2 = = 64 ⇒ n = 8. 5 0.212 r − / + e r a0 dr = = a2 r(n 8) 64a0. 0 4. In the classical model, 1 = 24a4 − 96a4 + 120a4 = 6a . μ 3 0 0 0 0 μ =− / ⇒ e =− e 8a0 e e (2me)l l 2me A similar calculation for 1/r yields is constant and independent of the principal quan- 1 1 = . tum number n. r 4a0 In the quantum-mechanical description the expectation values: 2. The excitation energy of Ee = 13.3 eV can popu- late the upper levels En with energies eh¯ μz =−ml , −l ≤ ml ≤+l , Ry∗ 2me E = IP− ≤ E 2 2 n 2 e e h¯ n μ2 = l(l + 1) . Ry∗ 13.6 e 4m2 ⇒ n2 ≤ = = e − − 45.3 IP Ea 13.6 13.3 Although the number of possible components μ ⇒ ≤ z n 6, depend on l, the values are still independent of n. Chapter 5 541

5. a) The velocity of the electron on the lowest Bohr and shows that the model of a charged sphere for orbit is v1 = c/137. Its relative mass is then the electron is not correct. The deﬁnition of the 2 electron spin as the angular momentum of a sphere m0 1 v m(v) = ≈ m0 1 + with mass me is wrong! − v2/ 2 2 c2 1 c b) The rotational energy is: = m + × −5 0 1 2.66 10 . = 1 ω2 = 1 2ω2 = 1 v2 Erot I mer me eq . Δ = − = × 2 5 5 The mass increase m1 m m0 2.66 − − −5 = × 15 ⇒ = × 9 10 m0.Forn = 2, because of v ∝ 1/n, v(2s) = For r 1.4 10 m Erot 6 10 J, − = 2 = × −14 3.65 × 10 3c while the mass energy E0 mec 8 10 J is much smaller. This shows again that the mechan- −6 ⇒ m(v) = m0 1 + 6.6 × 10 ical model of the electron can not be correct. 2 −6 7. The Zeeman splitting of the 2 S1/2-state is, accord- ⇒ Δm2 = 6.6 × 10 m0 . ingto(5.7), The difference is: ΔEs = gj μB B , δm = Δm1 − m2 with g = g ≈ 2. For the 32 P / state it is (see = × −5 = × −35 j s 1 2 2.0 10 m0 1.8 10 kg . Fig. S.7): b) The energy difference is: ΔE p = gj μB B − ΔE = E(2s) − E(1s) = 10 eV = 1.6 × 10 18 J. with =− The potential energy Epot 2Ekin. Since 1 3 + 1 3 − × 2 2 2 2 1 2 2 gj = 1 + = . 1 1 3 3 E = E + E = E ⇒ E = 2 · E . 2 2 2 pot kin 2 pot pot The four Zeeman lines (Fig. S.7) appear as two The difference in potential energy corresponds to pairs, where the smaller distance is a mass difference 2 2 Δm = m1 − m2 = 2ΔE/c Δν1 = μB B/h − 3 =−3.6 × 10 35 kg . and the larger distance is Both effects are opposite in sign. Because the rel- 4 ative velocity effect mv(1s) > mv(2s)isonly1/2 Δν = μ B/h . 2 3 B of the potential energy effect, the mass of the atom is larger in the 2s state than in the 1s state. 6. a) The angular momentum of a spherical body, rotating around an axis through its mass center is 2 |s|=I ω = m r 2ω = 3/4h¯ . 5 e The velocity at the equator is 5√ v = rω = 0.75h¯/(m r). equator 2 e −15 −31 For r = 1.4 × 10 m, me = 9.1 × 10 kg we obtain v = 1.8 × 1011 m/s c ! For r = 10−18 m ⇒ v = 2.5 × 1014 m/s. This is a contradiction to the special theory of relativity Fig. S.7. 542 Solutions to the Exercises

−24 For B = 1Tesla, μB = 9.27 × 10 J/T we ob- corresponding to the transition between the HFS tain components has a frequency − ΔE c Δν = 9.3 × 109 s 1 , ν = = 1 h λ 10 −1 Δν = 1.86 × 10 s . −25 2 with ΔE = 9.46 × 10 J. With μK = 5.05 × −27 a) In order to resolve these components with 10 J/T the magnetic field is a central frequency ν = 4.5 × 1014 s−1 the spectral ΔE 9.46 × 10−25 B = = −27 T resolving power has to be 5.58μK 5.58 × 5.05 × 10 ν λ 4.5 × 1014 = 33.5 T . | |=| |= = 4.8 × 104 . Δν Δλ 9.3 × 109 The magnetic field produced by the electron in the 1s state of hydrogen at the location of the pro- The resolving power of a grating spectrograph is ton is therefore with 33.5 T much larger than fields λ | |≤mN obtained in the lab. Δλ , 9. The frequencies of the transitions are given by where m is the interference order and N the num- ∗ 1 1 3 ∗ hν = Ry − = Ry . ber of illuminated grooves. For m = 2wethen 1 4 4 obtain: The Rydberg constant × 4 4 4.8 10 ∗ e m m N ≥ = 24,000 . Ry = μ with μ = n e 2 ε2 2 + 8 0h mn me b) The FabryÐPerot interferometer with mirror depends on the mass mn of the nucleus. For the separation d = 1 cm has a free spectral range Hatom 8 1 c 3 × 10 − − μ = m = m · 0.999456 . δν = = s 1 = 1.5 × 1010 s 1 . e + 1 e 2d 2 × 10−2 1 1836 = 2 With a finesse For the D 1H isotope √ π 1 ∗ R μ = me = me · 0.999728 . F = 1 + 1 1 − R 3672 = 3 the minimum separation of two resolvable lines is: For the isotope T 1Hitis ∗ μ = m · 0.999818 . Δν = δν/F . e The wavenumber and frequency of the Lyman = ⇒ ∗ = With R 0.95 F 61 and α-lines are then: 10 1 −1 1.5 × 10 − ν H = 82,258.2 cm Δν = ≈ 2.5 × 108 s 1 . 1 − 61 ⇒ ν = 2.466039 × 1015 s 1 , In order to resolve all four Zeeman lines, the ν 2 = ν 1 = −1 magnetic field has to be at least B ≥ 0.026 T. 1D 1.00027 1H 82,280.6 cm . 8. The potential energy of a magnetic dipole in a mag- The difference is netic field is: Δν = ν 2 − ν 1 1 1D 1H =−μ · E B . = −1 22.4 cm − The magnetic moment of the proton is ν 3T = 1.00036ν 1H = 82,288.0 cm 1 1 1 μ = 2.79μ . p K Δν = ν 3 − ν 1 2 1T 1H The separation of the two hyperfine components − Δ = μ · λ = = 29.8 cm 1 . is E 5.58 K B. The line with 21 cm, Chapter 6 543

3/2 The hyperfine-splittings are: Z − / ψ = Zr1 a0 a) 1H: E =±μ · B ⇒ΔE = 2|μ|·|B| with 1 s √ / e . 1 HFS πa3 2 B = 35 T (see Problem 5.8) and μ = 2.79μK 0 ⇒ Δ = μ · = × −25 =ˆ × E 5.58 K B 9.43 10 J 5.9 Inserting these relations with Z = 2, into the inte- −6 10 5.9 × 10eV. grand we obtain for the integral 2 b) 1D: The internal magnetic field is caused by |ψ |2 the electron and therefore the same for all three = 1 s 2 ϑ ϑ ϕ I r1 sin dr1 d d isotopes. The two hyperfine components have the r⎡12 energy + r2 − / r 2 r1 Z 3π ⎢ e Z2r1 a0 r 2 A = ⎣ 1 r r 3 d 1 d 12 EHFS = · [F(F + 1) − J(J + 1) − I (I + 1)] . πa r1r2 2 0 = r =r −r r1 0 12 2 1 ⎤ = / = / ∞ r 1+r2 With F 3 2, J 1 2 we obtain −Z2r /a e 1 0 r1 ⎦ + dr1 dr12 . A r2 EHFS = . = = − 2 r1 r2 r12 r1 r2 (2) With F = 1/2 ⇒ FHFS =−A. The splitting is Δ = 3 ϑ = then: F 2 A with For 0 we have (Fig. 5.8) A = g μ · B / J(J + 1) r − r for r < r , I K J √ r = 2 1 1 2 12 − > = 2gIμK BJ / 3. r1 r2 for r1 r2 .

3 = / ⇒ = = For ϑ = π ⇒ r12 = r1 + r2. The integration of the c) 1T: Here is I 3 2; F 2 and F 1 ﬁrst term in (2)gives 3 5 E(F = 2) =+ A ; E(F = 1) =− A 2 3 4 4 r2a0 a a − / √ I = − − 0 − 0 2Zr2 a0 1 2 3 e ⇒ ΔE = 2A = 4gIμK BJ / 3. Z Z 2r2 Z a3 + 0 3 Chapter 6 2r2 Z and for the second term: 1. The potential experienced by the second electron a r a2 in the He atom is (see Fig. 6.8) 0 2 0 −2Zr2/a0 I2 = + e . Z 2Z 2 |ψ |2 φ =− Ze + e 1 s(r1) τ (r2) d . = 4πε0r2 4πε0 r12 This gives, with r r2, the total potential felt by τ = 2 ϑ ϕ ϑ ϕ the second electron: d 1 r1 sin 1 sin dr1d 1d 1 (1) 2 = 2 + 2 − ϑ (Z − 1)e e Z 1 − / r12 r1 r2 2r1r2 cos φ r =− − + 2Zr a0 ( ) πε πε e ⇒ r12 dr12 = r1r2 sin ϑ dϑ 4 0r 4 0 a0 r with Z = 2fortheHeatom. 2. The charge density of the two 1s-electrons is approximately

2e − / η =−2e · ψ2(1s) =− e 2R b b3 · π with b = a0/Zeff . Fig. S.8. 544 Solutions to the Exercises ⎡ r0 4e 2R2 =− ⎣ −2R/b R 3 e d b r0 R=0 ⎤ ∞ ⎥ + · −2R/b + 2R e dR⎦ .

R=r0 The integrals can be solved analytically η 2e − / r0 dτ =− 1 + e 2r0 b 1 + . Fig. S.9. r r0 b The potential then becomes with b = a0/Zeff and when we rename r0 → r: The potential, experienced by the 2s electrons in −e − · / −rZeff the ﬁeld of the nucleus and the two 1s-electrons is φ(r) = 1 + 2e 2Zeff r a0 1 − . (Fig. S.9): 4πε0r a0

−3e 1 η 3. The mean distance between the atoms is d = φ = + dτ . −1/3 λ 4πε r 4πε r n . The de Broglie-wavelength dB becomes 0 0 0 larger than d for The integral can be solved as follows: h h > d ⇒ v< . η mv mn−1/3 dτ r The mean velocity is ∞ π π 2 − / 2e e 2R b 8kT =− R2 sin ϑ dR dϑ dϕ v = . b3π r πm R=0 ϑ=0 ϕ=0 This gives the condition for the temperature T : With r 2 = R2 + r 2 − 2r R cos ϑ ⇒ rdr = r R · / 0 0 0 πh2n2 3 sin ϑ dϑ the integral then becomes: T < . 8k · m ∞ = 12/ 3 = 18/ 3 = η 4e R − / Example: n 10 cm 10 m and m τ =− 2R b R r −27 d 3 e d d . 23 AMU = 23 × 1.66 × 10 kg r b r0 R=0 r − ⇒ T < 3.3 × 10 7 K = 330 nK . Is the point A (location of 2s-electron) outside the Below this temperature the particles are no longer charge distribution of the 1s-electrons (r > R)the 0 distinguishable because their location can only be integration over r extends from r − R to r + R. 0 0 deﬁned within a volume λ3 . The atoms form For r < R (2s-electron inside the core) it extends dB 0 a BoseÐEinstein condensate of identical particles. from R − r0 to R + r0. This gives: ⎡ 4. The potential energy of the two electrons is r0 r 0+R (Fig. S.10) η 4e ⎢ R τ =− ⎣ −2R/b R r 2 2 d 3 e d d 4e e r b r0 Epot =− + = = − πε πε R 0 r r0 R ⎤ 4 0r1 4 0(2r1) ∞ R +r 7 e2 7 e2 0 =− =− , R − R/b ⎥ + e 2 dR dr⎦ 8 πε0r1 4 πε0a0 r0 with r1 = a0/2. R=r0 r=R−r0 Chapter 6 545 Fig. S.10. 1 1 Z − / × + + e 2Zr a0 r r a 0 2 r − / × 2 − e r a0 dτ a0

∞ 4 4 r − / = − + r a0 2 e r a0 a0 The kinetic energy of the two electrons is: r=0 v2 4 4(Z − 1) (4Z − 1)r Zr2 = 2me = v2 + + − + Ekin me . r a 2 3 2 0 a0 a0 With π 2π − + / × e (2Z 1) r a0 r 2 dr sin ϑ dϑ dϕ v = Zh = h ϑ= 2πma0 πma0 0 0 ∞ 2 3 we obtain: 4r r − / = 4π 4r − + e r a0 2 2 h a0 a0 E = . r=0 kin 2 2 π ma 2 3 4 0 4r 7r 2r − / + 4r + − + e 5r a0 dr The total energy of the system is then: a a2 a3 0 0 0 7 e2 h2 4 8 E =− + . = 4π 4a2 − 8a2 + 6a2 + a2 + a2 πε π 2 2 0 0 0 25 0 125 0 4 0a0 ma0 42 48 Inserting the numerical values yields − a2 + a2 625 0 3125 0 − E =−1.30 × 10 17 J =−82 eV . 6788 = 4π a2 The experimental result is E =−78.9 eV. This 3125 0 shows that our simple model approaches the real 2 ⇒ =− e situation quite well. The small difference comes Epot(2s) 0.272 4πε0a0 partly from the fact that we have neglected the =−7.39 eV . relativistic mass increase. 5. The expectation value of the potential energy e · A similar calculation for the 3s state gives: φ(r) of an electron in the 2s state is =− Epot(3s) 3.19 eV ⇒ Δ = = ψ∗φ ψ τ Epot 4.2 eV Epot e 2 (r) 2 d 2 . ≈−1 because Ekin 2 Epot we obtain Assuming a spherically symmetric wave func- ΔE = ΔE + ΔE tion ψ2(2s) as in the H atom (because the electron pot kin moves in a potential that is essentially a Coulomb 1 ≈ ΔEpot ≈ 2.1 eV . potential with the effective charge Zeff = 1) we can 2 write: This agrees fairly well with the experimental value 2 Δ = 1 r Eexp 2.3 eV, obtained from the difference of 2 −r/a0 |ψ2| = 2 − e the wavenumbers ν(3s ↔ 1s) − ν(2s ↔ 1s)ofthe 32πa3 a 0 0 two-photon allowed transition. e2 ⇒ E =− 6. The largest mean distance between the electrons pot πε × π 3 4 0 32 a0 is realized if the total spin of all electrons has 546 Solutions to the Exercises

μ its maximum allowed value. Since for this case The radius rn of the myon is the spin function is symmetric, the spatial part of n2 a the wave function is antisymmetric with respect μ = 0 rn . to exchanging two electrons. This means that the Z 206.6 ψ = el = / wave function (r1, r2) has nodes for r1 r2. The smallest radius of the electron is r1 a0 Z. Therefore the potential energy of the mutual repul- el = μ The condition r1 rn gives: sion is minimized, which gives a minimum for the n2 total energy. = 1 ⇒ n ≈ 14 . 7. With the screening constant S, the potential energy 206.6 μ of a Rydberg electron can be written as The radius r14 of the myon is about the same as the 2 el (Z − S)e lowest radius r1 of the electron orbit. Epot =− . 9. The potential energy of the electron with the wave 4πε r 0 function ψ(r) is (see Problem 6.1) 1 For the total energy (note that Ekin =− Epot)we 2 2 obtain Epot =+e |ψ(r)| φ(r)dr . (Z − S)e2 E =− For the 3s electron the probability to ﬁnd the elec- πε 8 0r tron inside the n = 2 shell is larger than for the = Ekin + Epot . 3p electron, because the 3p-functions has a node = According to the Rydberg formula (6.32)the at r 0. Therefore the shielding of the nuclear energy can be also expressed by charge is smaller for the 3s electron than for the 3p electron. The potential energy is lower and −(Z − S)2 Ry∗ Ry∗ E =− =− . therefore its total energy. n2 (n − δ)2 10. The potential energy of the second electron in the − The comparison yields the relation H ion is n =+ φ S = Z − Epot(r2) e (r), n − δ where the potential φ(r) has been calculated in between the screening constant S and the quantum Problem 6.1 and where we have to insert Z = 1. defect δ.Forδ → 0 we obtain S = Z − 1, which e 1 1 − / = − = ⇒ φ r =− + 2r a0 means a Coulomb potential with Zeff Z S 1. ( ) πε e . 8. For the myonic atom the reduced mass is 4 0 a0 r mμ · mN The wave functions of the second electron can be μ = . approximated by the hydrogen wave function mμ + mN 1 −r/a With mμ = 206.76 me, mN = 140 × 1836 me we ψ(r) = √ e 0 π 3/2 obtain μ = 206.6 me. a0 ∗ ∗∞ ⇒ Ryμ = 206.6 · Ry when we neglect the repulsion between the two electrons. The potential energy is then 206.6Ry∗∞ Z 2 ⇒ E =− n 2 min =+ |ψ |2φ τ n Epot e (1s) (r)d = → The energy of the photon on a tansition n 2 ∞ = 2 n 1is 4πe 1 1 − / =− + e 4r a0 r 2 dr ∗∞ 1 1 πε π 3 a r hν = 206.6Ry Z 2 − 4 0 a0 0 1 4 0 e2 3 3 e2 = 3 2 × × =− a2 =− 60 206.6 13.6 eV πε 3 0 πε a 4 0a0 32 8 4 0 0 = 7.59 × 106 eV = 7.59 MeV . =−10.2 eV . Chapter 7 547

The crude approximation gives a binding energy shells is spherically symmetric. The valence elec- = + =− tron moves in a spherical potential, which is nearly EB Epot Ekin 5.1 eV a Coulomb potential (∼ 1/r) for larger r,but of the second electron. It is higher than the ex- deviates from it for small r. perimental value of −2.5 eV, because we have 13. The potential energy of the electron is neglected the repulsion between the two electrons. e2 11. The energy of the state with principal quantum E x =− − eE x pot( ) πε 0 number n is 4 0r = α 2 2 with x r cos . ∗ Z n E E =−Ry eff ⇒ Z 2 =− n ; E < 0. n n2 eff Ry∗ n In the x-direction α = 0 ⇒ cos α = 1. We than can ∗ write: With Ry = 13.6 eV one obtains: 2 × dEpot e 2 4 5.39 = − eE . Z (n = 2) = = 1.58 πε 2 0 eff 13.6 dr 4 0r ⇒ = Zeff 1.26 . The maximum of the potential barrier is at / = The nuclear charge Ze with Z = 3 is screened by dEpot dr 0 the two 1s electrons by 1.74e. 1/2 = e For the Rydberg level with n 20 we obtain: ⇒ r = m πε × 4 0 E0 2 400 0.034 Z (n = 20) = ≈ 1 3 eff 13.6 e E0 ⇒ Epot(rm) =− . ⇒ Zeff = 1. πε0

For high Rydberg levels the screening of the nu- Without the external ﬁeld Epot(r)⇒0forr →∞. clear charge Ze is nearly complete by the (Z − 1) The lowering of the ionization potential is, there- electrons of the atomic core. fore, 12. For all alkali atoms there is a single electron in the = valence shell with principal quantum number n e3 E = = = = Δ IP =− 0 2(Li),n 3(Na),n 4(K),n 5(Rb),n 6 ( ) πε . (Cs). The larger the n is, the better the shielding 0 of the nuclear charge Ze by the Z − 1 electrons Due to the tunnel effect, the effective decrease of in the core. This implies that Zeff decreases with IPis even slightly larger. increasing n and the binding energy of the valence electron (this is the ionization energy of the atom) decreases with increasing n. Chapter 7 The experimental technique for the determination of the binding energy is, for instance, the photoion- 1. a) The total emitted energy is ization of the atom. + − = 2 ν = 8 × × −19 A(En) + hν → A + e (Ekin). WFl N 3 P3/2 h 10 3.4 10 J = × −11 One can measure the frequency νg, where 3.4 10 J. − Ekin(e ) = 0 ⇒−En = hν . The time dependent ﬂuorescence power can be calculated as The approximate calculation of the binding energy −t/τ is based on the Hartree method (see Sect. 6.4.2), PFl = P0e , which converges rapidly because the charge distribution of the core electrons, which form closed where excitation at t = 0 is assumed. 548 Solutions to the Exercises

The emitted energy is related to the power by ∞ ∞ −t/τ WFl = PFl dt = P0 e dt = τ · P0 0 0 −11 −8 ⇒ P0 = 3.4 × 10 J/1.6 × 10 s − = 2.1 × 10 3 W. Fig. S.11. b) The angular distribution is 2 W(ϑ) = W0 sin ϑ , b) The collimation ratio of the atomic beam is 2π + π/2 (Fig. S.11) W = W sin2 ϑ dϑ dϕ total 0 b 1 ϕ=0 ϑ=−π/2 ε = = . +π/ 2d 200 1 1 2 = 2πW0 ϑ − sin 2ϑ Since the nozzle diameter is small compared to b 2 4 −π/2 we can regard the nozzle as a point-like source of 2 v = π W0 atoms. The transverse velocity distribution f ( x )is determined by |v |

9 −1 The mean velocity of the Ca atoms is at an oven we obtain with δνD = 6.77 × 10 s for ν1 − temperature T = 900 K ν = 0.1 δν : 0 D − × × −23 × × 2 α(ν ) = α(ν )e 0.028 ln 2 v = 8kT = 8 1.38 10 9 10 m 1 0 − = α ν π m π × 40 × 1.66 × 10 27 s 0.98 ( 0). = 690 m/s. 8 −1 For a Lorentzian profile with δνn = 1.14 × 10 s The minimum interaction zone is then we have: 8 −1 Δ = vΔ = × 2 × × −5 ν1 − ν0 = 0.1 δνD = 6.77 × 10 s s T 6.9 10 5.3 10 m − 7 2 = 3.7 × 10 2 m = 37 mm . 5.7 × 10 ⇒ α(ν1) = α(ν0) × 8 2 + × 7 2 4. a) The wavelength λ of a transition between energy 6.77 10 5.7 10 −3 levels with term values Ti , Tk is = 7 × 10 α(ν0). 1 1 For ν1 − ν0 = δνD we obtain for the Gaussian λik = = cm = 501.7 nm . Ti − Tk 19,932 profile: b) The natural line width is: α(ν1) = 0.146 α(ν0) 9 3 δν ≤ 1 + 1 = 10 + 10 and for the Lorentzian profile: n πτ πτ π π 2 i 2 k 2 1.4 2 −5 − α(ν ) = 7 × 10 α(ν ). = 1.14 × 108 s 1 = 114 MHz . 1 0 For ν − ν = 10 δν c) The Doppler width is 1 0 D ⇒ α ν = α ν −278 = × −120 α ν −7 ( 1) ( 0)e 2 10 ( 0) δνD = 7.16 × 10 ν0 T/M[mol/(g K)] 8 for the Gaussian profile and c 3 × 10 − ν = = s 1 0 λ × −7 2 5.017 10 5.7 × 107 14 −1 α ν = α ν = 5.98 × 10 s ( 1) ( 0) 6.77 × 1010 2 3 T = 10 K, M = 4g/mol − = 7 × 10 7 α(ν ). 9 −1 0 ⇒ δνD = 6.77 × 10 s = 6.77 GHz . for the Lorentzian profile. δν 5. a) The Lorentzian line profile with half-width n Here the absorption by the Gaussian profile is is: negligible compared to the absorption by the (δν /2)2 Lorentzian profile. α(ν) = α(ν ) n . 0 2 2 b) The relative absorption at ν = ν1 of the two (ν − ν0) + (δνn/2) profiles becomes equal for: A wavelength difference Δλ = 0.1 nm corre- 2 sponds to a frequency difference (δνn/2) − ν −ν / δν 2 = e [( 1 0) 0.6 D] ν − ν 2 + δν / 2 c − ( 1 0) ( n 2) Δν = (ν − ν ) = Δλ = 1.2 × 1011 s 1 1 0 λ2 ⇒ δν / 2 + ν − ν 2 − (δν / )2 ln ( n 2) ( 1 0) ln n 2 × 8 2 0.57 10 = ν − ν / δν 2 ⇒ α(ν1) = α(ν0) ( 1 0) 0.6 D . 1.2 × 1011 2 + 0.57 × 108 2 The numerical solution (which can be obtained = × −7 α ν 2.25 10 ( 0). with the program “Mathematica”) depends on the = δν /δν = ⇒ ν − ν = For a Doppler-broadened absorption profile ratio x n 0.Forx 0.01 1 0 439.6 δνn.Forx = 0.1 ⇒ ν1 − ν0 = 27.8 δνn. 2 2 − (ν1−ν0) /(2δνD) ln 2 α(ν) = α(ν0)e For x = 1 ⇒ ν1 − ν0 = 0.895 δνn. 550 Solutions to the Exercises

6. The frequency of the Kα lines of silver can be es- The natural line width of the Lyman α-line is timated for an effective charge Zeff = Z − 1 with 1 7 −1 Z = 47 from the relation: δνn = = 7.4 × 10 s 2πτ(2p) ∗ 1 1 τ = hν = Ry (Z − 1)2 − . since (2p) 2.1 ns. 2 2 Δν = n1 n2 The recoil shift is therefore only recoil ∗ 0.35 δν . The Doppler width at T = 300 K is: For n1 = 1, n2 = 2 and Ry = 13.6 eV we obtain: n 3 δν = × 10 −1 ⇒ Δν δν hν = 13.6 × 462 × eV = 21.6 keV D 3.06 10 s recoil D . 4 − = 3.45 × 10 15 J 8. The effective lifetime is determined by − ⇒ ν = 5.22 × 1018 s 1 1 1 = + nσv c τ τ r , λ = = × −11 = eff n ν 5.75 10 m 0.575 Å . where vr is the mean relative velocity. From the The experimental value is hν = 21.9 keV, λ = relation p = nkT we obtain for p = 1 mbar = 0.562 Å. The ionization energy of molybdenum is 102 Pa (see, e.g., the American Handbook of Physics) 2 p 10 Pa − n = = m 3 . IP 42Mo = 20.0 keV . kT 1.38 × 10−23 × 500 − ⇒ n = 1.45 × 1022 m 3 The kinetic energy of the photoelectron is then 8kT Ekin = hν − IP = (21.9 − 20.0) keV = 1.9 keV . v = r πm The velocity is with 7 −2 v = 2Ekin/me = 2.6 × 10 m/s = 8.6 × 10 c . m m m = 1 2 = m = × −26 + 0.55 (Na) 2.1 10 kg 7. The recoil momentum is m1 m2 2π − p = h¯ k with |k|= . 8 × 1.38 × 10 23 × 500 m λ ⇒ v = = 915 m/s. r 2.1 π × 10−26 s The recoil energy is 2 This gives p2 h¯ 2k2 (hν)2 1 E phot 9 Ekin = = = = . 1 10 2m 2m 2mc2 2 mc2 = + × 22 × × −19 × −1 τ 1.45 10 4 10 961 s For the transition n = 2 → n = 1intheHatomis eff 16 − hν = 10.2 eV and mc2(proton) = 938.8 MeV = 6.81 × 107 s 1 2 ⇒ τ = 1 10.2 − eff 14.7 ns . ⇒ E = eV = 5.5 × 10 8 eV . kin 2 938.8 × 106 14.7 = The effective lifetime is smaller by a factor 16 The velocity of the H atom after emission of the 0.919 than the natural lifetime τ = 16 ns. For p = photon is: 10 mbar the second term becomes p hk¯ hν σv = × 7 −1 v = = = c n r 5.3 10 s m m mc2 10.2 and the effective lifetime shortens to = × 3 × 108 m/s 9.38 × 108 τeff = 8.7 ns = 0.54 τn . = 3.3 m/s. For p = 100 mbar ⇒ τeff = 1.7 ns = 0.11 τn. Δν v − ⇒ recoil = = 1.09 × 10 8 9. The residual Doppler width in the collimated ν c atomic beam is ⇒ Δν = × −8 × × 15 −1 recoil 1.09 10 2.47 10 s a) (δνD)res = sin εδνD < 190 MHz . 7 −1 √ = 2.7 × 10 s . −7 With δνD = 7.16 × 10 ν0 T/M mol/(g K) = Chapter 7 551

9 −1 14 −1 2 × 10 s with ν0 = 5.09 × 10 s , M = The absorption coefﬁcient is 23 g/mol, T = 695 K, v = 800 m/s 1 190 α = N(F = 0) − N(F = 1) · σabs ⇒ sin ε ≤ = 0.095 . 3 2000 = 0.006 · N(F = 0) · σabs . b) If the residual Doppler width should be equal to δν = ⇒ −26 2 6 −3 the natural line width n 10 MHz With σabs = 3 × 10 m , N = 10 m and L = 7 × 16 δν 10 − 3 10 m we obtain sin ε = n = = 5 × 10 3 . δν × 9 − D 2 10 α · l = 5.4 × 10 6 . = × 10. a) From the Einstein coefﬁcient Aik 3 The absorption of the 21 cm line is therefore −15 −1 = 1 10 s τ we obtain the natural line width of negligible. For the Lyman α-line the situation is λ = the HFS transition with 21 cm different: Aik − − δν = = 5 × 10 16 s 1 . α = σ · = × −15 · 6 −1 = −9 −1 n 2π N 1 10 10 m 10 m . 9 The natural line width of the HFS transition is After 10 m the intensity has decreased to (1/e)I0. therefore extremely narrow. The Doppler width is The interstellar cloud with L = 3 × 1016 mis α × 8 √ therefore completely opague. Lyman- -light is −7 3 10 −1 δνD = 7.16 × 10 10 s = 3.2 kHz . completely absorbed. 0.21 c) The natural line width of the methane transition The collision broadening is: is σ v n r 1 −1 −1 δνcoll = . δν = s = 7.96 s . 2π n 2π × 2 × 10−2 v = 8kT = / = m1m2 This extremely small line width is used for stabi- With r πm 460 m s with m m +m 1 2 lizing the wavelength λ = 3.39 μmoftheHe-Ne −20 −1 ⇒ δνcoll = 7.3 × 10 s . laser (see Chap. 8). The collisional broadening is therefore completely −7 c δνD = 7.16 × 10 T/M(mol/(g K)) negligible. For the Lyman α-line we obtain: λ with λ = 3.39 μm, M = 16 g/mol, T = 300 K Aik 8 −1 δνn = = 1.6 × 10 s , 2π ⇒ δν = 9 −1 D 274 MHz . δνD = 5.6 × 10 s , − − δν = 7.3 × 10 13 s 1 . The mean velocity of the CH4 molecules is coll The Doppler width is by far the prominent broad- 8kT v = = 630 m/s. ening. πm b) At a temperature T = 10 K the ratio of the ⇒ Transit time populations is = 0.01 m −5 N(F 1) −hν/kT t = = 1.6 × 10 s = 3 · e ≈ 3 · 0.994 . trans 630 m/s N(F = 0) ⇒ δν = 1 ≈ 4 −1 The population difference is trans 10 s . 2πttrans 1 ΔN = N(F = 0) − N(F = 1) The only way to reduce transit-time broadening 3 is the enlargement of the laser beam diameter or = · = 0.006 N(F 0) . a reduction of the velocity by cooling the gas. 552 Solutions to the Exercises

11. According to [7.13] the matrix element for the with Δm =+1, Mx − iMy with Δm =−1 and ↔ Δ = transition 1s 2s is Mz with m 0. The total transition probability is proportional to = ψ ψ τ Mik e (2s)r (1s)d | |2 =| |2 + 2 + 2 Mik (Mik)x (Mik)y (Mik)z , 1 | + |2 = 2 + 2 = √ Mx iMy Mx My . 4π 2a3 0 Using the wave functions of Table 5.2 we obtain: r −r/(2a0) −r/a0 1 × 2 − e re dτ −r/a0 Mx + iMy = e (x + iy)r a0 8πa4 0 r ϑ ϕ r −3r/(2a ) = − 0 − / − ϕ C 2 e r × r 2a0 ϑ i 2 ϑ ϕ ϑ a0 e sin e r sin d d dr . r ϑ ϕ With x = r sin ϑ cos ϕ; y = r sin ϑ sin ϕ one ob- 2 × r sin ϑ dr dϑ dϕ , tains x + iy = r sin ϑ eiϕ + π/2 2π ∞ r − / 1 − / (M ) = C 2 − e 3r (2a0) ⇒ M + iM = r 4e 3r (2a0) dr ik x x y π 4 a0 8 a0 ϑ=−π/2 ϕ=0 r=0 π π × xr2 sin ϑ dr dϑ dϕ . 2 × sin3 ϑ dϑ dϕ . With x = r · sin ϑ cos ϕ the integration over ϕ yields ϑ=0 ϕ=0 5/ 2π The ﬁrst integral has the value 256 a0 81, the second integral is 4/3 and the third is 2π.This cos ϕ dϕ = 0. gives: ϕ= 0 2 2 256 A similar result is obtained for (M ) with y = (Mx + iMy) = a0 ik y 243 r sin ϑ sin ϕ. For the third component (Mik)z we = ϑ 2 e2ω3 a2 256 2 obtain with z r cos ⇒ A (Δm =±1) = ik 0 . ik ε 3 2π 3 0c h 243 ω = π × 15 −1 dϕ = 2π With ik 2 2.47 10 s the transition probability becomes: 0 − A (Δm =±1) = 1.25 × 1010 s 1 . but ik An analogous calculation for M with z = r cos ϑ + π/2 z +π/2 1 2 gives sin ϑ cos ϑ dϑ = sin ϑ = 0. ∞ −π/ 2 2 1 ϑ=−π/2 4 −3r/(2a0) Mz = √ r e dr 4π 2 a4 12. The transition probability is, according to (7.17), 0 r=0 π 2π Aik(1s → 2p) 2 2 3 2 × cos ϑ sin ϑ dϑ dϕ 2 e ω ∗ = ik ψ (2p)rψ (1s)dτ 3 i k ϑ=0 ϕ=0 3 ε0c h 1 256 2 The 2p level has three components with m = 0, ±1 = √ a5 2π π 4 81 0 3 that are degenerate without external magnetic ﬁeld. 4 2 a0 There are therefore three degenerate transitions = 256√ → a0 with different polarizations on the 1s 2p transi- 243 2 tion. If the quantization axis is chosen as the z-axis, 1 + ⇒ Aik(Δm = 0) = Aik(Δm ± 1) . the matrix element Mx iMy describes transitions 2 Chapter 7 553

13. The 3s level can only decay into the 2p level. Since p = α(ω) · E we obtain for the polarizability Therefore the transition probability for the transi- e2 tion 3s → 2p is α(ω) = (4) m ω2 − ω2 + iγω 9 0 1 10 −1 7 −1 Aik = = s = 4.3 × 10 s . The wave equation for a wave travelling through τ(3s) 23 a medium is: The natural line width is ∂2 E 1 ∂2 E 1 1 1 ΔE = μμ εε = ,(5) δν = + 0 0 ∂t2 v2 ∂t2 n 2π τ(3s) τ(2p) ph 1 − where = 4.3 × 107 + 4.76 × 108 s 1 2π 1 c vph = √ = √ = 83 MHz . μμ0εε0 με The Doppler-width is is the phase velocity of the wave. In dielectric me- − dia is the magnetic susceptibility μ ≈ 1 and we δν = 7.16 × 10 7 ν T/M[mol/(g K)] . D 0 obtain: With c c √ vph = √ = ⇒ n = ε .(6) 1 1 1 ε n ν = Ry∗ − = × 14 −1 0 2 2 4.57 10 s , h 2 3 Inserting the relation: D = ε0 E + P into (1)we M = 1g/mol , T = 300 K , obtain the equation, equivalent to (5): this gives 1 ∂2 E 1 ∂2 P ΔE = + 2 2 2 2 .(7) 9 −1 c ∂t ε0c ∂t δνD = 5.67 × 10 s = 5.67 GHz δν The dielectric polarization is for a wave with its ⇒ n = 0.014 E-vector in x-direction (E ={Ex ,0,0}): δνD = · α · = · α · · i(ωt−kz) for T = 300 K and 0.008 for T = 1000 K. Px N Ex N E0 e ,(8) · i(ωt−kz) 14. The incident wave E0 e induces the elec- where N is the number of oscillating dipoles per trons of the medium to forced oscillations. If the m3 and α is the polarizability. amplitude of this oscillation is within the range of Inserting (8)into(7)gives: linear optics (the restoring force is proportional to ω2 ω2 Nα the elongation of the atomic electron) we can de- −k2 E =− E − E x 2 x 2 x scribe the atomic electron by a harmonic oscillator. c ε0c ω2 Its equation of motion is: 2 ⇒ k = (1 + Nα/ε0)(9) i(ωt−kz) c2 mx¨ + bx˙ + Dx =−eE0e .(1) with v = c/n = ω/k ⇒ n = ck/ω, where n is With D/m = ω2, γ = b/m we can make the ph 0 the refractive index. ansatz ⇒ 2 = + α/ε iωt n 1 N 0 . (10) x = x0e . Inserting (4)into(10) yields Inserting this into (1)gives 2 / Ne eE0 m n2 = 1 + . x0 =− .(2) ω2 − ω2 + γω ω2 − ω2 + γω m 0 i 0 i 2 2 2 iωt with n = n − iκ ⇒ n = n − κ − 2in κ, The induced dipole moment p = p0 · e with where κ is the absorption coefﬁcient and n the p0 = e · x0 real part of the complex index of refraction. For e2 E γω ω2 ⇒ p = .(3)small damping ( ) the summation over all ω2 − ω2 + γω m 0 i atomic resonances gives the result (7.102). 554 Solutions to the Exercises

In order to transport a signal, the monochromatic where δν is the full half-width of the absorption wave has to be modulated. The envelope of this profile. With αki = Nkσki we obtain: modulated wave represents a wave packet with the ν 2 λ2 v = ω/ h c Aik Aik group velocity g d dk. σki = Bki/δν = = . c 8πν2δν 8πδν 9 −1 Inserting the Doppler width δνD ≈ 10 s we Chapter 8 obtain: −15 2 −11 2 σki = 10 m = 10 cm 1. a) The ratio of the populations Ni , Nk is: − ⇒ A = 1010 × 10 11 × 1 = 0.1 . Ni = gi −hν/kT e This means that 10% of the incident light power is Nk gk g = 2J + 1 = 3, g = 2J + 1 = 1 absorbed. i i k k c) For the compensation of 10% total losses the N − /λ / − − ⇒ i = 3e (hc ) kT = 3e 96 = 6.6 × 10 42 . condition −2αL ≥ 0.1 must be met. Nk ⇒ ( g /g N − N ) σ L ≥ The thermal population of the upper level |i is ( k i ) i k ki 0.05 . −11 2 therefore completely negligible. With gk = 1, gi = 3 and σki = 10 cm this b) The relative absorption of the incident wave gives: with intensity I0 is 1 −11 I − I Ni − Nk 10 × 20 ≥ 0.05 . A = 0 t , with 3 I0 10 3 −α With N = 10 /cm the upper state population I = I e L ≈ (1 − αL)I for αL 1 k t 0 0 density then has to be ⇒ A ≈ αL = N σ L . k ki 10 3 Ni ≥ 3.075 × 10 /cm . The population density N = 10−6 N can be ob- k = tained from the equation p = NkT For equal statistical weight factors gi gk the minimum population N would be ⇒ = −6 / i Nk 10 p kT . − N ≥ 1.025 N = 1.025 × 1010 cm 3 . For a pressure p = 102 Pa we obtain: i k −6 2 This would require a population in the upper state, 10 × 10 − N = m 3 which is 2.5% higher than in the lower state in k × −23 × 1.38 10 300 order to reach laser threshhold. − − = 1016 m 3 = 1010 cm 3 . 2. a) The Doppler width is The absorption coefficient α is related to the c −7 −1 δνD = 7.16 T/M10 s . Einstein coefficient Bki as follows: The absorbed λ power is: With λ = 632.8 nm, T = 600 K, M = 20 g/mol dWki this gives: = B hνwν(ν). ki 9 −1 dt δνD = 1.86 × 10 s = 1.86 GHz . The spectral energy density wν(ν) is related to the b) The mode separation is spectral intensity Iν(ν)bywν(ν) = Iν(ν)/c.The power absorbed by one atom is c 3 × 108 δν = = = 150 MHz . ∞ 2d 2 × 1 1 dWki 1 = I (ν) α(ν)dν The number of longitudinal modes within the full Nk dt Nk 0 half-width of the Doppler profile is then ≈ 1 ν α ν δν 1.86 × 109 I ( 0) 0( 0) , m = = 12 . Nk 1.5 × 108 Chapter 8 555

3. a) The frequency separation of the transmission we obtain √ maxima of the etalon is π c 2 = ⇒ ∗ = F = δν = . 0.075 F 28.5 . E 2nt F 2 ∗ If this should be larger than the Doppler width The finesse F must be at least 28.5 in order ∗ to√ select a single longitudinal mode. With F = −7 c 5000 9 −1 π / − δνD = 7.16 × 10 = 5 × 10 s R (1 R) we obtain for the reflectivity R of λ 40 the FabryÐPerot at λ = 488 nm, the thickness t of the etalon should > be R 0.89 . c c ν t = ≤ 4. a) The frequencies m of the resonator modes are δν δν 2n E 2n D c c 8 ν = = · = λ/ 3 × 10 − m because L m 2. = m = 2 × 10 2 m = 2cm. λ 2L × × × 9 2 1.5 5 10 With L =1m,ν =5×1014s−1 ⇒ m =3.33×106, b) The intensity Iz transmitted through the FabryÐ = + α Perot interferometer is: L L0(1 T ), 1 ΔL = L − L0 = L0αΔT It = I0 − 1 + F sin2(πΔsν/c) = 1 × 12 × 10 6 × 1m − with F = 4R/(1 − R)2. The finesse = 1.2 × 10 5 m, √ √ Δν Δ π R π F Δν = L = × −5 F∗ = = = m 1.2 10 − Δν ν L 1 R 2 E − − ⇒ Δν = 1.2 × 10 5 × 5 × 1014 s 1 equals the ratio of frequency separation Δνm of = × 9 −1 transmission maxima to the half-width ΔνE of the 6 10 s . transmission peaks. The transmission of the FPI Since the mode distance is only Δν = 150 MHz = = L has decreased from the maximum It I0 to It the laser frequency jumps during the temperature 1 I 3 0 for change after a shift of about Δν/2 ≈ 75 MHz back 2 πΔ ν/ = to the next mode. Temperature tuning alone does F sin ( s c) 2 πΔs not allow a continuous frequency shift of more than = F sin2 (ν + Δν ) 75 MHz, unless the resonator contains additional c 0 L frequency selective optical elements. πΔs 2 b) The refractive index n can be written as ⇒ sin Δν = . c L F n = 1 + aN ⇒ n − 1 ∝ N , If ν0 is the frequency of the transmission peak, the condition where N is the number density of air molecules. If the air pressure changes by Δp = 10 mbar = 2 πΔ ν / = ⇒ πΔ ν / = sin ( s 0 c) 0 s 0 c m (integer) 1%, the same relative change occurs for (n − 1). must be met. For λ = 488 nm, Δs = 2nt = 6cm For air at atmospheric pressure is n − 1 = 2.7 × −4 ⇒ m = 122,950. The frequency condition ΔνL of 10 . The change for Δp = 1% is then Δ(n − 1) longitudinal modes is = 2.7 × 10−6. The optical path length inside the c resonator changes then by Δν = = 125 MHz . L − − 2d Δ(nL) = 2.7 × 10 6 × 0.2 m = 5.4 × 10 7 m With Δ (nL) 8 −1 πΔs 6π ⇒ Δν = ν0 = 2.7 × 10 s . sin Δν = sin 1.25 × 108 L c L 3 × 1010 Also in this case the laser frequency jumps back to = 0.078 the next resonator mode, if Δν > Δνm/2. 556 Solutions to the Exercises

5. a) The classical diffraction theory yields for the Since a considerable fraction of the absorbed en- diffraction of a plane wave with wavelength λ by ergy is lost by heat conduction into the cold sur- a circular aperture with diameter d the angular roundings, one needs, in fact, about ten times as divergence of the central diffraction maximum long, i.e. t = 2.6 ms. λ 6. a) The minimum spectral width is Δα = 1.2 . Δν ≥ /Δ d 0.5 T . Δ = −14 ⇒ Δν ≥ × 14 −1 For a laser beam with a Gaussian intensity profile With T 10 s 0.5 10 s . For a wavelength λ = 600 nm this corresponds to with half-width w0 at the focusing lens with focal length f we obtain for the half-width of the local a spectral width − spot: Δλ = 6 × 10 8 m = 60 nm . λ b) The spatial pulse length is in the beginning: w0 = f . πws cΔT − Δs = = 2 × 10 6 m = 2 μm. 0 n With ws = d/2 we obtain: After passing through a dispersive medium with λ length L, the pulse length has broadened. The w0 = f = 42 μm. πd/2 difference of optical path length for the different λ This differs from the result for a plane wave by the wavelengths within the spectral pulse profile is factor 1.2π/2 ≈ 1.9. dn − Δ(nL) = L Δλ = L · 4.5 × 104 × 6 × 10 8 b) The intensity is obtained from dλ −3 ∞ = 2.64 × 10 L . = π · P 2 r I (r)dr In order to keep Δs ≤ 4 μm, the condition 0 ∞ − / 2 1 = π · · (r r0) = πw2 · Δ = Δ − Δ ≤ μ 2 I0 r e dr 0 I0 (nL) ( s s0) 2 m 0 n × 12 has to be fulfilled. P 10 10 W 9 2 I0 = = = 1.8 × 10 W/m . −6 πw2 π · 422 m2 n 2.0 × 10 m − 0 ⇒ L = = 1.1 × 10 3 m 2.64 × 10−3 = c) Only 10% 1 W can be used for the evapo- = 1.1 mm , ration. The mass of the evaporated material with thickness D is for n = 1.5. This means that after the passage through a 1.1-mm thick glass plate the pulse length = · πw2 M 0 D has already doubled from 2 to 4 μm. − − = 8 × 103 π × 422 × 10 12 × 10 3 kg c) Because of the nonlinear, intensity-dependent − part of the refractive index the wavelength λ is red- = 4.4 × 10 11 kg . hifted during the pulse rise time (dI/dt > 0).Itis / > The heat of evaporation is blue-hifted at the trailing edge (dI dt 0). If the material is chosen in such a way that the 6 We = 6 × 10 J/kg . linear part of n in the surrounding of λ0 (central wavelength of the spectral pulse profile) shows The necessary energy for evaporation is anomalous dispersion (dn/dλ>0), the red part is −4 delayed more than the blue part of the pulse. This W = We · M = 2.6 × 10 J. compensates the opposite effect of the nonlinear The time needed for the evaporation of the mass M part of n. Complete compensation demands is: d (n (λ) + n I ) = 0. − λ 0 2 t = W/P = 2.6 × 10 4 s = 0.26 ms . d Chapter 8 557

7. The resonator quality factor is defined as With a net gain −(2αd + γ ) = 0.05 we obtain P Wk 1 = 0.05 = Qk =−2πν , e 1.05 . dWk/dt P0 where The time T for a round trip is −γ 2d 2 − W (t) = W (0)e k t T = = 10 8 s. k k c 3 is the energy stored in the kth mode of the res- The time-dependent power then becomes: γ onator. If k are the losses per round trip, the power 2αd + γ after one round trip has decreased to P(t) = P exp − t . 0 2d/c −γ = L P P(0)e . With a mirror transmission T = 0.02 the output power of 1 mW demands a power of 50 mW inside The loss factor γL is composed of reflection losses and other losses (diffraction, scattering, absorp- the resonator. The initial power P(0) is given by tion). The reflection losses per round trip are one photon, i.e., hνc − γ =−ln(R R ) = 0.02 . P(0) = = 4.5 × 10 10 W. R 1 2 2d If the other losses together are also 0.02, we obtain a) γ = P / × −8 L 0.04 . = e0.05 (0.666 10 s)t P0 Since the round trip time of a light wave in a res- 1s P ⇒ t = onator with mirror separation d is 6 ln 7.5 × 10 P0 −2 2d − 5 × 10 T = = 1.3 × 10 7 ln s c 4.5 × 10−10 − the losses per sec and are = 25 × 10 7 s = 2.5 μs. c γ = γ /T = γ . b) If we take into account saturation effects, the L 2d L gain depends on the laser power inside the res- − onator. We obtain: With d = 1m ⇒ γ = 1.5 × 108 × 0.04 s 1 = × 6 −1 dP(t) gain per round trip 6 10 s . The quality factor is then for a fre- = P(t) quency ν = 5 × 1014 s−1: dt round trip time 1 π × 14 =− (−2α d + γ ) − 2adP(t) P(t). 2 5 10 8 0 Qk = = 5.2 × 10 . T 6 × 106 With T = 2d/c,(−2α0d + γ ) = 0.05 this gives Per oscillation period of the light the power de- dP 1 creases by the fraction = (0.05 − 2adP)P . dt T 2π − η = = 1.2 × 10 8 . This is a nonlinear differential equation × 8 5.2 10 y˙ − Ay + By2 = 0 /γ ≈ × −7 It takes 1 1.7 10 s until the power in the with resonator mode has decreased to 1/e of its initial 0.05 2ad value. A = , B = . 8. After one round trip the power has increased by the T T factor Division by y2 gives: P y˙ A 1 = −(2αd+γ ) − + B = e . 2 0. P0 y y 558 Solutions to the Exercises

Substitution: Chapter 9 1 1 z(t) = ⇒˙z =− y˙ , y(t) y2 1. The potential energy of the Coulomb repulsion of ⇒˙z + Az − B = 0. the two protons is 2 The solution of the homogeneous equation (B =0) e −18 Epot = = 2.3 × 10 J = 13.6 eV . gives: 4πε0 · 2a0 − z = Ce At . The potential energy of the electron in the state |φ+ is The solution of the inhomogeneous equation 2 e + 1 1 (B = 0) is: E =− |φ |2 + dτ . pot πε − − 4 0 rA rB z = C(t)e At ⇒˙z = C˙ − CA e At . Thewavefunctionis Inserting this into the inhomogeneous equation + φA + φB yields φ = √ 2 + SAB C˙ − CA+ AC = BeAt Inserting φA and φB from (9.9)gives: B − / − / − + / ⇒ C = eAt + D 1 e 2rA a0 + e 2rB a0 + 2e (rA rB) a0 A |φ+|2 = π 3 + B − 2 a 1 SAB ⇒ z = + De At 0 A e2 E =− 1 pot 2 3 ⇒ = 8π ε0a (1 + SAB) y −At 0 B/A + De − / − / − + / e 2rA a0 + e 2rB a0 + 2e (rA rB) a0 1 ⇒ P(t) = . / + −0.05t/T rA 2ad 0.05 De − / − / − + / e 2rA a0 + e 2rB a0 + 2e (rA rB) a0 For t = 0 ⇒ P = P0 + dτ . rB 1 ⇒ D = − 40ad With elliptical coordinates P0 r + r P0 μ = A B ⇒ P t = ; ( ) −0.05t/T . R 40adP0 + (1 − 40adP0)e rA − rB 8 ν = ; With P(t1) = 50 mW ⇒ P(t1)/P0 ≈ 10 be- R −10 cause P0 = 4.5 × 10 W. The denominator ϕ = arctan (y/x) therefore has to be 10−8. With a = 0.4 W−1 m−1, d = 1 m we obtain where the two protons are sitting in the focal points. With these coordinates and × × × −10 + −0.05t/T = −8 40 0.4 4.5 10 e 10 3 − / − R 2 2 ⇒ e 0.05t T = 2.8 × 10 9 dτ = μ − ν dμ dν dϕ , 8 T − ⇒ t =− ln 2.8 × 10 9 the integral can be solved analytically. With 0.05 T = R μ + ν ⇒ t =+ ln 3.57 × 108 rA ( ), 0.05 2 R = 20 T 19.7 = 394 T . r = (μ − ν), B 2 With T = 2 10−8 s ⇒ 263 × 10−8 s = 2.63 μs. 3 the overlap integral SAB becomes: With a = 0.55 W−1 m−1 t increases to 1 − + / S = e (rA rB) a0 dτ t = 20 T ln 1010 s = 30.7 μs. AB π 3 a0 Chapter 9 559 ⎡ ∞ +1 2π For = 2 ⇒ SAB = 0.586 ⇒ 3 R ⎢ − μ/ = ⎣ μ2 · e R a0 dμ dν · dϕ e2 8πa3 2Ca2 = 0 μ= ν=− ϕ= 0 πε · 1 1 ⎤0 4 0a0 1.586 ∞ 1 2π 27.2 eV ⇒ Epot =− · 1.879 =−32.2 eV . − μ/ ⎥ − e R a0 dμ ν2 dν dϕ⎦ . 1.586 The kinetic energy of the electron can be obtained μ= ν=− ϕ= 1 1 0 from the energy relation The integration yields − − Ekin(e ) + Epot(e ) + Epot(protons) = E(H) + E , R R2 B −R/a0 SAB = e 1 + + . + 2 where EB is the binding energy of H and E(H) is a0 3a0 2 the energy in the groundstate of the H-atom. This gives for the potential energy − ⇒ Ekin(e ) =−13.6 eV − 2.65 eV + 32.2 eV

· 2 − 13.6 eV C R − (μ+ν) − (μ−ν) Epot =− e + e = 2π 2.35 eV μ ν ϕ − μ This approximate method gives a too small val- +e μ dμ dν dϕ ue. The real value is Ekin = 12 eV, compared to = with Ekin(H) 13.6 eV. 2. For R → 0theH2-molecules converge towards the e2 He-atom 2He. Therefore the energy of the electrons C = ; = R/a 2 3 0 has to converge towards the groundstate energy of 8πε0a (1 + SAB) 0 the He-atoms. (Note that the two missing neutrons 2 − (μ+ν) − (μ−ν) ⇒ Epot =−C · R μ e + e do not affect this energy.) This is, according to μ ν Sect. 6.1: + − μ μ μ ν e d d E(He) =−78.9 eV . ∞ − μ =−C · R2 μe dμ This value can be composed of the energy =− × × =− μ=1 E1 2 4 13.6 eV 108.8 eV +1 − ν + ν without repulsion between the two electrons, and × e + e + 1 dν the Coulomb energy ν=−1 E =+29.9 eV 1 2 =− CR2 + − − 2 1 e e of this repulsion. ∞ The energy of the electron is: × μ · − μ μ 2 e d el e E (H2, R = Re) = 2 · E(H) + EB(H2) − , 1 4πε0 Re 1 =− Ca2 + − − where E(H) =−13.6 eV is the electronic energy 2 0 1 e e of the H-atom, EB(H2) =−4.7 eV is the bind- × + − 2/ πε = ( 1)e ing energy of the H2-molecule and e (4 0 Re) × −17 = 1 − 0.3 10 J 19.4 eV is the nuclear Coulomb =−2Ca2 1 + 1 − e 2 0 repulsion at a distance Re = 0.074 nm between the two protons. The result is: − + (1 + )e . el E (H2, R = Re) =−51.1 eV . 560 Solutions to the Exercises

This energy is composed of negative potential en- and the vibrational-rotational energy is then ergy, due to electronÐproton attraction, positive ∞ potential energy of electronÐelectron repulsion and Evib, rot = N(v, J) · E(v, J) positive kinetic energy of the two electrons. v,J=0 3. a) The total energy of the rigid H2-molecule 1 with E = n + · h¯ω ; (including Coulomb repulsion between the two vib 2 vib nuclei) is, according to (9.2) Erot = J(J + 1)hc · Brot . The results do not differ much from the estimation E(H2) = 2E(H) + EB(H2) above. =−27.2 eV − 4.7 eV 4. In the product e2 = Eel(H , R ) + ψ (r, R) = χ(R) · ψel(r, R) 2 e 4πε R 0 e of nuclear wave functions χ(R) and electronic ⇒ Eel(H , R ) =−19.4 eV − 31.9 eV 2 e wave function ψel(r, R) the parameter R is not =−51.3 eV . a variable. The wave function ψ(r, R) is a function of r, which can be calculated for any arbitrary In order to separate the H2-molecule into 2 elec- but ﬁxed value of R. trons and 2 protons one has to put the ener- The Schrödinger equation is: gy E = Eel(H , R ) − e2/(4πε R ) = 31.9 eV in- 2 e 0 e h¯ 2 2 to the molecule. − Δk(χ · ψ )(1) 2M el b) At a temperature of 300 K the mean vibrational k=1 = + energy is: Evib kT (potential kinetic energy of h¯ 2 the vibrating nuclei). The rotational energy (2 de- − Δe(χ · ψel) + Epot · χ · ψel = Eχψel . 2me grees of freedom for rotations around two possible ψ∗ axis perpendicular to the internuclear axis) is Multiplying the equation with el and integrating over the electron coordinates, gives with 1 ψ∗ ψ τ = E = 2 · kT = kT el el d 1 rot 2 ⇒ Evib + Erot = 2kT . the equation: ¯ 2 2 ¯ 2 − h Δ χ − ψ∗ h Δ ψ τ · χ The relation between kT and E is: k el e el d el 2M 2me k=1 =ˆ = 1eV kT for T 11,604 K . + χ · ψ∗ ψ τ = · χ el Epot el d E

⇒ at T = 300 K: because the operator Δk acts only upon χ and Δe only upon ψel. 300 2kT =ˆ 2 · eV = 52 meV . The time-averaged potential energy of the nuclei, 11.604 averaged over the motion of the electron is e2 This is very small compared to the electronic ener- E (R) = Eel + Eel + el ≈ pot kin pot 4πε R gy difference E 10 eV between the ﬁrst excited 0 2 electronic state and the groundstate of H2. ∗ h¯ =− ψ Δeψe dτe The correct calculation of Evib and Erot has to take el 2me into account the quantization of the energy levels. e2 v + ψ∗ el + ψ τ The population of a level ( , J)is: el Epot el d el 4πε0 R −(E +E )/kT ¯ 2 (2J + 1)e rot vib ⇒−h Δ χ + χ = χ N(v, J) = ∞ k Epot(R) E . + −(Erot+Evib)/kT 2M v,J=0(2J 1)e k Chapter 9 561

The equation for the electrons in the rigid molecule of molecular vibrations around the equilibrium nucl = is obtained from (1) with k Ekin 0. distance Re we obtain: 5. The Schrödinger equation (9.80) is for the non- 2 2ω2 2 1 h c e 1 rotating molecule (J = 0), M = M1 − M2/(M1 + E(v) = h · c · ω v + − v + . e E M2): 2 4 D 2 For the harmonic oscillator the second term is zero. 1 d 2 dS 2M R + E − Epot(R) S = 0. This quadratic term has the consequence, that the R2 dR dR h¯ 2 vibrational energy levels are no longer equidistant, (2) but the distance ΔE = E(v + 1) − E(v) decreases Introducing the function χ(R) = R · S(R)(2) be- with increasing v. comes: For d2χ 2M + E − E R χ = 1 2 2 pot( ) 0. hcω v + = 2E dR h¯ e 2 D With the relative elongation = (R − Re)/Re of the vibrating nuclei and the Morse-potential the dissociation limit is reached. The maximum possible quantum number v is then: −a 2 Epot(R) = ED 1 − e v = 2ED − 1 we obtain: max ω . hc e 2 d2χ 2M + E − E − −a 2 χ = 2 2 D 1 e 0. (3) 6. The ionization energy of H2 is (Fig. S.12): dR h¯ ion ion + With the trial solution: E (H2) = EB(H2) + E (H) − EB H √ 2 −ε − / = + − χ = z A 1 e z 2 · u , (4.48 13.6 2.65)eV = 15.43 eV . (4) − E z = 2A · e a ; ε = , ED + The binding energies EB(H2) and EB H2 are de- 2E · MR2 A2 = D e ﬁned as the energies from the minimum of the h¯ 2a2 potential curve to the dissociation limit. The mea- Equation (3) becomes the Laguerre differential sured values of the dissociation energy refers to equation: the lowest vibrational level. In this case the energy √ relation is: d2u du 2A 1 − ε + 1 + − 1 ion v = = v = + ion dz2 dz z E (H2, 0) ED(H2, 0) E (H) + √ − E H , v = 0 ,(5) A − 1 − A 1 − ε D 2 + u · 2 = 0. z This equation has the eigenvalues (see text books on differential equations) v + 1/2 2 ε = 1 − 1 − A 2 1 1 1 2 = v + − v + , A 2 A2 2 where v = 0, 1, 2 is the vibrational quantum number. With the vibrational frequency ω = 2ED e hc · A Fig. S.12. 562 Solutions to the Exercises

2 which differs from (4) by the difference of the zero- With Epot = k · (R − Re) point energies ⇒ − = / Δ = v = − + v+ = R Re Epot k . EZP Evib(H2, 0) Evib H2 , 0 . The energy of the vibrational level v = 1is 7. With Re = 1.2745 Å ⇒ h¯ 3 B = , Evib = h · ν . e π μ 2 2 4 c Re m1 · m2 μ = At the turning point is Ekin = 0 ⇒ Evib = m + m = 3 ν 1 2 Epot 2 h 1 · 35 = = 0.9722 AMU for H35Cl 36 ⇒ · − 2 = 3 · ν · k (R Re) h 1 37 37 2 = = 0.9737 AMU for H Cl 1/2 38 3 − ⇒ R − R = · 6.6 × 10 34 · 9 × 1013 513 e 2 35 −1 ⇒ Be H Cl = 10.68 cm , −11 = 1.32 × 10 m = 0.132 Å . − ⇒ B H37Cl = 10.659 cm 1 . e The vibrational amplitude in the level v = 1is For H35Cl we obtain: therefore only about 10% of the internuclear distance Re: νrot(J = 0 → J = 1) = 2cBe − − − / = 6 × 1010 · 10.68 s 1 = 64.1 × 1010 s 1 (R Re) Re 0.104 . = 641 GHz , νrot(J = 4 → J = 5) = 10cBe = 3204 GHz = 3.204 THz . Chapter 10 For H37Cl we obtain: 1. The three principal rotational axes are perpendic- 10 −1 νrot(J = 0 → J = 1) = 6 × 10 · 10.66 s ular to each other and intersect in the center of = ⇒ λ = /ν = mass S (Fig. S.13), which divides the heights h of 639.6 GHz c 4.7 cm , the triangle in the ratio 2 : 1. If the sides of the tri- Δν H35Cl − H37Cl = 1.4 GHz rot angle are denoted as s we obtain the three moments νrot(J = 4 → J = 5) = 3.198 THz of inertia for rotations around the axis i: ⇒ λ = 0.94 cm . 2 2 = = h + 2h = Ia I1 2m m The rotational energy Erot(J 5) is: 3 3 E /hc = J · (J + 1) · B = 30 B 2 2 rot e e = mh2 = ms2 cos2(α/2) , −1 35 3 3 = 320.4 cm ⇒ Erot = 39.7 meV for H Cl −1 37 = 319.8 cm ⇒ Erot = 39.6 meV for H Cl 8. The vibrational frequency for the model of a harmonic oscillator is: 1 ν = k/μ ⇒ k = 4π 2ν2 · μ . 2π For H35Cl is μ = 0.9722 AMU − ⇒k = 4π 2 · 9 × 1013 · 0.9722 ·1.66 × 10 27kg/s2 = 513 kg/s2 . Fig. S.13. Chapter 10 563

where m is the mass of the Na-atom. I = I = 2mx2 = 2ms2 sin2(α/2) b 2 2 2 I = I = 2mr2 + m h c 3 3 2 = 2 + h 2 with r x 3 we obtain:

2 2 2 I3 = 2mx + mh 3 1 Fig. S.14. = 2ms2 sin2(α/2) + cos2(α/2) . 3 For α = 80◦ this gives: 2 2 h¯ −47 2 I1 = 0.39 m · s ; I2 = 0.83 m · s I = = 3.50 × 10 kg · m a 4πc · A = · 2 − − I3 1.22 m s I = 64.3 × 10 47 ks m2 10 47 ⇒ = + b I3 I1 I2 . −47 2 −47 Ic = 68.3 × 10 kg m 10 The last equation is true for all planar molecules. From Fig. S.14 we ﬁnd: The Na3-molecule represents an asymmetric rotor = = = with I1 I2 I3 I1. y1 = m(N) = 14 = = × × −27 = = 0.4375 With m 23 1.66 10 kg; s 3.24 Å y2 2m(O) 32 −10 3.24 × 10 m = 2 + 2 Ia 2m(0)y1 m(N)y2 −45 2 ⇒ Ia = I1 = 1.56 × 10 kg m 2 2 32 −45 2 = y 2m(O) + m(N) Ib = I2 = 3.32 × 10 kg m 1 14 −45 2 Ic = I3 = 4.85 × 10 kg m = + · 2 (32 73) AMU y1 The rotational constants are deﬁned as: = 2 = × −25 · 2 105 AMU y1 1.74 10 kg y1 h¯ − − = · 2 A = = 17.85 m 1 = 0.1785 cm 1 Ib 2m(O) x Inserting the numerical value I1 4πc · I − − a = 3.5×10 47 ks·m2 yields y =1.42×10 11m h¯ 1 = = −1 = −1 − B 8.388 m 0.0839 cm ⇒ y = y /0.438 = 3.246 × 10 11 m 4πc · Ib 2 1 −11 h¯ − − ⇒ y = 4.67 × 10 m C = = 1 = 1 π · 5.742 m 0.0574 cm 4 c Ic = 32 AMU · x2 − One can prove, that within the accuracy of round- = 0.53 × 10 25 x2 · kg · m2 ing up and down the relation Inserting the value of Ib gives 1 1 1 − + = x = 11.10 10 m ◦ A B C With tan(α/2) = x/y ⇒ α = 134 is valid for the planar Na3-molecule. This is, From S = x2 + y2 ⇒ S = 1.19 Å however, only strictly valid for the non-vibrating 3. The linear molecule C2H2 with 4 atoms has 3 · molecule. 4 − 5 = 7 normal vibrations, where the two vibra- 2. The values of the moments of inertia can be ob- tions ν4 and ν5 are degenerate (Fig. S.15). There tained from the rotational constants. are therefore 5 different vibrational frequencies −1 A = 800 m ν1, ν2, ..., ν5. Transitions from the (0, 0, 0, 0, 0)- B = 43.4 m−1 vibrational ground state to excited vibrational C = 41.0 m−1 states are infrared active, if the dipole moment of 564 Solutions to the Exercises

Fig. S.15. −8 is then Fr =−1.3 × 10 N. For the bending vibration ν2 the amplitude Δy of the C-atom is 2m(O) 32 Δy (C) =− · Δy (O) =− Δy . 1 m(C) 2 12 2 The time dependent distance d(t) between the C-atom and one O-atom of the vibrating molecule is (Fig. S.17): 1/2 = 2 + Δ − Δ 2 d d0 ( y1 y2) 1/2 = 2 + Δ 2 d0 (1.375 y1) / 2 1 2 Δy1 = d0 1 + 1.9 d0 the upper state is different from that of the ground state. These are the vibrations ν and ν . Δ 2 3 5 ≈ + y1 The Raman active transitions are: ν1, ν2 and ν4 d0 1 0.95 . d0 (Fig. S.15). 4. In the approximation of harmonic oscillations the The change (d − d0) during the bending vibration frequencies ν are related to the force constants k is then by 2 d − d0 = 0.95Δy /d0 . 1 1 ν = k/μ , 2π The restoring force is μ =− − =− · Δ 2/ where is the reduced mass. F k2(d d0) k2 0.95 y1 d0 . For the normal vibration ν1 the C-atom remains at rest and the two O-atoms oscillate with opposite The energy is then phases (Fig. S.16). The restoring forces for the two 1 E = h · ν = h · ν · c = 2 · · k (d − d )2 . C−O bonds are equal with opposite directions. The vib 2 2 0 reduced mass is On the other side is 2m(O) · m(C) 2 · 16 · 12 μ = = AMU = π 2 2ν2 · μ = −2 2m(O) + m(C) 44 k2 4 c 2 231 kg s . = 8.73 AMU . This gives a change 1/2 The restoring force constant k1 is then: d − d0 = (hν · c/k2) = π 2ν2 · μ = π 2 2ν2μ = × −10 = k1 4 1 4 c 1 7.6 10 m 0.076 Å . = 4π 2 · 9 × 1020 · 13882 × 8.73 AMU − = 9.96 × 102 kg/s2 ≈ 1000 kg s 2 .

The restoring force Fr =−k · (R − Re)atthe −11 maximum elongation (R − Re) = 1.3 × 10 m

Fig. S.16. Fig. S.17. Chapter 11 565

This should be compared with the average distance Therefore about 4Ð5 vibrational levels are simul- d0 ≈ 1.2 Å. The vibrational amplitude is therefore tanously excited. For v = 15 the nuclear vibra- only 6% of the C−O-distance. tions can be treated classically. 5. a) The ground state configuration of BH2 is − 2 2 1 2 ⇒ ΔE = h · ν = hc · 123 cm 1 ...(2a1) (1b2) (3a1) X A1. According to the vib Walsh diagram in Fig. 10.11 the energy of the 10 −1 ⇒ νvib = 3 × 10 · 123 s (2a )-orbital and of the (1b )-orbital decreases 1 2 = × 12 −1 from α = 90◦ to α = 180◦, while that of the 3.7 10 s (3a1) strongly increases. The total energy has 1 −13 ⇒ Tvib = = 2.7 × 10 s = 270 fs . a minimum (from an estimation of the curves in νvib Fig. 10.11)atα ≈ 125◦Ð135◦. The correct value is α = 131◦. b) In the first excited state the electron in the Chapter 11 (3a1)-orbital is excited into the (1b1)-orbital. The energy of this orbital is independent of α 1. The net absorption coefficient α is, according to and therefore the excited state leads to a lin- (11.2) given by α = ◦ ear configuration ( 180 ) which is labelled as α = − / σ 2 2 1 2 Nk (gk gi )Ni ki .(1) (2σg) (2σu) (1πu) A πu. 6. In the approximation of the Morse-potential the In our case is energy levels are gk = 2Jk + 1 = 3; gi = 2Ji + 1 = 5. The transition wave number is 2 E(v) = h · c · 15.5ωe − ωexe(15.5) ν = + − + Be[Ji (Ji 1) Jk(Jk 1)] −1 −1 = hc 2176.2 cm = 4Be = 42.36 cm . − = 4.3 × 10 20 J = 0.27 eV . The frequency of the absorption line is: − ν = c · ν = 1.27 × 1012 s 1 . In this case is h · ν kT The energy difference between neighbouring vibrational levels is: N N − ν/ N hν ⇒ i = k · e h kT ≈ k 1 − gi gk gk kT hν ΔE = E(v = 15) − E(v = 14) ⇒ α ≈ N σ .(2) k kT ki = hc[ωe − 30ωexe] At a pressure p = NkT the density of molecules − − = hc (159 − 36) cm 1 = hc · 123 cm 1 is N = p/kT. With p = 1 mbar = 102 Pa = 0.015 eV . 102 m−3 ⇒ N = . 1.38 × 10−23 · T/K − The Fourier-limited frequency width of the 30 fs For T = 100 K ⇒ N = 7.25 × 1022 m 3.The laser pulse is population density of the absorbing level Nk(Jk) is then: (2Jk + 1) · N − / = · Erot kT 1 − Nk(Jk) e ,(3) Δν = ≈ 15.8 × 1012 s 1 Z 2πΔt − where ⇒ ΔE = h · Δν = 11.5 × 10 21 J − / − Z = g e En kT = 6.3 × 10 2 eV = 0.063 eV . n n 566 Solutions to the Exercises

is the partition function, which is a normalization the real cosinus-Fourier-transform of gc(ω)is = factor, that assures n Nn N. ∞ For 2 f t = g ω ωt ω c( ) π c( ) cos d ΔErot = Erot(J + 1) − Erot(J) kT 0 ∞ 2 we can approximate the partition function Z by the gc(ω) = fc(t) cos ωt dt . integral over a continuous variable J: π 0 ∞ The computer, used to perform these transforma- − · + / kT Z ≈ (2J + 1)e hc Be J(J 1) kT dJ = . tions in Fourier-Sepctroscopy subtracts the con- hcBe 0 stant background term in (11.74a), which is independent√ of the time t. Substituting fc(t)by ¯ For our example of the HCl-molecule is Jk = π/2S(t) and gc(ω)byI (ω · v/c) one obtains = −1 = × −2 · / 1, Be 10.59 cm and Z 6.56 10 T K. ∞ For T = 100 K this becomes Z = 6.56 and the ¯ ω · v/ = ω · v/ ratio I ( c) S(t) cos( c)dt . 0 N 3 − k ≈ · e 0.3 ≈ 0.34 . N 6.56 3. The grating equation is d(sin α + sin β) = λ . This means that 34% of all molecules are in the For our example is level with Jk = 1. Inserting (3)in(2)gives 1 d = mm = 0.833 μm, 1200 (2J + 1)h2cν B k e −Erot/kT ◦ 1 α = e · σ · N . α = 30 ⇒ sin α = , (kT)2 2 1 588.9 = × 18 −3 sin β1 =− + =+0.2067 With the numerical values: N 7.25 10 m , 2 833.3 B = 1059 m−1, J = 1, T = 100 K one abtains ◦ e k ⇒ β1 = 11.93 = 0.2082 rad , − − 1 589.5 α = 1.9 × 10 3 m 1 . sin β =− + = 0.2074 rad 2 2 833.3 ⇒ β = ◦ = For T = 300 K the absorption coefﬁcient becomes 2 11.97 0.2089 rad .

− − The angular difference is α = 2.1 × 10 4 m 1 . ◦ − Δβ = 0.04 = 7.36 × 10 4 rad . 2. The two functions In the focal plane of the imaging mirror with focal +∞ length f = 1 m is the lateral distance between the 1 − ω f (t) = √ g(ω) · e i t dω two spectral lines: 2π −∞ − Δs = f · Δβ = 1 · 7.36 × 10 4 m = 0.736 mm . +∞ 1 + ω g(ω) = √ f (t) · e i t dt With a slitwidth d < 360 μm of the entrance slit 2π −∞ the two lines can be separated. 4. a) The shift of the Raman line against the Rayleigh form a Fourier-pair. With the relation line is for Δv = 1, ΔJ = 0fortheH2-molecule Δν = 4395 cm−1. The wave number of the argon- ω ei t = cos ωt + isinωt ion laser line is Chapter 11 567

7 For a temperature T is ν = 10 = −1 L 20,492 cm 488 −7 −1 −1 ΔνD(cell) = 7.16 × 10 ν0 T/M . ⇒ νR = (20,492 − 4395)cm = 16,097 cm ⇒ λ = R 490.9 nm . With T = 500 K, M = 23 g/Mol, ν0 = 5.09 × 14 −1 b) The term value of the rotational level J = 1is 10 s −1 with Be = 60.8 cm 9 −1 ⇒ ΔνD = 1.7 × 10 s − T(J = 1) = J · (J + 1)B = 121.6 cm 1 . ◦ 9 −1 e ⇒ ΔνD(beam) = (sin 2 ) · 1.7 × 10 s − The difference between the Rayleigh and Raman = 5.9 × 107 s 1 . line is −1 2 a) In order to resolve the hyperfine structure of the Δν = 121.6 cm ⇒ Δλ = λ · Δν = 2.9 nm . 2 3 P1/2 level, the condition The spectral resolving power of the spectrometer 9 −1 should be at least ΔνD(beam) = 1.7 × 10 · sin ε s < 190 MHz λ 488 R ≥ = = 168 , Δλ 2.9 must be fullfilled. This gives ◦ which can be achieved already with a small grating sin ε<0.11 ⇒ ε<6.4 . or prism spectrometer. 5. According to Beer’s absorption law the transmitted b) laser power is 16 − ◦ = · −αx ≈ − α α sin ε< = 9.4 × 10 3 ⇒ ε = 0.55 . Pt P0 e P0 (1 x)for x 1. 1700 The power, absorbed per cm path length is Here one has to take into account, that the natural −6 −1 −7 Δν = ΔP = α · P0 = 10 · 10 W = 10 W. line width is already n 10 MHz. The absorption profile is the convolution of Lorentzian and λ = For a wavelength 500 nm is Gaussian profiles and the total linewidth is − h · ν = 2.48 eV = 3.97 × 10 19 Ws. Δν ≈ Δν2 + (Δν sin ε)2 The absorbed power ΔP corresponds to n D / 1 2 − Δ −7 ⇒ Δν · sin ε ≤ (Δν)2 − Δν2 s 1 = P = 10 W D n N − h · ν 3.97 × 10 19 Ws − = 106 · (16)2 + (10)2 s 1 = × 11 / √ 2.5 10 absorbed photons s − = 106 · 156 s 1 × 11 which generate 2.5 10 fluorescence photons. ≤ 12.5 MHz The photo detector receives 12.5 × 106 ΔΩ 0.2 ⇒ sin ε ≤ = 0.0074 · N = · 2.5 × 1011 = 4 × 109 photons/s. 1700 × 106 π π ◦ 4 4 ⇒ ε ≤ 0.43 . With a quantum efficiency η =0.2 of the detector this gives 0.2 · 4 × 109 = 8 × 108 photo- 7. a) The transverse force Fx acting on the atoms electrons/s. If the amplification of the photomul- flying into the z-direction is tiplier is G = 106 the output current is Fx =−|pm · grad B| . 8 −19 6 IA = 8 × 10 · 1.6 × 10 · 10 = 0.13 mA . The magnetic moment in the 2S / state is main- 6. With a collimation angle ε = 2◦ the residual 1 2 ly due to the electron spin, i.e. pm = μB = 9.27 × Doppler width of an absorption line is − 10 24 J/Tesla. The deflection angle α of the sodi- ΔνD(beam) = sin ε · ΔνD(cell) . um atoms is 568 Solutions to the Exercises v tan α = x . which implies that the mean cycle time absorption- vz fluorescence should be τc = 1/Rabs = 16 ns. The spontaneous life time is τsp = 16 ns. Since the min- The velocity vx is after a flight time t = L/vz imum mean cycle time is τc ≈ 2 τsp the maximum μ 7 −1 t B L rate can be only Rmax = 3.15 × 10 s .Thelaser vx = μB ·|grad B|· = | grad B|· m m vz beam diameter therefore has to be enlarged to ◦ d = 2cm. For α = 3 ⇒ tan α = 0.052. With L = 0.2 m, 8. The effective life time is vz = 600 m/s ⇒ vx = 0.052 · 600 m/s 1 1 2 = + n · σ · v . m · vx · vz m · v tan α τ τ r ⇒ grad B = = z eff spont μB · L μB · L 1 · × −27 · × 4 · If τeff = τspont we obtain = 23 1.66 10 36 10 0.052 2 9.3 × 10−24 · 0.2 · σv = 1 = × 2 / n r . 3.8 10 T m. τspont b) The photon transfers the momentum With p = n · k · T ⇒ Δ = · ν/ p · σ · v 1 k · T p h c . r = ⇒ p = . k · T τspont σ · vr · τspont The momentum of the sodium atoms in z-direction is The mean relative velocity vr is = · v · pz m z . v = 2 8 kT r π · m The deflection angle for the absorption of one m · m 23 · 28 m = 1 2 = = photon is with + 12.6 AMU √ m1 m2 51 Δ · ν α = p = h π · m · kT tan ⇒ p = pz c · m · vz σ · τ √ 4 spont with h · ν = 2.1 eV . π ·12.6·1.66 × 10−24 ·1.38 × 10−23 ·400 = Pa For n absorbed photons it is: 4 · 10−19 · 1.6 × 10−8 = × 5 = h · ν 0.93 10 Pa 0.93 bar . α = n · tan · · v . c m z 9. Assume the level Ei is selectively excited by a cw In order to reach a deflection of α = 3◦ one needs laser with an excitation rate Rexc. If the total c · m · v deactivation rate is n = z · tan α h · ν D = Ni Ai + n · σq · v , 3 × 108 · 23 · 1.66 × 10−24 · 600 = where n is the density of collision partners. We · × −19 2.1 1.6 10 obtain D = R under stationary conditions. This = × 3 1 10 photons , yields the stationary population density With a laser beam diameter d = 1 cm the time of D R Ni = = , flight of the atoms through the laser beam is Ai + n · σq · v Ai + nσq v = /v = × −5 t d z 1.6 10 s. where σq is the total deactivation cross section for nonradiative transitions. The fluorescence power, The minimum absorption rate is then emitted by level Ei is = / = 3/ × −5 = × 7 −1 Rabs n t 10 1.6 10 6.3 10 s , Pi = Ni · Ai . Chapter 12 569

E 7 −1 2π 7 −1 If level m is populated by collisional energy With δ =2π × 10 s , k = λ = 0.94 × 10 m , transfer from level Ei we obtain: 1 7 −1 6 γ = τ ≈ 6 × 10 s , R0 = 10 dNm ⇒ a = ni · n · σi→m · v − Nm · Am = 0 dt 16 · 2π · 107 · 1.06 × 10−34 · 0.942 · 1014 · 106 Nm Am = ⇒ σi→m = · 4π 2 · 1014[1 + (4π/6)2]2 N n · v i −7 15 × 10 − Pm Ai = = 8.2 × 10 23 Ns/m. = · . × 14 Pi n · v 182 10 −23 8.2 × 10 − ⇒ a/M = = 2.1 × 103 s 1 Measuring the relative total ﬂuorescence powers 23 × 1.66 × 10−24 Pm/Pi yields σi→m,ifAi = 1/τi is known from M −3 E ⇒ τdamp = = 0.48 × 10 s = 480 μs. lifetime measurements of level i . a 3. The restoring force in z-direction is

Chapter 12 Fz =−D · z , where the force constant is 1. The lifetime of the 3 2P / level is τ = 16 ns, 3 2 16kδ the optimum absorption-emission cycle period is D = R0 · pm · b · . 2τ = 32 ns, the absorption rate R = 1/(2τ) = γ 2 1 + 4δ2/γ 2 2 109/64 ≈ 1.6 × 107 s−1. Each absorbed photon With transfers the momentum Δp = h · ν/c onto the Δv = 2π − Na-atom and decreases its velocity by k = = 0.94 × 107 m 1 , Δp/m = h · ν/(m · c) ≈ 3cm/s. λ = · × 1 − The velocity decreases per s is then a 3 1.6 γ = = 6 × 107 s 1 , 107 = 4.8 × 107 cm/s2 = 4.8 × 105 m/s2.Inor- τ 6 −1 7 −1 der to bring the Na atom to rest, N = 700/0.03 = R0 = 10 s and δ = 2π · 10 s , 23,333 absorptions are necessary. This takes a time −24 pm ≈ μB = 9.28 × 10 J/T, = −2 / 23,333 − b 10 T m T = ≈ 1.5 × 10 3 s = 1.5 ms . 1.6 × 107 we obtain − − The deceleration path length is D = 106 · 9.28 × 10 24 · 10 2 16 · 0.94 × 107 · 2π · 107 1 2 × N/m Δz = v T − aT 2 0 2 36 × 1014 1 + 16π 2/36 − = × 3 −2 1.5 10 = 3.7 × 10 N/m. 1 − × 700 − · 4.8 × 105 · 1.5 × 10 3 m The oscillation frequency is: 2 −3 Ω = D/M = 1.5 × 10 (700 − 360) m − = 0.51 m . = 3.7 × 10−2/(23 · 1.66 × 10−24)s 1 10 −1 2. The net force on the atoms is = 3 × 10 s . The damping constant is F =−a · v with a 3 −1 16δhk¯ 2 β = = 2.1 × 10 s a = . M 2 2 2 γ 1 + (2δ/γ ) ⇒ τdamp = 480 μs. 570 Solutions to the Exercises

The atoms perform 3 × 1010 · 4.8 × 10−4 = 1.4 × ΔE = h · Δv = 0.44h/ΔT . 7 10 oscillations, before the oscillation amplitude With ΔT =2 × 10−14 s ⇒ ΔE =1.5 × 10−20 J = / has decreased to 1 e of its initial value. 0.09 eV. The energy separation of the vibrational 4. The mean distance d is related to the density N by levels is = −1/3 ⇒ = 3 −13 1 d N d 10 cm ΔE = (E(v + 1) − E(v − 1)) − − = 0.46 × 10 4 cm = 4.6 × 10 7 m. 2 1 = hc [2ω − (4v + 2)ω x ] . The de Broglie wavelength is 2 e e e h v = ω = −1 ω = λ = √ = = × −7 For 20 this gives with e 30 cm , exe dB d 4.6 10 m −1 3mkBTc 0.04 cm h 1 ⇒ T = Δ = ω − ω c 2 E hc [2 e 82 exe] 3mk d 2 B − − × −34 = 5.61 × 10 22 J = 3.5 × 10 3 eV . = 6.6 10 3·23×1.66 × 10−27 ·1.38 × 10−23 ·6.72 ×10−14 The laser pulse can therefore excite = × −7 = 9.3 10 K 930 nK . 0.09 n = = 26 BEC occurs under these conditions at Tc = 0.0035 930 nK. The atoms in the trap can reach a region, vibrational levels of the Cs2-molecule. =−μ · where the magnetic field energy EM m B 6. The intensity at the output of the Michelson inter- = 1 v2 v2 = equals their kinetic energy Ekin 2 m with ferometer is: 3kT . m I0 2 It = · cos (Δϕ/2) , 3 3 − − 2 ⇒ E = kT = · 1.38 × 10 23 · 9.3 × 10 7 J. kin 2 2 where Δϕ = 2π · Δs/λ is the phase difference be- Δ The magnetic field is tween the two interfering beams. If s changes by δ ⇒ Δϕ changes by δϕ = 2πδ/λ. The intensity 3 B Em 2 kT change B = b · r ⇒ r = = = . b μm · b μm · b Δϕ + δϕ Δϕ − ≤ −8 2 −24 −1 It It 10 I0 With μm Na S1/2 ≈ μB = 9.27 × 10 JT 2 2 −23 −7 −8 1.5 × 1.38 × 10 · 9.3 × 10 should be smaller than 10 I0. With Δϕ = π ⇒ ⇒ r = m 9.27 × 10−24 · 10−3 Δϕ + δϕ Δϕ δϕ −3 cos = cos cos = 2.08 × 10 m = 2.08 mm . 2 2 2 13 −3 13 Δϕ With a density of 10 cm there are N = 10 · − sin sin(δϕ/2) 4 πr 3 = 3700 atoms in the trap. 2 3 =− δϕ/ ≈−δϕ/ 5. The energy of a vibrational level with quantum sin( 2) 2 number v is: I0 − ⇒ I = · δϕ ⇒ δϕ ≤ 4 × 10 8 rad . t 4 1 1 2 E(v) = hc ωe v + − ωexe v + . The phase of a plane wave is sensitive to the po- 2 2 sition of the plane mirror surface, averaged over The energy width of the femtosecond pulse with the whole surface. Deviations from an ideal plane, Gaussian time profile, for which Δν·ΔT =0.44 is caused by the atomic structure, are averaged out. References

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aÐb-diagram 62 ÐMO 336 autoionization 233, 282, 378 absolute mass 53 ÐMO-LCAO 339 Autrecourt, N. 8 absorption Ð self-consistent field 223 average number of photons per mode Ð continuous 280 Argon laser 311 249 Ð Doppler-free two-photon 186 Aristoteles 8 Avogadro, A. 11 Ð effective coefficient 282 asymptotic solution 162 Avogadro constant 12, 14, 17 Ðlosses 290 atom Ð minimum coefficient 422 Ð alkali 218 B2 molecule 345 Ð nonlinear 456 Ð anti-hydrogen 240 Balmer’s formula 116 Ðrelative 440 Ð carbon 229 Balmer series 172 Ðspectrum 112 Ð doubly excited 282 band 374 absorption coefficient 273 Ð exotic 237 Ð fundamental 405 absorption edges 276 Ð helium 201 Ð head 375 abundance Ð images 20 Ðorigin 372 Ð isotopic 64 Ð interferometry 495 Ð parallel 406 actinides 216 Ð kaonic 239 Ð perpendicular 406 Airy formulas 431 Ð laser 495 Ð system 374 α-decay 137 Ð multielectron 221 Ð vibrational 372 amplifier Ð myonic 238 beam Ð optical 306 Ð pionic 239 Ð atomic 44 amplitude Ð quantum structure 111 Ð molecular 453 Ð quadrature 504 Ð stability 117 beam profile for the fundamental angular momentum 152 Ð thermal He 101 modes 300 Ð conservation 254 Ð tin 228 BEC 493 Ð coupling 224 atomic beam 44 Beer 456 Ð electronic 341 atomic clock 503 Beer’s law 456 Ð operator 152 atomic force microscope 29 bending vibration 402 Ð quantum number 153 atomic interference pattern 99 benzene 392 Ð rotational 359 atomic mass unit (AMU) 10 Bernoulli, D. 9 Ð vibrational 402 atomic orbital 334 beryllium 211 anti-proton 240 Ð1s deformation 339 Berzelius, J.J. 10 anti-Stokes radiation 461 Ð deformation 334 binding apodization 442 atomic polarizability 354 Ð covalent 356 approximation atomic spectrum 112 Ð ionic 356 Ð adiabatic 357 atomic unit 335 Ð types 356 Ð BornÐOppenheimer 378 atomic weight 10 binding energy 335, 338, 364 Ð breakdown of the attenuation coefficient 273 BiotÐSavart’s law 178 BornÐOppenheimer 403 Auger blue-shadowed 376 Ð HeitlerÐLondon 337 Ð effect 234 Bohr magneton 169 ÐLCAO 331 Ð process 233 Bohr radius 115, 162 582 Subject Index

Bohr’s atomic model 113 Ð charge exchange 40 Ðintegral 66, 69 bolometer 434 Ð cross section 421 Ð integral scattering 65 Boltzmann, L. 8 Ð elastic 268 Ð threefold differential 467 Boltzmann distribution 249 Ð inelastic 269 crystal bond Ð phase-changing 269 Ð birefringent 310 Ð chemical 349, 383 Ð quenching 480 Ð body-centered cubic 32 Ð covalent 350 Ð radius 267 Ð face-centered cubic 32 Ð homopolar 350 Ð reactive 421, 479 Ð liquid 501 bonding Ð time 267 Ð uniaxial birefringent 315 Ð hydrogen 356 Ð uenching 269 cyclotron frequency 63 Ð localized 394 collision broadened emission line Ð localized σ -type 393 268 Dalton, J. 9, 10 Born’s statistical interpretation 123 collision complex 414 damping constant 262, 493 boron 211 collision-induced rate 261 De Broglie wavelength 97 BoseÐEinstein condensation 493 colour-center laser 305 deflection angle 72 boson 205 Compton effect 91, 274 degenerate 145, 165 bound-bound transition 282 Compton wavelength 93 delocalized 393 boundary conditions 133 concept of nature 124 Ð π orbitals 394 Bragg diffraction 280 condition Democritus 7, 8 Bragg reflection 100 Ð phase-matching 462 density bremsstrahlung 271, 280 configuration Ð effective 438 Brownian motion 20 Ð electron 224 Ðenergy 81 Ð interaction 224 Ð mode 85 carbon 211 Ð two-electron 231 Ð momentum 82 CARS 462 contribution Ð packing 32 cathode rays 34 Ð monopole 351 Ð population 398 CCD array 424 conversion efficiencies 315 Ðpower 83 center of mass 159 coordinate Ð probability 106, 120 centrifugal distortion 361 Ð elliptical 328 derivative chaotic movement 120 Ðinversion 341 –first 438 charge distribution 70 Ð mass-weighted 400 detector 433 chemical laser 311 Coriolis force 403 Ð LangmuirÐTaylor 480 clamped nuclei 336 correlation diagrams 348 Ð phase-sensitive 445 classical particle paths 120 correspondence principle 194 Ðthermal 434 classical wave description 95 Coulomb force 34 Ð time constant 434 Clausius, R.J. 8, 9 Coulomb-repulsion 337 determinant cloud chamber 24 coupling Ð Slater 223 cluster 408 Ð between vibrations and rotations device Ð mass distribution 410 402 Ð photoconductive 435 Ð melting temperature 411 Ð fine structure energies 225 Ð photovoltaic 435 Ð size 409 Ð intermediate cases 229 diagram Ð temperature 411 Ð jÐ j 228 Ð Slater 230 CO2-laser 312 Ð LÐS 225 ÐWalsh 389 CO2 molecule 388 Ð of angular momenta 224 diatomic molecules coherence length 304 Ð schemes 224 Ð vibrations 363 coherent control 499 Ð spin-orbit 207, 342 difference-frequency generation coincidence experiments 467 Ðterms 402 316 CO-laser 311 Ð vector diagram 228 diffraction collision covalent binding 356 Ð atoms 98 Ð broadening 267 cross section ÐBragg 280 Ðcentral 68 Ð differential 65, 66, 69, 75 Ð electron 97, 108 Subject Index 583

Ðlosses 290, 296, 300 Ð independent 203, 221 Epicurus 8 Ð X-ray 15 Ðmass 41 equation diffusion 30 Ð mechanical model 197 Ð uncoupled 401 Ð coefficient 14 Ð motion 361 equilibrium distances 360 Ðprocess 22 Ð multiplier 39 Erbium laser 305 dipole Ð optics 47 excess 432 Ð approximation 252 Ð Rydberg 235 excimer 312, 347 Ð electric 351 Ðshell 209 excitation Ð Hertzian 250 Ðspin 204 Ð inner-shell 233 dipole moment electronic transition 372 Ð one-electron 231 Ð induced 352 electron microscope 25, 52, 102 Ð simultaneous 232 Ð momentary 353 electron model 195 Ð single electron 232 Ðtotal 404 electron radius exotic atom 237 Dirac equation 159, 190 Ð classical 196 expansion discharge electron shell 168 Ð adiabatic 410 Ð hollow cathode 448 electron spin 176 Ð Dunham 366 dispersion element expansion of the universe 44 Ð angular 424 Ð alkali 216 expectation value 149, 166 Ð linear 424 Ð alkaline earth 216 Ð for the location x 149 Ð of FPI 433 Ð chemical properties 217 Ð of the kinetic energy 150 Ð of the wave packet 109 Ð piezo 27 experiment displacement Ð trans-uranium 216 Ð ideal 478 Ð mass weighted 401 elliptical coordinate 328 dissociation energy 364 elliptical mirror arrangement 452 FabryÐPerot-etalon 310 distribution emission factor Ð angular 467 Ð coefficient 37 Ð Hönl–London 404, 406 Ð function 21 –field 37 fadenstrahlrohr 42 Ð Gaussian 73 Ð induced 248 Fano Ð radial of atomic electrons 209 Ð secondary electron 38 Ðprofile 283 Ð temperature 463 Ðspectrum 112 Fano parameter 283 Doppler broadening 264 Ð spontaneous 248 Fano profile 283, 378, 379 Dunham coefficients 366 Ðthermal 37 Faraday’s law 14 duo-plasmatron 41 Empedocles 8 Faraday polarization rotator 310 energetic order femtosecond 316 eigenfunction 150 Ð of the orbitals 344 Fermi-contact interaction 186 Ð common 155 energy fermion 205 eigenvalue 150 Ð binding 335 field eigenvolume 29 Ð density 81 Ð axially symmetric 47 Einstein coefficient 248 Ð dissociation 364 field ionisation 137 Ð for absorption 248 Ð equipotential curves 416 finesse 433 EinsteinÐde Haas effect 177 Ð fine structure coupling 225 fine structure 178 EinsteinÐPodolskiÐRosen paradox Ð ionization 117, 212, 411 Ð component 179, 344 512 Ð mean rotational 364 Ð constant 181 electric field Ð of Rydberg levels 236 Ð coupling energies 225 Ð perpendicular 46 Ð rotational 362 Ð labeling of component 225 electric quadrupole moment 258 Ðsurface 383 Ð Sommerfeld’s constant 180 electron 35 Ð transfer 291 Ð splittings 343 Ð configuration 224, 344 Ð zero-point 139 fine-structure component 278 Ð delocalized 392 energy density fingerprint region 440 Ð diffraction 97, 108 Ðspectral 249 flashlamp Ð free 37 energy pump 289 Ð helical 293 584 Subject Index

floating ball 45 gain factor 290 Ð sp2 386 fluctuation gas Ð sp3 386 Ð of the laser frequency 303 Ð discharges 34 hydrogen atom 164 Ð phase 304 Ð effusion 44 hyperfine component 185 Ð technical 303 gas constant 12 hyperfine constant 185 fluorescence 260 gas laser 310 hyperfine structure 184, 455 Ð laser-induced 450 gate function 442 focal length Gay-Lussac, J.L. 11 imaging Ð of a magnetic lens 51 Gedanken-experiment 122 Ð of electron beams 47 Ð of a magnetic sector field 52 generator impact parameter 67, 77 focusing Ð photocurrent 436 induced dipole moment 352 Ð double 57 gerade 341 induced emission 248 Ð velocity-independent 55 giant pulse 317 infrared-active 405, 461 force globar 440 inner-shell excitation 233 Ð Coriolis 403 Gordon 289 integral Ð Coulomb 34 grating Ð overlap 387 Ð nuclear 76 Ð concave 428 intensity 82, 95 Ðrestoring 400 Ð equation 427 Ð of a rotational line 397 formaldehyde molecule 392 ÐRowland 465 Ð spectral 82 formula grating equation 279 intensity distribution of the Ð modified Rydberg 221 grating pair 322 fundamental modes 299 Ð Rydberg 234 gravitational redshift 94 interaction Ð SchawlowÐTownes 304 greenhouse effect 369 Ð configuration 224 Fortrat-diagram 372, 375 ground state Ð dipoleÐdipole 187 Fourier limitation 310 Ð helium 206 Ð exchange 349 Fourier transformation 441 Ð of the H atom 172 Ð Fermi-contact 186 FranckÐCondon factor 373 gyromagnetic ratio 176 Ð multipole 350 FranckÐCondon principle 376 Ð octupole 354 FranckÐHertz experiment 118 H2 molecule 335 Ð rotation-vibration 364 free-electron laser 305 H2AB molecule 392 ÐvanderWaals 353 free spectral range 431 H2O molecule 383 Ð virtual 191 frequency halogens 218 interference Ð cyclotron 63 hard spheres 18, 70 Ð effects 477 Ð modulation 438 HartreeÐFock method 224 Ð fractional order 432 Ð tunnel 391 Hartree method 222 Ð multiple-beam 430 frequency selective optical elements He2 molecule 345 interference pattern 96 302 2+ He2 345 interference phenomena 121 frequency spectrum heat interferometer Ð of optical resonators 301 Ð conduction 30 Ð FabryÐPerot 431 frequency spectrum of induced Ðspecific 13 Ð MachÐZehnder 17 emission 295 Heisenberg’s uncertainty principle Ð Michelson 429, 440 Fresnel number 297 106 interferometry full-width at half-maximum HeÐNe laser 293 Ð atom 495 261 Hermitian polynomials 141 Ðneutron 100 function Heraclitus 8 Ð X-ray 16 Ð amplitude 304 hidden parameter 513 inversion 290 Ð deflection 477 Holmium laser 305 inversion tunneling 138 Ð hyperbolic 413 Hönl–London factor 373 ion 33 Ð partition 398 Humboldt, A.v. 11 Ð free 39 Ð spherical harmonic 147 hybridization 384 + ÐH2 327 Ð symmetric spin 204 Ð sp 385 Ð optics 45 Subject Index 585

Ð sources 40 Ð neodyniumm-glass- 305 magic numbers 410 Ð trapped 503 Ð pulsed 291 magnetic moment ionic contribution 339 Ð Q-switched 316 Ð nuclear 185 ionization Ð ruby flashlamp-pumped 292 magnetization 177 Ð electron impact 39 Ð semiconductor 307 magnetron Ðenergy 117 Ð single mode 301 Ð motion 64 –field 234 Ð solid-state 305 Ð movement 63 Ðthermal 40 Ð three level 293 Maiman 289, 293 ion multiplier 452 Ð Titan-Sapphire 305, 320 maser 391 ionosphere 40 Ð tunable diode 445 mass isomers 390 Ð ultrashort pulses 422 Ð absolute 53 isotope 64 lattice constant 16 Ð photon 94 isotopic abundance 64 LCAO Ð relative atomic 53 Ð improvements 334 mass absorption coefficient 275 jellium model 410 learning algorithm 501 mass spectrometer 53 Legendre’s polynomials 147 Ð quadrupole 61 kanalstrahlen 34 LenardÐJones potential 33, 268, 354, Ð time-of-flight 58 Kerr lens 320 476 mass spectrum 58 Ð mode-locking 319 lens ÐNan 60 kinetic gas theory 17 Ð converging 47 Mathieu’s differential equations 62 Kirchhoff’s diffraction theory 298 Ð electro-optic 47 matrix elements 250, 368 Kirchhoff’s law 83 Ð magnetic 50 matter wave 102 Kolos 339 leucht-electron 218, 247 Ð function 104 K-shell 209 Leucippus 7 Maxwell, J.C. 8, 9 level MaxwellÐBoltzmann distribution 265 Lamb, Willis 457 Ð vibronic 306, 407 molecule 345 mean free path length 29, 65 Lamb dip 457 Li2 lifetime 232 mean quadratic deviation 150 Lamb shift 191 Ð effective 261 measuring process 123 Landé factor 178, 183 Ð mean 260, 413 medium lanthanides 216, 218 light Ð active 289 Laplace operator 130 Ð circular polarized 254 mercury lamp 250 laser line method Ð argon 311 Ðkernel 262 Ð ab-initio 222 Ð atom 495 Ðprofile 261 Ð Hartree 222 Ð chemical 311 Ð wings 262 Ð HartreeÐFock 224 ÐCO 311 linewidth metrology ÐCO2 312 Ð color-center 305, 306 Ð natural 262, 263 Ð optical 501 Ð diode 307 liquid crystal 501 microscope Ð dye 308 LISA 510 Ð atomic force 29 Ð erbium- 305 lithium 210 Ð electron 25, 52, 102 Ðexcimer 312 Littrow-grating 312 Ð scanning electron 26 localization of the particle 106 Ð transmission electron 25 Ð fixed-frequency 304 Ð flashlamp-pumped dye 309 lone pairs 388 Ð tunneling 27 Ð free-electron 305 Lorentzian profile 263 microwave range 360 Lorentz transformation 179 mode 84 Ðgas 310 ÐHeÐNe 293 L-shell 209 Ð density 85 α Ð Holmium 305 Lyman -line 266 Ð fundamental 298 Ð of open resonator 297 Ð line width 303 Ð mode-locking 318 Mach, Ernst 7 Ð phase-coupled 319 Ð neodymium 305 macroscopic measurements 422 Ð spectral density 85 586 Subject Index

Ð TEM 298 Na D-line 266 Ð harmonic 141, 363 Ð transverse 299 neodynium-glass-laser 305 Ðmicrowave 438 model neutron 65 Ostwald, W. 7 Ð jellium 410 Ð interferometry 100 overlap integral 332 Ð semiclassical 116 Ð spectrometer 100 overtone-transitions 371 Ð shell-structure 410 NH3 molecule 390 modified Rydberg formula 221 noise equivalent input power NEP P-transition 372 modulation 433 pair formation 274 Ð frequency 438 noncrossing rule 348 para-helium 209 Ð phase 445 nonlinear absorption 456 parity 256 modulator nonlinear optic 313 Parmenides 8 Ð electro-optic 445 normalization constant 339 particle mole 12 normal vibration 399, 401 Ð free 130, 131 Ð volume 12 nuclear force 76 Ðinabox 138 molecular beam 453 nuclear g-factor 185 particle aspects of light 95 molecular dynamics 416 nuclear magneton 185 particle model 81 molecular polarizability 474 particle wave 101 molecule 11 observable 150 partition function 86 Ð acetylene 386 octupole interaction 354 PaschenÐBack effect 187 Ð aromatic 392 oil droplet 35 path Ð asymmetric rotor 399 one-dimensional box Ð elliptical 168 ÐB2 345 Ðenergy 139 Pauli principle 205 Ð butadiene 392 operator 149, 150 penetration depth 134 ÐCO2 388 Ð commutable 151 periodic system 216 Ð diatomic 327 Ð Hermitian 151 Perrin, J.B. 13 Ð formaldehyde 392 Ð of the angular momentum 152 perturbations 421 ÐH2AB 392 Ð reflection 341 phase grating 99 ÐH2 335 optic phase matching 314 ÐH2O 383 Ð electron 47 Ð angle 315 ÐHe2 345 Ð nonlinear 313 phase velocity 102 ÐLi2 345 optical amplifier 306 photo-association 495 ÐNH3 390 optical fiber bundle 452 photodiode 435 Ð polyatomic 383 + optical frequency mixing 316 photodissociation 498 Ð rigid H2 328 optical pulse photoeffect 37, 274 Ð symmetric top 397 Ð compression 321 photoelectric effect 89 moment Ð ultrashort 322 photoionization 40, 280, 281 Ð of inertia 394, 395 optical reflection coefficient 134 photomultiplier 436 Ðprincipal 395 optical trapping 491 photon 86, 93 Ð quadrupole 351 orbital Ðmass 94 Morse, P.M. 355 Ð antibonding 345 Ðspin 93 Morse potential 355, 365 Ð bonding 345 photon avalanche 291 MOT 491 Ð hybrid 384 photon number per mode 295 M-shell 209 Ð molecular 329, 331 photon recoil 191, 487 multielectron atom 221 Ð sp-hybrid 385 photosphere 40 multiplet 224 orbital angular momentum 161 π-meson 239 Ð structure 226 orbital magnetic moment 169 Planck’s constant 86 multiplicity 206, 231, 343 order Planck’s formula 249 multipole interaction 350 Ð energetic 340 Planck’s radiation law 84, 86 multipole transition 257 ortho-helium 209 planetary atom 236 myonium 241, 242 oscillator planetary model 113 Subject Index 587 plasma 40 profile Raman active 461 plasma-frequency 71 ÐFano 283 Raman effect Plato 8 Ð Gaussian line 266 Ð resonance 461 Pockels cell 317 Ð line 269 Ramsauer effect 140 point-like charge 197 Ð Lorentzian 263 random walk 73 pointing vector 82 Ðspectral 263 rate constant 412 Poisson distribution 95 Ð Voigt 267 Rayleigh 423 polarizability property RayleighÐJeans radiation law 85 Ð atomic 354 Ð optical 412 reaction Ð molecular 474 Ð symmetry 341 Ð absolute rates 415 polarization proton 65 Ð chemical 412 Ð beam splitter 317 Proust, J.L. 9 Ð constant 412 Ð dielectric 313 pump and probe experiments 417 Ð coordinate 414 polarized Ð endothermic 415 Ð circularly 93 Q-factor 296 Ð exchange 415 Ð linearly 93 Q-line 372 Ð exothermic 415 population inversion 292 quality factor 295 Ð first order 412 positronium 241 quantization axis 154, 188 Ðinverse 415 potential quantum bits 513 Ðorder 412 Ð barrier 391 quantum defect 221 Ðrate 412 Ð difference 377 quantum electrodynamics 124, 159 Ð second order 413 Ð effective 222 quantum nondemolishing experiment recombination Ð interaction 268 124 Ð radiation 284 Ð LenardÐJones 33, 268, 354, 476 quantum number 163, 330 Ð radiative 284 Ð monotonic 68 Ð angular momentum 153 Ðrate 284 ÐMorse 355, 365 Ð magnetic 153, 254, 255 Ð three-body 284 Ð parabolic 355 Ð projection 342 Ð two-body 284 Ð spherically symmetric 145 Ðspin 256 red-shadowed 376 ÐvanderWaals 352 Ð spin projection 330 reduced mass 360 potential barrier 132 quantum structure of atoms 111 reflection potential curves 333 quark 185 ÐBragg 100 potential energy qubit 514 Ð loss factor 296 Ðcurves 329 Ðlosses 290, 296 potential step 134 R-transition 372 reflection coefficient 134, 136 potential well 140 radiation reflector power Ð anti-Stokes 461 Ð cylindrical 293 Ð resolving 442 Ð blackbody 82 reflectron 60 Ð spectral resolving 423, 442 Ð cavity 83 refractive index 319 predissociation 366 Ðpower 264 relative absorption 440 pressure broadening 269 Ð recombination 284 relative atomic mass 53 principle Ðspectral 83 relative motion 159 Ð building-up 209 Ð Stokes 460 Ð of electron and nucleus 161 ÐPauli 205 Ð synchrotron 463 relativistic correction 173 probability Ðthermal 249 relativistic energy shift 173 Ð density 106, 120 radiation field relativistic mass correction 174 process Ð isotropic 253 repulsive potential 333 Ðdiffusion 22 radiative recombination 280 resolving power 442 Ð electrolytic 15 radical resonator Ð photographic 412 Ð enhancement 315 ÐNH2 389 Ð visual 463 raisin cake model 71 Ð open optical 296 product-ansatz 161 Raman, Chandrasekhara 460 Ð optical 289, 295 588 Subject Index

restoring force 400 signal-to-noise ratio 423 Ðiron 228 ring-resonator 310 silver halide grains 412 ÐofCO 439 rotation 394 single mode laser 301 Ð of our sun 285 rotational barrier 366 size Ð photoelectron 467 rotational constants 361 Ðcluster 410 spin rotational energy 362 Slater determinant 336 Ð absolute value 176 rotational term values 359 Slater orbitals 340 Ð electron 204 rotation-vibration interaction 364 solar constant 87 Ð magnetic moment 176 rotor spatial distribution spin-orbit coupling 178 Ð asymmetric 396, 399 Ð of the electron 166 Ð constant 179 Ð rigid 359 spatial resolution 109 splitting Ð symmetric 395 spectral gain profile 301 Ðinversion 391 Ð vibrating 364 spectral intensity 82 squeezing 504 Rowland arrangement 171 spectral radiation 83 stable solutions 62 Rutherford spectral resolving power 423, 442 star Ð atomic model 73 spectral response 433 Ð binary 508 Ð scattering formula 74 spectrograph Stark-effect Rydberg atom Ð parabola 54 Ð linear 473 Ð doubly excited 378 spectrometer 423 Ð quadratic 474 Rydberg constant 115, 116, 194 Ð grating 426 Stark shift 439 Rydberg formula 234 Ð infrared 440 Stark-spectroscopy 473 Rydberg state 234, 236 Ð ion-cyclotron-resonance 63 state Ðneutron 100 Ð doubly excited atomic 232 scattering Ðprism 425 Ð electronic 340, 341 Ð classical 66 spectroscopy Ð entangled 512 Ð electron 465 Ð cavity-ringdown 448 Ð excited atomic 231 Ð inelastic 478 Ð electron 465, 468 Ð excited molecular 346 Ðlosses 290 Ð excitation 451 Ð metastable 232 Ðplane 67 Ð Fourier transform 440 Ð metastable excited 264 Ð potential 67 Ð infrared 440 Ð metastable He 294 Ð rainbow 477 Ð ionization 452 Ð Rydberg 234, 236 Ð reactive 479 ÐLambdip 457 Ð singlet 205 Ð super-elastic 460 Ð laser-absorption 444 Ðtriplet 205 scattering coefficient 273 Ðmicrowave 437 statistical interpretation 105 scattering experiment 70 Ð nonlinear laser 455 statistical weight 249, 422 Schawlow, A.L. 289, 391 Ð optoacoustic 445 Stefan–Boltzmann’s law 88 SchawlowÐTownes formula 304 Ð optogalvanic 447 StefanÐBoltzmann constant 89 Schrödinger’s cat 513 Ð photoacoustic 447 stellar atmospheres 267 Schrödinger equation 129 Ð photoelectron 467 SternÐGerlach experiment 175 Ð time-dependent 130 Ð Raman 460 SternÐVollmer plot 261 secondary emission coefficient 38 Ð saturation 456 Stokes radiation 460 second harmonic 314 Ð time-resolved 497 Stokes’ law 36 selection rule 247, 253 Ð two-photon 459 storage ring DORIS 465 Ð for harmonic oscillator 371 Ð ZEKE 469 structure Ð parity 255 spectrum Ð bipyramidal 387 semiclassical model 116 Ð absorption 112 Ð rotational 406 sensitivity 422 Ð atomic 112 subshell 210 Ð absolute 433 Ð continuous 280, 377 sum frequency 316 shell-structure model 410 Ð continuous emission 285 supra-liquid 495 shielding constant 202, 216 Ð emission 112 surface short laser pulses 316 Ð helium 208 Ð equipotential 48 Subject Index 589 susceptibility Ð electronic 372 vibration period 143 Ð electric 313 Ð forbidden 252 virtual interaction 191 symmetric top 362 Ð fundamental 371 viscosity Ð molecules 397 Ð magnetic dipole 257, 259 Ðofagas 30 Ðnear 399 Ð matrix elements 367 Voigt profile 267, 455 Ð oblate 396, 397 Ð multipole 257 volume Ð prolate 396, 397 Ð overtone 405 Ð atomic 212 symmetry Ð probability 247, 250 Ðaxis 394 Ð quadrupole 257 wave Ð group 390 Ð radiationless 480 Ð gravitational 508 Ð operations 390 Ð radiative 480 wave description 81 Ð selection rules 407 Ð spontaneous 248 wave equation 130 system Ð state dynamics 498 wavefunction 163 Ð four-level 294 Ð two-photon 186, 259 Ð normalized 164 Ð π-electron 392 Ð vibrational-rotational Ð of the harmonic oscillator 142 Ðtwo-level 293 369 Ð parity 341 transport wave packet 103, 416, 497 technique Ð coefficients 29 Ð dispersion 109 Ð Doppler-free 453 trap Ð spread of the 132 temperature Ð magneto-optical 491 Ð width 108 Ð absolute 19 tunnel effect 135, 366 wavelength-tunable light source term assignment 188 two-dimensional box 144 304 term values 344 which way experiment 510 theory ultrashort optical pulse 322 Wien’s law 88 Ð nonlocal 513 ultraviolet catastrophe 85 Wigner, Eugene 348 theory of the photo effect 90 uncertainty relation for energy and work-function 37, 89 thermal dissociation time 110 ÐofH2 171 ungerade 341 X-ray 270 thermopile 434 Ð characteristic 271 Thomas factor 179 vacuum ultraviolet 171 Ð characteristic fluorescence 272 Thomson’s atomic model 71 valence method 336 Ð diffraction 15 threshold condition 290 van der Waals constant 354 Ð fluorescence 278 Titan-Sapphire laser 320 van der Waals equation 29 Ð scattering 273 Titan-Sapphire-laser 305 van der Waals interaction 353 Ð total reflection 279 TOF 58 van der Waals potential 352 Ð wavelength 278 Ð broadening 487 velocity 235 Ð design 60 Ð group 104 YAG 305 total dipole moment 404 velocity selector 476 Young’s double slit experiment 121 Townes, C.H. 289, 391 vibration transformation Ð asymmetric stretch 402 Zeeman component 169, 255 Ð Hadamar 514 Ð bending 402 Zeeman effect 168 transition Ð normal 399, 401 Ð anomalous 181 Ð bound-bound 282 Ð symmetric stretch 402 Zeeman splitting 170 Ð combination 405 vibrational constants 363 Zeiger 289