Atomic Physics Chapter 4

Chapter 4 Fine structures of Atoms

11/20/1316/11/2018 Jinniu Hu 1 !1 7.4 Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect 253

7.4 Magnetic Effects on Atomic Spectra— The Normal Zeeman Effect It was shown as early as 1896 by the Dutch physicist Pieter Zeeman that the spectral lines emitted by atoms placed in a magnetic field broaden and appear to split. The splitting of an energy level into multiple levels in the presence of an external magnetic field is called the Zeeman effect. When a spectral line is split into three lines, it is called the normal Zeeman effect. But more often a spectral line is split into more than three lines; this effect is called the anomalous Zeeman effect. The normal Zeeman effect, discussed here, can be understood by considering the atom to behave like a small magnet. We will return to our discussion of the anomalous Zeeman effect, which is more complicated, in Section 8.3. By the 1920s considerable fine structure of atomic spectral lines from hydrogen and other elements had been observed. Fine structure refers to the splitting of a spectral line into two or more closely spaced lines. Meggers AIP/Emilio Collection. Segrè Visual F. Archives, W. As a rough model, think of an electron circulating around the nucleus as a circular current loop. The current loop has a magnetic moment m ϭ IA where A The Dutch physicist Pieter is the area of the current loop and the current I ϭ dq/dt is simply the electron Zeeman (1865– 1943) studied at ϭ Ϫ the University of Leiden under the charge (q e) divided by the period T for the electron to make one revolution famous physicists H. Kamerlingh (T ϭ 2pr/v). Thus Onnes and H. A. Lorentz and 2 received his degree in 1890. While q Ϫe pr Ϫe r v e at Leiden he showed that atomic m ϭ IA ϭ A ϭ ϭ ϭϪ L (7.25) T 2pr /v 2 2m spectral lines were split under the 1 2 influence of an applied magnetic where L ϭ mvr is the magnitude of the orbital angular momentum. Both the field. After this discovery he left magnetic moment m and angular momentum L are vectors so that Leiden in 1897 to go to the Univer- sity of Amsterdam, where he re- e mained until 1935. He shared the m ϭϪ L (7.26) 1902 Nobel Prize for Physics with 2m his mentor Lorentz.

The relationship between m and L is displayed in Figure 7.4. In the absence of an external magnetic field to align4.1Magnetic them, the magneticmoment of electron moments m of atoms point in random directions. In classical electromagnetism, As a rough model, think of an electron circulating around if a magnetic dipole having a magnetic moment m is placed in an external magnetic field, the dipole will experience a torque t ϭ m ϫthe B tendingnucleus asto aligna circular the current loop. The current loop has a magnetic moment dipole with the magnetic field. The dipole also has a potential energy VB in the L field given by q µ = IA = A T VB ϭϪm # B (7.27) e⇡r2 erv = = Proton According to classical physics, if the system can change its potential2⇡r/v energy,2 the magnetic moment will align itself with the external magnetic fielde to minimize L e ؊ = energy. 2m m Note the similarity with the case of the spinning top in a gravitational field. A child’s spinning top is said to precess about the gravitationalwhere L=mvrfield; thatis the is, magnitudethe of the orbital angular Figure 7.4 Representation of axis (around which the spinning top is rotating) itself rotatesmomentum. about the direction the orbital angular momentum L of the gravitational force (vertical). The gravitational field is not parallel to the and magnetic moment m of the 16/11/2018 Jinniu Hu angular momentum, and the force of gravity pulling down on the spinning top hydrogen atom due to the elec- results in a precession of the top about the field direction. Precisely the same thing tron orbiting the proton. The di- happens with the magnetic moment of an atom in a magnetic field. The angular rections of L and m are opposite momentum is aligned with the magnetic moment, and the torque between m because of the negative electron and B causes a precession of m about the magnetic field (see Figure 7.5), not an charge.

03721_ch07_241-271.indd 253 9/29/11 4:44 PM Magnetic moment of electron

Both the magnetic moment and angular momentum are vectors so that e µ~ = L~ 2m Gyromagnetic ratio e = 2m In classical electromagnetism, if a magnetic dipole having a magnetic moment, is placed in an external magnetic

ﬁeld, the dipole will experience a torque ~⌧ = µ~ B~ ⇥ which aligns the dipole with the magnetic ﬁeld.

16/11/2018 Jinniu Hu Magnetic moment of electron

The dipole also has a potential energy in the ﬁeld given by V = µ~ B~ B · According to classical physics, if the system can change its potential energy, the magnetic moment will align itself with the external magnetic ﬁeld to minimize energy.

Precession in gravitational ﬁeld

16/11/2018 Jinniu Hu 254 Chapter 7 The Hydrogen Atom alignment. The magnetic ﬁeld establishes a preferred direction in space along which we customarily deﬁne the z axis. Then we have

e U m ϭ m ϭϪm m (7.28) z 2m / B /

Bohr magneton where mB ϭ e U/2m is a unit of magnetic moment called a Bohr magneton. Be- 1 cause of the quantization of Lz and the fact that L ϭ / / ϩ 1 U Ͼ m/ U, we cannot have Έm Έ ϭ m ; the magnetic moment cannot align itself exactly in the z z 1 2 direction. Just like the angular momentum L, the magnetic moment m has only certain allowed quantized orientations. Note also that in terms of the Bohr

magneton, m ϭϪmBL / U.

Magnetic moment of electron EXAMPLE 7.5 An atom with magnetic moment in an external magnetic field has Larmor precession. Determine the precessional frequency of an atom having B z magnetic moment m in an external magnetic field B. This precession is known as the Larmor precessionPrecessional. frequency m u d 1 dL ! = = Strategy We have already seen that the torqueL t is dtequal L sin ✓ dt dL = L sin ✓d to m ϫ B, but we also know from classical mechanics that the torque is / . The torque in Figure 7.5 is perpendicular to dL dt Newton’s second law in angular form u m, L, and B and is out of the page. This must also be the dL~ L direction of the change in momentum dL as seen in Figure~⌧ = 7.5. Thus L and m precess about the magnetic field. The dt Larmor frequency vL is given by df/dt. So L sin u f e eB ! = µB sin ✓ = dL df L 2mµ sin ✓ 2m Solution The magnitude of dL is given by L sin ✓u df (see ◆ Figure 7.5), so vL is given by 16/11/2018 Figure 7.5Jinniu An Hu atom having magnetic moment m feels a torque df 1 dL t ϭ m ϫ B due to an external magnetic field B. This torque must v ϭ ϭ (7.29) L dt L sin u dt also be equal to dL/dt. The vectors m and L are antiparallel, so the vector dL/dt must be perpendicular to m, B, and L. As shown in We now insert the magnitude of L ϭ 2mm/e from Equation the figure, dL/dt requires both m and L to precess (angle f) about (7.26). The value of dL/dt, the magnitude of m ϫ B, can be the magnetic field B. determined from Figure 7.5 to be mB sin u. Equation (7.29) becomes e eB vL ϭ mB sin u ϭ (7.30) 2m m sin u 2m a b

What about the energy of the orbiting electron in a magnetic ﬁeld? It takes work to rotate the magnetic moment away from B. With B along the z direction, we have from Equation (7.27)

VB ϭϪmzB ϭϩmBm/B (7.31) The potential energy is thus quantized according to the magnetic quantum number m/; each (degenerate) atomic level of given / is split into 2/ ϩ 1 different energy states according to the value of m/. The energy degeneracy of a given n/ level is

03721_ch07_241-271.indd 254 9/29/11 4:44 PM Magnetic moment of electron

Orbit angular momentum operator Lˆ =ˆr pˆ ⇥ The components of orbit angular moment ˆ @ @ Lx =ˆypˆz zˆpˆy = i~ y z @z @y ✓ ◆ ˆ @ @ Ly =ˆzpˆx xˆpˆz = i~ z x @x @z ✓ ◆ ˆ @ @ Lz =ˆxpˆy yˆpˆx = i~ x y @y @x ✓ ◆ The square of orbit angular momentum operator ˆ2 ˆ2 ˆ2 ˆ2 L = Lx + Ly + Lz

16/11/2018 Jinniu Hu Magnetic moment of electron

Eigenstate and eigenvalue Qfˆ = f f is the eigenstate and ! is the corresponding eigenvalue.

The eigenstate and eigenvalue of orbit angular moment ˆ2 2 2 L Ylm = L Ylm = l(l + 1)~ Ylm ˆ LzYlm = ml~Ylm

l is the quantum number of orbit angular momentum

ml is the magnetic quantum number

16/11/2018 Jinniu Hu 7.3 Quantum Numbers 251

Magnetic Quantum Number mO The orbital angular momentum quantum number / determines the magnitude of the angular momentum L, but because L is a vector, it also has a direction. Classically, because there is no torque in the hydrogen atom system in the absence of external ﬁelds, the angular momentum L is a constant of the motion and is conserved. The solution to the Schrödinger equation for f(u) speciﬁed that / must be an integer, and therefore the magnitude of L is quantized. The angle f is a measure of the rotation about the z axis. The solution for

g(f) speciﬁes that m/ is an integer and related to the z component of the angular momentum L.

Lz ϭ m/ U (7.23)

The relationship of L, Lz, /, and m/ is displayed in Figure 7.3 for the value / ϭ 2. The magnitude of L is fixed L ϭ 1/ / ϩ 1 U ϭ 16U . Because Lz is quantized, only certain orientations of L are possible, each corresponding to a different m 3 1 2 4 / (and therefore Lz). This phenomenon is called space quantization because only Space quantization certain orientations of L are allowed in space. We can ask whether we have established a preferred direction in space by choosing the z axis. The choice of the z axis is completely arbitrary unless there is an external magnetic field to define a preferred direction in space. It is cus- tomary to choose the z axis to be along B if there is a magnetic field. This is why

m/ is called the magnetic quantum number. Will the angular momentum be quantized along the x and y axes as well? The answer is that quantum mechanics allows L to be quantized along only one direction in space. Because we know the magnitude of L, the knowledge of a second component would imply a knowledge of the third component as well 2 2 2 2 Magnetic momentbecause of of electron the relation L ϭ Lx ϩ Ly ϩ Lz . The following argument shows that this would violate the Heisenberg uncertainty principle: If all three components n =1, 2, of3, L4 ,...were known, then the direction of LInteger would also be known. In this case we would have a precise knowledge of one component of the electron’s position in l =0, 1, space,2, 3,...,n because the1 electron’s orbital motion Integer is conﬁned to a plane perpendicular to L. But conﬁnement of the electron to that plane means that the electron’s m = l, l +1,...,0, 1,...,l 1,l Integer l

L z

2ប m ᐉ ϭ 2

ប m ᐉ ϭ 1 L ϭ w ᐉ(ᐉ ϩ1)ប 0 ϭ w 6ប m ᐉ ϭ 0

Ϫប m ᐉ ϭ Ϫ1

Figure 7.3 Schematic diagram Ϫ2ប m ᐉ ϭ Ϫ2 of the relationship between L and

Lz with the allowed values of m/. 16/11/2018 Jinniu Hu

03721_ch07_241-271.indd 251 9/29/11 4:44 PM Magnetic moment of electron

Magnetic moment operator e ˆ µ~ˆ = L~ 2m e µˆ = Lˆ z 2m z The eigenvalue of magnetic moment e~ µ = l(l + 1) l 2m e~ p µ = m z 2m l Bohr magneton e~ µ = B 2m

The energy of the orbiting electron in a magnetic ﬁeld E = µ B = µ m B B z B l 16/11/2018 Jinniu Hu 4.2Stern-Gerlach experiment

In 1922 O. Stern and W. Gerlach reported the results of an experiment that clearly showed evidence for space

quantization. 7.4 Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect 257

Screen m South mᐉ ϭ Ϫ1

mᐉ ϭ 0 p-state! m ϭ ϩ1 North ᐉ Atomic! atoms beam! oven

Figure 7.8 Schematic diagram of expected result of Stern-Gerlach experiment if atoms in a p state are used.16/11/2018 Three patterns of atoms, dueJinniu to m/ Huϭ Ϯ1, 0, are expected to be observed on the screen. The magnet poles are arranged to produce a magnetic ﬁeld gradient as shown in Figure 7.7. The experiment performed by Stern and Gerlach reported only two lines, not three (see Sec- tion 7.5).

If we pass an atomic beam of particles in the / ϭ 1 state through a magnetic field along the z direction, we have from Equation (7.31) that VB ϭ ϪmzB, and the force on the particles is Fz ϭ Ϫ(dVB/dz) ϭ mz(dB/dz). There will be a different force on atoms in each of the three possible m/ states. A schematic diagram of the Stern-Gerlach experiment is shown in Figure 7.8. The m/ ϭ ϩ1 state will be deflected down, the m/ ϭ Ϫ1 state up, and the m/ ϭ 0 state will be undeflected. Stern and Gerlach performed their experiment with silver atoms and observed two distinct lines, not three. This was clear evidence of space quantization, although the number of m/ states is always odd (2/ ϩ 1) and should have produced an odd number of lines if the space quantization were due to the magnetic quantum number m/.

EXAMPLE 7.8

In 1927 T. E. Phipps and J. B. Taylor of the University of the constant speed of the atom within the ﬁeld. The separa- Illinois reported an important experiment similar to the tion d is therefore found from

Stern-Gerlach experiment but using hydrogen atoms in- 2 1 2 1 Fz 2 1 dB ¢x stead of silver. This was done because hydrogen is the sim- d ϭ a zt ϭ t ϭ mz plest atom, and the separation of the atomic beam in the 2 2 m 2m dz vx inhomogeneous magnetic ﬁeld would allow a clearer inter- We know all the valuesa neededb to determinea ba d exceptb the pretation. The atomic hydrogen beam was produced in a speed vx, but we do know the temperature of the hydrogen discharge tube having a temperature of 663 K. The highly gas. The average energy of the atoms collimated along the x collimated beam passed along the x direction through an 1 2 3 direction is 2 m v x ϭ 2 kT. inhomogeneous ﬁeld (of length 3 cm) having an average gradient of 1240 T/m along the z direction. If the magnetic 8 9 moment of the hydrogen atom is 1 Bohr magneton, what is 2 the separation of the atomic beam? Solution We calculate v x to be Ϫ23 3kT 3 81.389 ϫ 10 J/K 663 K Strategy The force can be found from the potential en- v 2 ϭ ϭ x m ϫ Ϫ27 ergy of Equation (7.31). 1 1.67 10 kg21 2 7 2 2 dV dB ϭ 1.64 ϫ 10 m /s F ϭϪ ϭ m z dz z dz The separation d of the one atom is now determined to be

The acceleration of the hydrogen atom along the z direction 1 Ϫ24 ϭ ϫ d Ϫ27 9.27 10 J/T 1240 T/m is az ϭ Fz/m. The separation of the atom along the z direc- 2 1.67 ϫ 10 kg 2 tion due to this acceleration is d ϭ azt /2. The time that the 1 21 2 0.03m 2 atom spends within the inhomogeneous ﬁeld is t ϭ ⌬x/vx 1 2 Ϫ3 ϫ ϭ ϫ 7 2 2 0.19 10 m ⌬ ϫ where x is the length of the inhomogeneous ﬁeld, and vx is 1.64 1 10 m2 /s 1 2

03721_ch07_241-271.indd 257 9/29/11 4:44 PM Stern-Gerlach experiment

16/11/2018 Jinniu Hu 256 Chapter 7 The Hydrogen Atom

The international system of units has been used (T ϭ tesla where ⌬m/ ϭ 1 Ϫ 0 ϭ 1. Hence, we have for magnetic ﬁeld). The energy splitting is determined from E ϭ 9.27 ϫ 10Ϫ24 J/T 2.00 T ϭ 1.85 ϫ 10Ϫ23 J Equation (7.31) (see also Figure 7.6a): ¢ Ϫ4 1 21 2 ϭ 1.16 ϫ 10 eV ¢E ϭ mBB ¢m/ (7.33) An energy difference of 10Ϫ4 eV is easily observed by optical means.

will see in Section 7.6 that the selection rule for m/ does not allow more than AIP/Emilio Segrè Visual Archives, Segrè Collection. three different spectral lines in the normal Zeeman effect (see Problem 33). Efforts were begun in the 1920s to detect the effects of space quantization (m ) Otto Stern (1888– 1969) was / born in a part of Germany (now by measuring the energy difference ⌬E as in Example 7.7. In 1922 O. Stern and W. in Poland), where he was edu- Gerlach reported the results of an experiment that clearly showed evidence for cated. He worked in several uni- space quantization. If an external magnetic field is inhomogeneous—for example, Stern-Gerlachversities until he left Germany experiment in if it is stronger at the south pole than at the north pole—then there will be a net 1933 to avoid Nazi persecution force on a magnet placed in the field as well as a torque. This force is represented and emigrated to the United in Figure 7.7, where the net force on m (direction of S to N in bar magnet) is dif- States. He was educated and ferent for different orientations of m in the inhomogeneous magnetic field B. If antrained external as a theorist but magneticchanged field is inhomogeneous, the force to experimentation when he be- on thegan his particlesmolecular beam experi- is ments in 1920 at the University of Frankfurt with Walter Gerlach. He continued his dEdistinguishedB ca- dB Freerz =in Hamburg and later= at µz South Carnegie Institute dz of Technology dz B in Pittsburgh. He received the No- greater Whenbel Prize l=1 in 1943. Net! N S force 7.4 MagneticµFigurez =Effects 7.7 An1 on ,inhomogeneous Atomic0, 1µ Spectra—TheB Normal ZeemanS Effect 257N magnetic field is created by the Net! smaller south pole. Two bar mag- force B lower nets representingScreen atomic magnetic moments have m in opposite m North South directions. Becausemᐉ the ϭ forceϪ1 on the top of the bar magnets is greater than thatm onᐉ ϭthe 0 bottom, there will be a net translational p-state! force on the bar mmagnets ϭ ϩ 1 North ᐉ Atomic! atoms (atoms). beam! oven 16/11/2018 Jinniu Hu

Figure 7.8 Schematic diagram of expected result of Stern-Gerlach experiment if atoms in a p 03721_ch07_241-271.indd 256 9/29/11 4:44 PM state are used. Three patterns of atoms, due to m/ ϭ Ϯ1, 0, are expected to be observed on the screen. The magnet poles are arranged to produce a magnetic ﬁeld gradient as shown in Figure 7.7. The experiment performed by Stern and Gerlach reported only two lines, not three (see Sec- tion 7.5).

EXAMPLE 7.8

03721_ch07_241-271.indd 257 9/29/11 4:44 PM bei48482_Ch07.qxd 1/23/02 9:02 AM Page 232

Stern-Gerlach experiment 232 Chapter Seven

Oven

Magnet pole

Beam of silver atoms N Inhomogeneous magnetic

S Field off Classical Magnet pole pattern Field on Actual Photographic pattern plate

Figure 7.3 The Stern-Gerlach experiment.

16/11/2018 Jinniu Hu

The Stern-Gerlach Experiment

pace quantization was first explictly demonstrated in 1921 by Otto Stern and Walter Gerlach. S They directed a beam of neutral silver atoms from an oven through a set of collimating slits into an inhomogeneous magnetic field as in Fig. 7.3. A photographic plate recorded the shape of the beam after it had passed through the field. In its normal state the entire magnetic moment of a silver atom is due to the spin of only one of its electrons. In a uniform magnetic field, such a dipole would merely experience a torque tending to align it with the field. In an inhomogeneous field, however, each “pole” of the dipole is subject to a force of different magnitude and therefore there is a resultant force on the dipole that varies with its orientation relative to the field. Classically, all orientations should be present in a beam of atoms. The result would merely be a broad trace on the photographic plate instead of the thin line formed without any magnetic field. Stern and Gerlach found, however, that the initial beam split into two distinct parts that correspond to the two opposite spin orientations in the magnetic field permitted by space quantization.

An example is the great difference in chemical behavior shown by certain elements whose atomic structures differ by only one electron. Thus the elements that have the atomic numbers 9, 10, and 11 are respectively the chemically active halogen gas flu- orine, the inert gas neon, and the alkali metal sodium. Since the electron structure of an atom controls how it interacts with other atoms, it makes no sense that the chemical properties of the elements should change so sharply with a small change in atomic number if all the electrons in an atom were in the same quantum state. Stern-Gerlach experiment

Stern and Gerlach performed their experiment with silver atoms and observed two distinct lines, not three.

This was clear evidence of space quantization, although

the number of ml states is always odd and should have produced an odd number of lines if the space quantization

were due to the magnetic quantum number ml. 16/11/2018 Jinniu Hu 4.3 Intrinsic Spin

In 1925 Samuel Goudsmit and George Uhlenbeck, two young physics graduate students in Holland, proposed that the electron must have an intrinsic angular momentum and therefore a magnetic moment

16/11/2018 Jinniu Hu Intrinsic Spin

In order to achieve the angular momentum needed, Paul Ehrenfest showed that the surface of the spinning electron (or electron cloud) would have to be moving at a velocity greater than the speed of light!

If such an intrinsic angular momentum exists, we must regard it as a purely quantum-mechanical result.

To explain experimental data, Goudsmit and Uhlenbeck proposed that the electron must have an intrinsic spin quantum number s =1/2 16/11/2018 Jinniu Hu 258 Chapter 7 The Hydrogen Atom

Phipps and Taylor found only two distinct lines, as did Stern we just calculated! The total separation of the two lines (one and Gerlach for silver atoms, and the separation of the lines deﬂected up and one down) was 0.38 mm. The mystery re- from the central ray with no magnetic ﬁeld was 0.19 mm as mained as to why there were only two lines.

7.5 Intrinsic Spin It was clear by the early 1920s that there was a problem with space quantization and the number of lines observed in the Stern-Gerlach experiment. Wolfgang

Pauli was the first to suggest that a fourth quantum number (after n, /, m/) assigned to the electron might explain the anomalous optical spectra discussed in Section 7.4. His reasoning for four quantum numbers was based on relativity, in which there are four coordinates—three space and one time. The physical significance of this fourth quantum number was not made clear. In 1925 Samuel Goudsmit and George Uhlenbeck, two young physics graduate students in Holland, proposed that the electron must have an intrinsic angular momentum and therefore a magnetic moment (because the electron is charged). Classically, this corresponds in the planetary model to the fact that the Earth rotates on its own axis as it orbits the sun. However, this simple classical picture runs into serious difficulties when applied to the spinning charged electron. In order to achieve the angular momentum needed, Paul Ehrenfest showed that the surface of the spinning electron (or electron cloud) would have to be moving at a velocity greater than the speed of light! If such an intrinsic angular momentum exists, we must regard it as a purely quantum-mechanical result (see Problems 44 and 45). To explain experimental data, Goudsmit and Uhlenbeck proposed that the Intrinsic spin electron must have an intrinsic spin quantum number s ϭ 1/2. The spinning quantum number electron reacts similarly to the orbiting electron in a magnetic field. Therefore, we should try to find quantities analogous to the angular momentum variable L,

Lz, /, and m/. By analogy, there will be 2s ϩ 1 ϭ 2(1/2) ϩ 1 ϭ 2 components of Magnetic spin the spin angular momentum vector S. Thus the magnetic spin quantum number quantum number ms has only two values, ms ϭ Ϯ1/2. The electron’s spin will be oriented either “up” or “down” in a magnetic ﬁeld (see Figure 7.9), and the electron can never be spinningIntrinsic with itsSpin magnetic moment ms exactly along the z axis (the direction of the external magnetic ﬁeld B).

For each atomic state described by the three quantum numbers (n, /, m/) The magnetic spin quantum number ms has only two values discussed previously, there are now two distinct states, one with ms ϭ ϩ1/2 and one ms = 1/2 with ms ϭ Ϫ1/2. These states are degenerate± in energy unless the atom is in an externalThe magnetic electron’s field. spin In a willmagnetic be orientedfield these stateseither will “up” have ordifferent “down” energies in due ato magnetican energy separationfield, and like the that electron of Equation can (7.33). never We be say spinning the splitting of these energy levels by the magnetic field has removed the energy degeneracy.

with its magnetic moment ms exactly alongz the z axis

1! m ϭ ϩ s 2 Figure 7.9 (a) A purely classical 3! S ϭ ប schematic of the intrinsic spin w 4 angular momentum, S, of a spinning electron. (b) The quantiza- S tion of S, which can have only two 1! positions in space relative to z m ϭ Ϫ s 2 (direction of external magnetic ﬁeld). The z component of S is

Sz ϭ ϮU/2. (a) (b) 16/11/2018 Jinniu Hu

03721_ch07_241-271.indd 258 9/29/11 4:44 PM Intrinsic Spin

The intrinsic spin angular momentum

S~ = s(s + 1)~ = 3/4~ | | The magnetic momentp of intrinsic spinp ~ ~ µ~ s = gsµBS/~ = 2µBS/~ where gs is the Lande factor and

gs =2 The z component of spin magnetic moment

µs,z = 2msµB where m = 1/2 s ± 16/11/2018 Jinniu Hu 4.4 Total angular momentum of single electron If an atom has an orbital angular momentum and a spin

angular momentum, these angular momenta282 combine Chapter 8to Atomic Physics produce a total angular momentum, ~ ~ ~ Because L, L , S, and S are quantized, the total angular momentum and its J = L + S S z z z component Jz are also quantized. If j and mj are the appropriate quantum num- Because L, Lz, S, and Sz are quantized, the bers for the single electron we are considering, quantized values of J and Jz are, total angular momentum and its z component in analogy with the single electron of the hydrogen atom, J

Jz are also quantized J ϭ 2j j ϩ 1 (8.4a) L U J = j(j + 1) ~ Jz ϭ1mj U 2 (8.4b) J = m z pj~ Because m/ is integral and ms is half-integral, mj will always be half-integral. Just as the value of m ranges from Ϫ/ to /, the value of mj ranges from Ϫj to j, and Because ml is integral and ms is half-integral,j ϭm ᐉj ϩ will s always / therefore j will be half-integral. be half-integral. ϭ 1 ϩ 1! ϭ 3! 2 2 The quantization of the magnitudes of L, S, and J are all similar. 16/11/2018 Jinniu Hu L ϭ 2/ / ϩ 1 U 2 S ϭ s 1s ϩ 1 2U (8.5)

S J ϭ 2j 1 j ϩ 1 2U

L The total angular momentum quantum 1number2 for the single electron can only have the values J j ϭ / Ϯ s (8.6)

j ϭ ᐉ Ϫ s which, because s ϭ 1/2, can only be / ϩ 1/2 or / Ϫ 1/2 (but j must be 1/2 if ϭ Ϫ 1! ϭ 1! 1 2 2 / ϭ 0). The relationships of J, L, and S are shown in Figure 8.5. For an / value of 1, the quantum number j is 3/2 or 1/2, depending on whether L and S are Figure 8.5 When forming the aligned or antialigned. The notation commonly used to describe these states is total angular momentum from nL (8.7) the orbital and spin angular mo- j menta, the addition must be done where n is the principal quantum number, j is the total angular momentum vectorially, J ϭ L ϩ S. We show quantum number, and L is an uppercase letter (S, P, D, and so on) representing schematically the addition of L the orbital angular momentum quantum number. and S with / ϭ 1 and s ϭ 1/2 to In Section 7.5 we brieﬂy mentioned that the single electron of the hydrogen form vectors J with quantum atom can feel an internal magnetic ﬁeld Binternal due to the proton, because in numbers j ϭ 1/2 and 3/2. the rest system of the electron, the proton appears to be circling the electron (see Figure 7.10). A careful examination of this effect shows that the spins of the Spin-orbit coupling electron and the orbital angular momentum interact, an effect called spin-orbit

coupling. As usual the dipole potential energy Vs/ is equal to Ϫ ms # Binternal. The spin magnetic moment is proportional to ϪS, and B internal is proportional to L, so that Vs/ ϳ S # L ϭ SL cos a, where a is the angle between S and L. The result of this effect is to make the states with j ϭ / Ϫ 1/2 slightly lower in energy than for j ϭ / ϩ 1/2, because a is smaller when j ϭ / ϩ 1/2. The same applies for the atom when placed in an external magnetic ﬁeld. The same effect leads us to

accept j and mj as better quantum numbers than m/ and ms, even for single- electron atoms like hydrogen. We mean “better” because j and mj are more directly related to a physical observable. A given state having a definite energy can no longer be assigned a definite Lz and Sz, but it can have a definite Jz. The wave functions now depend on n, /, j, and mj. The spin-orbit interaction splits the 2P level into two states, 2P3/2 and 2P1/2, with 2P1/2 being lower in energy. There are additional relativistic effects, not discussed here, that give corrections to the spin-orbit effect. In the absence of an external magnetic field, the total angular momentum has a fixed magnitude and a fixed z component. Remember that only Jz can be known; the uncertainty principle forbids Jx or Jy from being known at the same time as Jz.

03721_ch08_272-297.indd 282 9/29/11 10:12 AM Total angular momentum of 8.2 Total Angular Momentum 283 single electron z

Bext Bext The quantization of the magnitudes of various angular Jare ϭ L ϩ S all similar, mjប J L = l(l + 1)~

Figure 8.6 (a) The vectors L S = s(s + 1) and S precess around J. The total p ~ L angular momentum J can have S a ﬁxed value in only one direc- J = pj(j + 1)~ L tion in space—not shown in this ﬁgure. (b) However, with an

The total angular momentum quantum number for S the external magnetic field Bext along p the z axis, J will precess around single electron can only have the values, the z direction ( Jz is fixed), and both L and S precess around J. We j = l s have shown the case where L and ± (a) (b) S are aligned. 1 = l The vectors L and S will precess around J (see Figure 8.6a). In an external mag- ± 2 netic field, however, J will precess about Bext, while L and S still precess about J as The notation commonly used to describe these shownstates in Figure 8.6b. The motion of L and S then becomes quite complicated. Optical spectra are due to transitions between different energy levels. We 2S+1 have already discussed transitions for the hydrogen atom in Section 7.6 and gave LJ L =0, 1the, rules2, 3listed,... in Equation (7.36). For single-electron atoms, we now add the selection rules for ⌬j. The restriction of ⌬/ ϭ Ϯ1 will require ⌬j ϭ Ϯ1 or 0. The = S,allowed P, D, transitions F, .for . a . single-electron atom are ¢n ϭ anything ¢/ ϭϮ1 16/11/2018 Jinniu Hu (8.8) Single-electron atom ¢mj ϭ 0, Ϯ1 ¢j ϭ 0, Ϯ1 allowed transitions

The selection rule for ⌬mj follows from our results for ⌬m/ in Equation (7.36) and from the result that mj ϭ m/ ϩ ms, where ms is not affected. Figure 7.11 presented an energy-level diagram for hydrogen showing many possible transitions. Figure 8.7 is a highly exaggerated portion of the hydrogen energy-level diagram for n ϭ 2 and n ϭ 3 levels showing the spin-orbit splitting. All of the states (except for the s states) are split into doublets. What appeared

3D 3P3/2 5/2 3P1/2 nϭ3 3S 1/2 Figure 8.7 3D 3/2 (a) The unperturbed Ha line is shown due to a Energy Ha transition between the n ϭ 3 and n ϭ 2 shells of the hydrogen 2P3/2 atom. (b) The more detailed level nϭ2 2S 1/2 2P1/2 structure (not to scale) of the hydrogen atom leads to optical ﬁne Unperturbed Fine structure structure. The spin-orbit interac- (a) (b) tion splits each of the / ϶ 0 states.

03721_ch08_272-297.indd 283 9/29/11 10:12 AM 262 Chapter 7 The HydrogenTotal Atomangular momentum of single electron Energy! S! P! D! F! G! (eV) n! ᐉ ϭ 0 1 2 3 4 0 ∞ Ϫ0.8 4 Ϫ1.5 3 F series S series D series Ϫ3.4 2

P series

Figure 7.11 Energy-level diagram of hydrogen atom with no external magnetic ﬁeld. Also shown are allowed photon transitions between some levels. Ϫ13.6 1 16/11/2018 Jinniu Hu E(4S) Ͻ E(4P) Ͻ E(4D) Ͻ E(4F ). For hydrogen, the energy levels depend only on the principal quantum number n and are predicted with great accuracy by the Bohr theory. We have previously learned that atoms emit characteristic electromagnetic radiation when they make transitions to states of lower energy. An atom in its ground state cannot emit radiation; it can absorb electromagnetic radiation, or it can gain energy through inelastic bombardment by particles, especially electrons. The atom will then have one or more of its electrons transferred to a higher energy state.

Selection Rules We can use the wave functions obtained from the solution of the Schrödinger equation to calculate transition probabilities for the electron to change from one state to another. The results of such calculations show that electrons absorbing or emitting photons are much more likely to change states when ⌬/ ϭ Ϯ1. Such Allowed and transitions are called allowed. Other transitions, with ⌬/ % Ϯ1, are theoretically forbidden transitions possible but occur with much smaller probabilities and are called forbidden transitions. There is no selection rule restricting the change ⌬n of the principal quan-

tum number. The selection rule for the magnetic quantum number is ⌬m/ ϭ 0, Ϯ1. The magnetic spin quantum number ms can (but need not) change between 1/2 and Ϫ1/2. We summarize the selection rules for allowed transitions: ¢n ϭ anything Selection rules ¢/ ϭϮ1 (7.36)

¢m/ ϭ 0, Ϯ1 Some allowed transitions are diagrammed in Figure 7.11. Notice that there are no transitions shown for 3P S 2P, 3D S 2S, and 3S S 1S because those transitions violate the ⌬/ ϭ Ϯ1 selection rule. If the orbital angular momentum of the atom changes by U when absorption or emission of radiation takes place, we must still check that all conservation laws

03721_ch07_241-271.indd 262 9/29/11 4:44 PM Total magnetic moment 8.2 Total Angular Momentum 283

The total magnetic moment is, z

µ~ j = µ~ l + µ~ s, Bext Bext J ϭ L ϩ S ~ = gjµBJ/~ mjប J and,

µj,z = gjmjµB Figure 8.6 (a) The vectors L where, gj is the Lande factor of and S precess around J. The total L total angular momentum and angular momentum J can have S a fixed value in only one direc- 3 S(S + 1) L(L + 1) L tion in space—not shown in this gj = + figure. (b) However, with an S 2 2J(J + 1) external magnetic field Bext along the z axis, J will precess around 3 Sˆ2 Lˆ2 = + the z direction ( Jz is fixed), and 2 2Jˆ2 both L and S precess around J. We have shown the case where L and 16/11/2018 Jinniu Hu (a) (b) S are aligned.

The vectors L and S will precess around J (see Figure 8.6a). In an external magnetic ﬁeld, however, J will precess about Bext, while L and S still precess about J as shown in Figure 8.6b. The motion of L and S then becomes quite complicated. Optical spectra are due to transitions between different energy levels. We have already discussed transitions for the hydrogen atom in Section 7.6 and gave the rules listed in Equation (7.36). For single-electron atoms, we now add the selection rules for ⌬j. The restriction of ⌬/ ϭ Ϯ1 will require ⌬j ϭ Ϯ1 or 0. The allowed transitions for a single-electron atom are ¢n ϭ anything ¢/ ϭϮ1 (8.8) Single-electron atom ¢mj ϭ 0, Ϯ1 ¢j ϭ 0, Ϯ1 allowed transitions

03721_ch08_272-297.indd 283 9/29/11 10:12 AM Total magnetic moment

Lande factor of different state

States gj mjgj

2S1/2 2 ±1

2P1/2 2/3 ±1/3

2P3/2 4/3 ±2/3, ±6/3

2D3/2 4/5 ±2/5, ±6/5

2D5/2 6/5 ±3/5, ±9/5, ±15/5

16/11/2018 Jinniu Hu 4.5 Spin-orbit coupling

16/11/2018 Jinniu Hu bei48482_Ch07.qxd 1/23/02 9:02 AM Page 248 Spin-orbit coupling

This classical picture indicates that the orbiting proton

creates a magnetic ﬁeld at the position of the electron. 248 Chapter Seven

+ Ze – e

(a)(b)

Figure 7.13 (a) An electron circles an atomic nucleus, as viewed from the frame of reference of the nucleus. (b) From the electron’s frame of reference, the nucleus is circling it. The magnetic field the electron experiences16/11/2018 as a result is directed upwardJinniu from Hu the plane of the orbit. The interaction between the electron’s spin magnetic moment and this magnetic field leads to the phenomenon of spin-orbit coupling.

The potential energy Um of a magnetic dipole of moment ␮ in a magnetic field B is, as we know,

Um ϭϪ␮B cos ␪ (6.38) where ␪ is the angle between ␮ and B. The quantity ␮ cos ␪ is the component of ␮ parallel to B. In the case of the spin magnetic moment of the electron this component is ␮sz ϭϮ␮B. Hence

␮ cos ␪ ϭϮ␮B and so

Spin-orbit coupling Um ϭϮ␮B B (7.15)

Depending on the orientation of its spin vector S, the energy of an atomic electron will be higher or lower by ␮B B than its energy without spin-orbit coupling. The result is that every quantum state (except s states in which there is no orbital angular momentum) is split into two substates. 1 ᎏᎏ The assignment of s ϭ 2 is the only one that agrees with the observed fine-structure doubling. Because what would be single states without spin are in fact twin states, the 2s ϩ 1 possible orientations of the spin vector S must total 2. With 2s ϩ 1 ϭ 2, the 1 ᎏᎏ result is s ϭ 2.

Example 7.3

Estimate the magnetic energy Um for an electron in the 2p state of a hydrogen atom using the Bohr model, whose n ϭ 2 state corresponds to the 2p state. Solution A circular wire loop of radius r that carries the current I has a magnetic field at its center of magnitude ␮ I B ϭ ᎏ0 2r Spin-orbit coupling An electron moving through an electric field E experiences an effective magnetic field B given by 1 B~ = ~v E~ c2 ⇥ Since, the electric field is 1 @V ~r E~ = We have e @r r ~ 1 1 @V ~ B = 2 L mec er @r The potential of the electron’s magnetic moment with the

orbital ﬁeld gives U = µ~ B~ s · 1 @V ~ ~ = gsµB 2 L S mec er @r · 16/11/2018 Jinniu Hu Spin-orbit coupling

With Coulomb potential and relativistic correction that arises because we are calculating the magnetic ﬁeld in a frame of reference that is not stationary but rotates as the electron moves about the nucleus, 2 1 Ze ~ ~ U = 2 2 3 L S 4⇡"0 2mec r · The expectation value of this potential gives an energy change of ,

1 1 ~ ~ Eso = 2 2 3 L S 8⇡"0mec hr ih · i

16/11/2018 Jinniu Hu Spin-orbit coupling

The expectation value of spin-orbit coupling

j(j + 1) l(l + 1) s(s + 1) 2 L~ S~ = ~ h · i 2 The expectation value of radii 1 Z3 3 = 3 3 hr i n l(l +1/2)(l + 1)a1

Thus the spin–orbit interaction produces a shift in energy of

Z4e2~2 j(j + 1) l(l + 1) s(s + 1) Eso = 2 2 3 3 16⇡"0mec a1 n l(l +1/2)(l + 1)

16/11/2018 Jinniu Hu Spin-orbit coupling A single electron has s =1/2 so, for each l, its total angular

momentum quantum number j has two possible values: 1 j = l bei48482_Ch07.qxd 1/23/02 9:02 AM Page 249 ± 2

We ﬁnd that the energy interval between these levels:Many-Electron Atoms 249

4 2 2 +µBB Z e 1 2p ∆E = 2µBB ~ –µBB Eso = 2 2 3 3 8⇡"0mec a1 n l(l + 1) (↵Z)4 = 3 E0 2n l(l + 1) 1s

Figure 7.14 Spin-orbit coupling splits the 2p state in the hydrogen atom into two substates ⌬E apart. The result is a doublet (two closely spaced lines) instead of a single spectral line for the 2p → 1s transition. The energy interval of 2p state of hydrogen

The orbiting electron “sees” itself circled f times per second by the proton of charge ϩe that is the nucleus, for a resulting magnetic field of 5 ␮ fe B ϭ ᎏ0 Eso =4.53 10 eV 2r ⇥ The frequency of revolution and orbital radius for n ϭ 2 are, from Eqs. (4.4) and (4.14), ␷ f ϭϭᎏ 8.4 ϫ 1014 sϪ1 2␲r 2 Ϫ10 16/11/2018 Jinniu Hu r ϭ n a0 ϭ 4a0 ϭ 2.1 ϫ 10 m Hence the magnetic field experienced by the electron is

(4␲ ϫ 10Ϫ7 T и m/A)(8.4ϫ 1014 sϪ1)(1.6 ϫ 10Ϫ19 C) B ϭϭᎏᎏᎏᎏᎏᎏ 0.40 T (2)(2.1 ϫ 10Ϫ10 m)

which is a fairly strong field. Since the value of the Bohr magneton is ␮B ϭ eប͞2m ϭ 9.27 ϫ 10Ϫ24 J/T, the magnetic energy of the electron is

Ϫ24 Ϫ5 Um ϭ ␮BB ϭ 3.7 ϫ 10 J ϭ 2.3 ϫ 10 eV The energy difference between the upper and lower substates is twice this, 4.6 ϫ 10Ϫ5 eV, which is not far from what is observed (Fig. 7.14).

7.8 TOTAL ANGULAR MOMENTUM Both magnitude and direction are quantized

Each electron in an atom has a certain orbital angular momentum L and a certain spin angular momentum S, both of which contribute to the total angular momentum J of the atom. Let us first consider an atom whose total angular momentum is provided by a single electron. Atoms of the elements in group 1 of the periodic 4.6Zeeman effect

1896, Pieter Zeeman found that the splitting of spectral

lines by a magnetic ﬁeld

16/11/2018 Jinniu Hu 1.8 The Zeeman eﬀect 17

(a)

Fig. 1.7 (a) An apparatus suitable for the observation of the Zeeman effect. The light emitted from a discharge lamp, between the pole pieces of the electromagnet, passes through anarrow-bandfilterandaFabry– Perot étalon. Key: L1, L2 are lenses; F–filter; P–polarizer to discriminate between π-andσ-polarizations (op- tional); Fabry–Perot étalon made of arigidspacerbetweentwohighly- (b) (c) reflecting mirrors (M1 and M2); D – 1.016/11/2018 Jinniu Hu1.0 detector. Other details can be found in Brooker (2003). A suitable procedure is to (partially) evacuate the étalon cham- ber and then allow air (or a gas with a higher refractive index such as carbon 0.5 0.5 dioxide) to leak in through a constant- flow-rate valve to give a smooth linear scan. Plots (b) to (e) show the intensity I of light transmitted through the 0.0 0.0 Fabry–Perot étalon. (b) A scan over two free-spectral ranges with no mag- (d) (e) netic field. Both (c) and (d) show a Zee- 1.0 1.0 man pattern observed perpendicular to the applied field; the spacing between the π-andσ-components in these scans is one-quarter and one-third of the free- spectral range, respectively—the mag- 0.5 0.5 netic field in scan (c) is weaker than in (d). (e) In longitudinal observation only the σ-components are observed— this scan is for the same field as in (c) 0.0 0.0 and the σ-components have the same position in both traces. Zeeman effect

In a magnetic ﬁeld, then, the energy of a particular atomic state depends on the value of ml as well as on that of n E = µ B = µ m B B z B l Each (degenerate) atomic level of given l is split into 2l+1 different energy states according to the value of ml 7.4 Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect 255 The energy degeneracy of a given nl level is removed by a magnetic ﬁeld ᐉ ϭ 1 mᐉ mᐉ 1 1 ᐉ ϭ 1 ⌬E ϭ m B ᐉ ϭ 1 n ϭ 2 B 0 2p 0 ⌬E Ϫ1 Ϫ1 B ϭ 0 B ϭ B0ˆk 16/11/2018 Jinniu Hu Energy

0 E0

Ϫ1 E0 Ϫ mBB

CONCEPTUAL EXAMPLE 7.6

What is the lowest n/ state in the hydrogen atom that has a Solution We want to ﬁnd the lowest energy n/ state that degeneracy of 5? has ﬁve m/ states. This is true for a / ϭ 2 state, because 2/ ϩ 1 ϭ 5. The lowest possible / ϭ 2 state will be 3d, because n Ͼ / is required.

EXAMPLE 7.7

What is the value of the Bohr magneton? Use that value to Solution The Bohr magneton is determined to be

calculate the energy difference between the m/ ϭ 0 and m/ e U ϭ ϩ1 components in the 2p state of atomic hydrogen placed m B ϭ in an external ﬁeld of 2.00 T. 2m 1.602 ϫ 10Ϫ19 C 1.055 ϫ 10Ϫ34 J # s Strategy To ﬁnd the Bohr magneton we insert the known ϭ Ϫ31 2 9.11 ϫ 10 kg values of e, U, and m into the equation for mB [see text after 1 21 2 Ϫ24 Equation (7.28)]. The energy difference is determined m B ϭ 9.27 ϫ 10 1 J/T 2 (7.32) from Equation (7.31).

03721_ch07_241-271.indd 255 9/29/11 4:44 PM 7.4 Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect 255

ᐉ ϭ 1 mᐉ mᐉ 1 1 ᐉ ϭ 1 ⌬E ϭ m B ᐉ ϭ 1 n ϭ 2 B 0 2p 0 ⌬E Ϫ1 Ϫ1 B ϭ 0 B ϭ B0ˆk Energy

Figure 7.6 The normal Zeeman effect. (a) An external magnetic field removes the degeneracy of a 2p level and reveals the three different energy states. (b) There are now transitions with three dif- ᐉ ϭ 0 ferent energies between an ex- Zeeman effect 1s cited 2p level and the 1s ground B ϭ 0 B ϭ B0ˆk state in atomic hydrogen. The en- If the degenerate energy7.4 of Magnetic a state Effects is ongiven Atomic bySpectra—The E0, the Normal Zeeman Effect 255 ⌬ three different energies in a magnetic field B for a l=1 ergy E has been grossly exagger- (a) ᐉ ϭ 1 mᐉ (b) mᐉ ated along the energy scale. state are 1 1 ᐉ ϭ 1 ⌬E ϭ m B ᐉ ϭ 1 removed by a magneticn field ϭ 2 (see Figure 7.6a).B If the0 degenerate2p energy of a state0 ⌬E E = E0,E0 µBB Ϫ1 Ϫ1 is given by E0, the three different energies± in a magnetic field B for a / ϭ 1 state ˆ are B ϭ 0 B ϭ B0k Energy mO Energy 1 E0 ϩ mBB Figure 7.6 The normal Zeeman 0 E0 effect. (a) An external magnetic field removes the degeneracy of a Ϫ1 E0 Ϫ mBB 2p level and reveals the three different energy states. (b) There are now transitions with three dif- ᐉ ϭ 0 ferent energies between an ex- 1s cited 2p level and the 1s ground B ϭ 0 B ϭ B0ˆk state in atomic hydrogen. The en- ⌬ CONCEPTUAL16/11/2018 EXAMPLEJinniu 7.6 Hu ergy E has been grossly exagger- (a) (b) ated along the energy scale. removed by a magnetic field (see Figure 7.6a). If the degenerate energy of a state is given by E0, the three different energies in a magnetic field B for a / ϭ 1 state What is the lowest n/ stateare in the hydrogen atom that has a Solution We want to find the lowest energy n/ state that m Energy degeneracy of 5? O has five m/ states. This is true for a / ϭ 2 state, because 2/ ϩ 1 E ϩ m B 0 B 1 ϭ 5. The lowest possible / ϭ 2 state will be 3d, because 0 E0 n Ͼ / is required. Ϫ1 E0 Ϫ mBB

CONCEPTUAL EXAMPLE 7.6

What is the lowest n/ state in the hydrogen atom that has a Solution We want to ﬁnd the lowest energy n/ state that degeneracy of 5? has ﬁve m/ states. This is true for a / ϭ 2 state, because 2/ ϩ EXAMPLE 7.7 1 ϭ 5. The lowest possible / ϭ 2 state will be 3d, because n Ͼ / is required.

What is the value of the Bohr magneton? Use that value to Solution The Bohr magneton is determined to be calculate the energy difference between the m/ ϭ 0 and m/ e U ϭ ϩ1 components in the 2p state of atomic hydrogen placed m B ϭ in an external field of 2.00 T. EXAMPLE 7.7 2m 1.602 ϫ 10Ϫ19 C 1.055 ϫ 10Ϫ34 J s What is the value of the Bohr magneton? Use that value to Solution The Bohr magneton is determined to be # Strategy To find the Bohr magneton we insert the known ϭ Ϫ31 calculate the energy difference between the m/ ϭ 0 and m/ ϫ e U 2 9.11 10 kg values of e, , and m into ϭthe ϩ1 componentsequation in for the 2mp state [see of atomic text hydrogenafter placed m B ϭ1 21 2 U B 2m in an external field of 2.00 T. m ϭ 9.27 ϫ 10Ϫ24 J/T (7.32) Equation (7.28)]. The energy difference is determined B Ϫ191 Ϫ34 2 1.602 ϫ 10 C 1.055 ϫ 10 J # s ϭ from Equation (7.31). Strategy To find the Bohr magneton we insert the known Ϫ31 2 9.11 ϫ 10 kg values of e, U, and m into the equation for mB [see text after 1 21 2 Ϫ24 Equation (7.28)]. The energy difference is determined m B ϭ 9.27 ϫ 10 1 J/T 2 (7.32) from Equation (7.31).

03721_ch07_241-271.indd 255 9/29/11 4:44 PM 03721_ch07_241-271.indd 255 9/29/11 4:44 PM Zeeman effect

Because ml can have the 2l+1 values, a state of given orbital quantum number l is split into 2l+1 substates that

differ in energy by "BB when the atom is in a magnetic ﬁeld.

However we expect a spectral line from a transition between two states of different l to be split into only

three components for selection rules

m =0, 1 l ±

due to the spin of phonon as one.

16/11/2018 Jinniu Hu Zeeman effect

The normal Zeeman effect consists of the splitting of a

spectral line of frequency into three components whose frequencies are e ⌫ = ⌫ B 1 0 4⇡m ⌫2 = ⌫0 e ⌫ = ⌫ + B 3 0 4⇡m

where, the Larmor frequency is deﬁned as ! e ⌫ = L = B(T) = 14B(T) GHz L 2⇡ 4⇡m

16/11/2018 Jinniu Hu bei48482_ch06 1/23/02 8:16 AM Page 225

Quantum Theory of the Hydrogen Atom 225

leads to a “splitting” of individual spectral lines into separate lines when atoms radiate in a magnetic field. The spacing of the lines depends on the magnitude of the field. The splitting of spectral lines by a magnetic field is called the Zeeman effect after the Dutch physicist Pieter Zeeman, who first observed it in 1896. The Zeeman effect is a vivid confirmation of space quantization. Because ml can have the 2l ϩ 1 values of ϩl through 0 to Ϫl, a state of given orbital quantum number l is split into 2l ϩ 1 substates that differ in energy by ␮BB when the atom is in a magnetic field. However, because changes in ml are restricted to ⌬ml ϭ 0, Ϯ1, we expect a spectral line from a transition between two states of different l to be split into only three components, as shown in Fig. 6.17. The normal Zeeman effect consists of the splitting of a spectral line of frequency ␯0 into three components whose frequencies are B e ␯ ϭ ␯ Ϫ ␮ ᎏ ϭ ␯ Ϫ ᎏ B 1 0 B h 0 4␲m

Normal Zeeman ␯ ϭ ␯ (6.43) effect 2 0 B e ␯ ϭ ␯ ϩ ␮ ᎏ ϭ ␯ ϩ ᎏ B 3 0 B h 0 4␲m ZeemanIn Chap. effect 7 we will see that this is not the whole story of the Zeeman effect.

No magnetic field Magnetic field present

ml = 2 ml = 1 l = 2 ml = 0 ml = –1 ml = –2

h␯ – eh B h␯ + eh B ( 0 2m ( ( 0 2m (

h␯0

ml = 1 l = 1 ml = 0 ml = –1

∆ml = +1 ∆ml = –1

∆ml = 0

eB eB ␯ ␯0 – ␯0 + 0 ( 4πm ( ␯0 ( 4πm (

Spectrum without Spectrum with magnetic magnetic field field present Jinniu Hu 16/11/2018Figure 6.17 In the normal Zeeman effect a spectral line of frequency ␯0 is split into three components when the radiating atoms are in a magnetic field of magnitude B. One component is ␯0 and the others are less than and greater than ␯0 by eB͞4␲m. There are only three components because of the selection rule ⌬ml ϭ 0, Ϯ1. bei48482_Ch07.qxd 2/4/02 11:42 AM Page 229

Many-Electron Atoms 229

uantum mechanics explains certain properties of the hydrogen atom in an accurate, straightforward, and beautiful way. However, it cannot approach a Q complete description of this atom or of any other without taking into account electron spin and the exclusion principle. In this chapter we will look into the role of electron spin in atomic phenomena and into why the exclusion principle is the key to understanding the structures of atoms with more than one electron.

7.1 ELECTRON SPIN Round and round it goes forever

The theory of the atom developed in the previous chapter cannot account for a number of well-known experimental observations. One is the fact that many spectral lines actually consist of two separate lines that are very close together. An example of this fine structure is the first line of the Balmer series of hydrogen, which arises from transitions between the n ϭ 3 and n ϭ 2 levels in hydrogen atoms. Here the theoretical prediction is for a single line of wavelength 656.3 nm while in reality there are two lines 0.14 nm apart—a small effect, but a conspicuous failure for the theory. Another failure of the simple quantum-mechanical theory of the atom occurs in the Zeeman effect, which was discussed in Sec. 6.10. There we saw that the spectral lines of an atom in a magnetic field should each be split into the three components speci- fied by Eq. (6.43). While the normal Zeeman effect is indeed observed in the spectra of a few elements under certain circumstances, more often it is not. Four, six, or even more components may appear, and even when three components are present their spacing may not agree with Eq. (6.43). Several anomalous Zeeman patterns are shown in Fig. 7.1 together with the predictions of Eq. (6.43). (When reproached in 1923 for looking sad, the physicist Wolfgang Pauli replied, “How can one look happy when he is thinking about the anomalous Zeeman effect?”) In order to account for both fine structure in spectral lines and the anomalous Zeeman effect, two Dutch graduate students, Samuel Goudsmit and George Uhlenbeck, proposedAnomalous in 1925 that Zeeman effect

EverySoon electron after hasthe an discovery intrinsic angular of this momentum, effect by called Zeeman spin, whosein magni- tude1896, is the it same was forfound all electrons. that often Associated more withthan this three angular closely momentum is a magnetic moment. spaced optical lines were observed.

No magnetic field

Magnetic field present

Expected splitting

No magnetic field Magnetic field present

Expected splitting 16/11/2018 Jinniu Hu Figure 7.1 The normal and anomalous Zeeman effects in various spectral lines. 4.7Anomalous Zeeman effect

We shall see that the anomalous effect depends on the

effects of electron intrinsic8.3 spin. Anomalous Zeeman Effect 293 precession of J around B . Therefore, we first find the average m about J and m precesses! ext µ~ j = µ~ l + µ~ s,av Whole system! m Bext fast around J then find the interaction energy of av with Bext. We leave this as an~ exercise for precesses slowly! the student (see Problem 31). The result is = gjµBJ/~ around Bext z e UBext S

V ϭ gmJ ϭ mTheBBext genergymJ in magnetic (8.22)ﬁeld is, S 2m J where m is the Bohr magneton and EB = gjmjµBB B L J ϩ S Ϫm J J ϩ 1 ϩ S S ϩThe1 Ϫ frequencyL L ϩ 1 of splitting spectra , ( ) g ϭ 1 ϩ (8.23) 2J J ϩ 1 1 2 1 2 1 2 e ⌫ = ⌫0 +(m2g2 m1g1) B is a dimensionless number called the Landé1 g factor2 . The magnetic total angular4⇡ m momentum numbers mJ range from ϪJ to J in integral steps. The external ﬁeld Figure 8.14 Relationships be- ϩ ⌬ ϭ Bext splits each state J into 2J 1 equally spaced16/11/2018 levels separated by E V, withJinniu Hutween S, L, J, and m are indicated. V determined in Equation (8.22), each level being described by a different mJ. The vector Bext is in the z direc- In addition to the previous selection rules [Equation (8.14)] for photon tion. The magnetic moment m transitions between energy levels, we must now add one for mJ : precesses fast around J as J precesses more slowly around the mJ ϭϮ1, 0 (8.24) ¢ weak Bext. After J. D. McGervey, Intro- ϭ ϭ ⌬ ϭ duction to Modern Physics, New but mJ1 0 S mJ2 0 is forbidden when J 0. York: Academic Press (1983), p. 329. CONCEPTUAL EXAMPLE 8.10

Show that the normal Zeeman effect should be observed for mJ 1 1 transitions between the D2 and P1 states. ϩ2 Strategy Because 2S ϩ 1 ϭ 1 for both states, then S ϭ 0 ϩ1 and J ϭ L. The g factor from Equation (8.23) is equal to 1 (as 1 0 D 2 1 it will always be for S ϭ 0). The D2 state splits into ﬁve equally 1 Ϫ1 spaced levels, and the P1 state splits into three (see Figure e ប ⌬E ϭ Bext 8.15). We use the selection rules from Equations (8.14) and Ϫ2 2m (8.24) to determine which transitions are allowed.

1 Solution We start with every level in the D2 state and determine using the selection rules which transitions to levels in 1 the P1 state are allowed. We show in Figure 8.15 that there are only nine allowed transitions. The other transitions are disallowed by the selection rule for ⌬mJ. Even though there are nine different transitions, there are only three different Energy energies for emitted or absorbed photons, because transition energies labeled 1, 3, 6 are identical, as are 2, 5, 8, and also 4, 7, 9. Thus the three equally spaced transitions of the normal Zeeman effect are observed whenever S ϭ 0.

ϩ1 Figure 8.15 Examples of transitions for the normal Zeeman ef- ⌬E 1 fect. The nine possible transitions are labeled, but there are only 0 P1 three distinctly different energies because the split energy levels Ϫ1 1 1 are equally spaced (⌬E ) for both the D2 and P1 states. 123456789

03721_ch08_272-297.indd 293 9/29/11 10:12 AM 294 Chapter 8 AtomicAnomalous Physics Zeeman effect

mJ 3 ϩ Energy 2 2 ϩ 1 P3/2 2 Ϫ 1 2 Ϫ 3 2 2P Figure 8.16 Schematic diagram of anomalous Zeeman effect for ϩ 1 2 sodium (energy levels not to 1 2P Ϫ 1/2 2 scale). With Bext ϭ 0 for the unperturbed states, there is only one transition. With the spin-orbit interaction splitting the 2P state into two states, there are two possible transitions when Bext ϭ 0. Fi- 1 nally, Bext splits J into 2J ϩ 1 com- ϩ 2 2 ponents, each with a different mJ. S 2S 1/2 The energy splitting ⌬E for each Ϫ 1 major state is different because 2 ⌬E ϭ gm (e 2m)B and the J U ext With 2 Unperturbed With Landé g factor for g[⌬E( S1/2)] Ͼ states spin-orbit Bext g[⌬E(2P )] Ͼ g[⌬E(2P )]. All 3/2 1/2 Bext ϭ 0 interaction allowed transitions are shown. Bext ϭ 0 16/11/2018 Jinniu Hu The anomalous Zeeman effect is a direct result of intrinsic spin. Let us consider transitions between the 2P and 2S states of sodium as shown in Figure 8.16. 2 2 In a completely unperturbed state the P3/2 and P1/2 states are degenerate. How- 2 ever, the internal spin-orbit interaction splits them, with P1/2 being lower in en- 2 ergy. The S1/2 state is not split by the spin-orbit interaction because / ϭ L ϭ 0. When sodium is placed in an external magnetic ﬁeld, all three states are split into 2J ϩ 1 levels with different mJ (see Figure 8.16). The appropriate Landé g factors are 1 1 1 1 ϩ ϩ ϩ 2 2 2 1 2 2 1 S1/2 g ϭ 1 ϩ 1 1 ϭ 2 ϩ 1 2 # 22 2 11 2 1 1 1 1 ϩ ϩ ϩ Ϫ ϩ 2 2 2 1 1 2 2 2 1 1 1 1 P1/2 g ϭ 1 ϩ 1 1 ϭ 0.67 ϩ 1 2 2 #12 2 21 1 2 3 3 1 1 ϩ ϩ ϩ Ϫ ϩ 2 2 2 1 2 21 1 2 1 1 1 P3/2 g ϭ 1 ϩ 3 3 ϭ 1.33 2 ϩ 1 1 2 #12 2 2 1 2 All g factors are different, and the energy-splitting1 2⌬E calculated using Equation (8.22) for the three states are different. The allowed transitions using the selection rules are shown in Figure 8.16. There are four different energy transitions 2 2 2 2 for P1/2 S S1/2 and six different energy transitions for P3/2 S S1/2. If the external magnetic ﬁeld is increased, then L and S precess too rapidly about Bext and our averaging procedure for m around Bext breaks down. In that case, the equations developed in this section are incorrect. This occurrence, called the Paschen-Back effect, must be analyzed differently. We will not pursue this calculation further.*

*See H. E. White, Introduction to Atomic Spectra, New York: McGraw-Hill (1934) for more information.

03721_ch08_272-297.indd 294 9/29/11 10:12 AM bei48482_ch02.qxd 2/4/02 11:30 AM Page 53

Particle Properties of Waves 53

n our everyday experience there is nothing mysterious or ambiguous about the concepts of particle and wave. A stone dropped into a lake and the ripples that I spread out from its point of impact apparently have in common only the ability to carry energy and momentum from one place to another. Classical physics, which mirrors the “physical reality” of our sense impressions, treats particles and waves as separate components of that reality. The mechanics of particles and the optics of waves are traditionally independent disciplines, each with its own chain of experiments and principles based on their results. The physical reality we perceive has its roots in the microscopic world of atoms and molecules, electrons and nuclei, but in this world there are neither particles nor waves in our sense of these terms. We regard electrons as particles because they possess charge and mass and behave according to the laws of particle mechanics in such familiar de- vices as television picture tubes. We shall see, however, that it is just as correct to interpret a moving electron as a wave manifestation as it is to interpret it as a particle manifestation. We regard electromagnetic waves as waves because under suitable circumstances they exhibit diffraction, interference, and polarization. Similarly, we shall see that under other circumstances electromagnetic waves behave as though they consist of streams of particles. Together with special relativity, the wave-particle duality is central to an understanding of modern physics, and in this book there are few argu- ments that do not draw upon either or both of these fundamental ideas.

2.1 ELECTROMAGNETIC WAVES Coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behavior

In 1864 the British physicist James Clerk Maxwell made the remarkable suggestion that accelerated electric charges generate linked electric and magnetic disturbances that can travel indefinitely through space. If the charges oscillate periodically, the disturbances are waves whose electric and magnetic components are perpendicular to each other and to the direction of propagation, as in Fig. 2.1. From the earlier work of Faraday, Maxwell knew that a changing magnetic field can induce a current in a wire loop. Thus a changing magnetic field is equivalent in its effects to an electric field. Maxwell proposed the converse: a changing electric field has a magnetic field associated with it. The electric fields produced by electromagnetic induction are easy to demonstrate because metals offer little resistance to the flow of charge. Even a weak field can lead to a measurable current in a metal. Weak magnetic fields are much harder to detect, however, and Maxwell’s hypothesis was based on a symmetry argument rather than on experimental findings. Polarization of light Electric field

Direction of wave

Magnetic field

Figure 2.1 The electric and magnetic fields in an electromagnetic wave vary together. The fields are perpendicular to each other and to the direction of propagation of the wave.

16/11/2018 Jinniu Hu 1.8 The Zeeman eﬀect 17

1.8.1 Experimental observation of the Zeeman effect Figure 1.7(a) shows an apparatus suitable for the experimental observation of the Zeeman effect and Fig. 1.7(b–e) shows some typical experimental traces. A low-pressure discharge lamp that contains the atom to be studied (e.g. helium or cadmium) is placed between the pole pieces 1.8 The Zeeman effect 17 of an electromagnet capable of producing fields of up to about 1 T. In the arrangement shown, a lens collects light emitted perpendicular to 1.8.1 Experimental observationthe field (transverse of the Zeeman observation) and sends it through a Fabry–Perot effect étalon. The operation of such étalons is described in detail by Brooker (2003), and only a brief outline of the principle of operation is given Figure 1.7(a) shows an apparatus suitablehere. for the experimental observation of the Zeeman effect and Fig. 1.7(b–e) shows some typical experimental traces. A low-pressure discharge lamp that contains the atom to (a) be studied (e.g. helium or cadmium) is placed between the pole pieces of an electromagnet capable of producing fields of up to about 1 T. In the arrangement shown, aZeeman lens collects light emitted perpendiculareffect to the field (transverse observation) and sends it through a Fabry–Perot Polarizationétalon. The operation of ofZeeman such étalons is describedeffect in detail by Brooker (2003), and only a brief outline of the principle of operation is given Fig. 1.7 (a) An apparatus suitable here. Splitting and polarisation of spectral lines for the observation of the Zeeman effect. The light emitted from a discharge lamp, between the pole pieces (a) of the electromagnet, passes through anarrow-bandfilterandaFabry– Perot étalon. Key: L1, L2 are lenses; F–filter; P–polarizer to discriminate between π-andσ-polarizations (op- tional); Fabry–Perot étalon made of arigidspacerbetweentwohighly- (b) (c) reflecting mirrors (M1 and M2); D – 1.0 1.0Fig. 1.7 (a) An apparatus suitable detector. Other details can be found in for the observation of the Zeeman ef- Brooker (2003). A suitable procedure is fect. The light emitted from a dis- to (partially) evacuate the étalon cham- charge lamp, between the pole pieces ber and then allow air (or a gas with a of the electromagnet, passes through higher refractive index such as carbon 0.5 0.5anarrow-bandfilterandaFabry–dioxide) to leak in through a constant- Perot étalon. Key: L1, L2 are lenses; flow-rate valve to give a smooth linear F–filter; P–polarizer to discriminatescan. Plots (b) to (e) show the inten- between π-andσ-polarizations (op- sity I of light transmitted through the 0.0 0.0tional); Fabry–Perot étalon made of Fabry–Perot étalon. (b) A scan over arigidspacerbetweentwohighly-two free-spectral ranges with no mag- (b)(d) (c) (e) reflecting mirrors (M1 and M2); D – netic field. Both (c) and (d) show a Zee- 1.0 1.01.0 1.0detector. Other details can be found in man pattern observed perpendicular to Brooker (2003). A suitable procedure is the applied field; the spacing between σb π σr to (partially) evacuate the étalon cham- the π-andσ-components in these scans ber and then allow air (or a gas with a is one-quarter and one-third of the free- spectral range, respectively—the mag- 0.5 0.5higher refractive index such as carbon 0.5 0.5 dioxide) to leak in through a constant- netic field in scan (c) is weaker than flow-rate valve to give a smooth linear in (d). (e) In longitudinal observation scan. Plots (b) to (e) show the inten- only the σ-components are observed— sity I of light transmitted through26 the this scan is for the same field as in (c) 0.0 0.00.0 0.0Fabry–Perot étalon. (b) A scan over and the σ-components have the same two free-spectral ranges with no mag- position in both traces. (d) (e)Jinniu Hu netic field. Both (c) and (d) show a Zee- 16/11/20181.0 1.0 man pattern observed perpendicular to the applied field; the spacing between the π-andσ-components in these scans is one-quarter and one-third of the free- spectral range, respectively—the mag- 0.5 0.5 netic field in scan (c) is weaker than in (d). (e) In longitudinal observation only the σ-components are observed— this scan is for the same field as in (c) 0.0 0.0 and the σ-components have the same position in both traces. P6.2.7.3 LEYBOLD Physics Leaflets

Angular distribution and polarization Apparatus Depending on the angular momentum component ⌬MJ in the 1 cadmium lamp for Zeeman effect . . . . 451 12 direction of the magnetic field, the emitted photons exhibit 1 U-core with yoke ...... 562 11 different angular distributions. Fig. 2 shows the angular dis- 2 coils, 10 A, 480 turns ...... 562 131 tributions in the form of two-dimensional polar diagrams. They 1 pair of pole pieces with large bores . . . 560 315 can be observed experimentally, as the magnetic field is char- acterized by a common axis for all cadmium atoms. 1 Fabry-Perot etalon ...... 471 221 In classical terms, the case ⌬M = 0 corresponds to an in- 2 positive lenses with barrel, 150 mm . . . 460 08 J finitesimal dipole oscillating parallel to the magnetic field. No 1 quarter-wave plate ...... 472 601 quanta are emitted in the direction of the magnetic field, i.e. 1 polarization filter ...... 472 401 the ␲- component cannot be observed parallel to the magnetic 1 holder with spring clips ...... 460 22 field. The light emitted perpendicular to the magnetic field is 1 filter set, primary ...... 467 95 linearly polarized, whereby the E-vector oscillates in the direc- or tion of the dipole and parallel to the magnetic field (see Fig. 3) 1 holder for interference filter ...... 468 41 Conversely, in the case ⌬M = 1 most of the quanta travel in 1 interference filter, 644 nm ...... 468 400 J ± the direction of the magnetic field. In classical terms, this case 1 ocular with line graduation ...... 460 135 corresponds to two parallel dipoles oscillating with a phase 1 precision optical bench, standardized difference of 90Њ. The superposition of the two dipoles produces a circulating current. Thus, in the direction of the mag- cross section, 1 m ...... 460 32 netic field, circularly polarized light is emitted; in the positive 1 rider base with thread ...... 460 358 direction, it is clockwise-circular for ⌬M = +1 and anticlock- 7 optics rider 60/50 ...... 460 351 J wise-circular for ⌬MJ = −1 (see Fig. 3). 1 universal choke for 451 12 ...... 451 30 1 high current power supply ...... 521 55

Optical transitions between these levels are only possible in the form of electrical dipole radiation. The following selection rules apply for the magnetic quantum numbers MJ of the states involved:  = ±1 for ␴ components ∆MJ  = 0 for ␲ components (VI)  Thus, we observe a total of three spectral lines (see Fig. 1); the ␲ component is not shifted and the two ␴ components are shifted by Polarization of Zeeman effect ⌬E Fig. 2#M AngularJ=0 corresponds distributions to an of infinitesimalthe electrical dipole dipole oscillating radiation ⌬f = ± (VII) h parallel(⌬MJ :to angular-momentum the magnetic field. components No quanta areof the emitted emitted in pho- thetons direction in the directionof the magnetic of the magnetic field. The field) light emitted with respect to the original frequency. In this equation, ⌬E is the equidistant energy split calculated in (V). perpendicular to the magnetic field is linearly polarized ($).

Connecting leads with conductor cross-section 2.5 mm2

16/11/2018 Jinniu Hu Fig. 3 Schematic representation of the polarization of the Zeeman components (⌬MJ: angular-momentum components of the emitted photons in the direction of the magnetic field)

Spectroscopy of the Zeeman components

The Zeeman effect enables spectroscopic separation of the differently polarized components. To demonstrate the shift, however, we require a spectral apparatus with extremely high resolution, as the two ␴ components of the red cadmium line are shifted e.g. at a magnetic flux density B = 1 T by only ⌬f = 14 GHz , respectively ⌬␭ = 0,02 nm.

2 P6.2.7.3 LEYBOLD Physics Leaflets

Optical transitions between these levels are only possible in the form of electrical dipole radiation. The following selection rules apply for the magnetic quantum numbers MJ of the states involved:  = ±1 for ␴ components ∆MJ  = 0 for ␲ components (VI)  Thus, we observe a total of three spectral lines (see Fig. 1); the ␲ component is not shifted and the two ␴ components are Polarization of Zeeman effect shifted by Fig. 2 Angular distributions of the electrical dipole radiation ⌬E #MJ=1 corresponds to two parallel dipoles oscillating with ⌬f = ± (VII) (⌬MJ: angular-momentum components of the emitted pho- h 0 a phasetons indifference the direction of 90 of the . The magnetic superposition field) of the two with respect to the original frequency. In this equation, ⌬E is dipoles produces a circulating current (%+, %-). the equidistant energy split calculated in (V).

Connecting leads with conductor cross-section 2.5 mm2

16/11/2018 Jinniu Hu Fig. 3 Schematic representation of the polarization of the Zeeman components (⌬MJ: angular-momentum components of the emitted photons in the direction of the magnetic field)

Spectroscopy of the Zeeman components

2 格罗春图Grotrian Diagram

2 -3/2 -1/2 1/2 3/2 D3/2

2 P3/2 -3/2 -1/2 1/2 3/2

σ σ

π z

1 1 16/11/2018 Jinniu Hu (m g m g ) J 2 2 J1 1 L 0 m g m (g g ) J 2 J1 2 1 L Homework

The Physics of Atoms and Quanta

12.1, 12.3, 12.5, 13.1, 13.2, 13.3, 13.4

16/11/2018 Jinniu Hu Exercise class

1.Treat the hydrogen atom as a one-dimensional entity of

length 2a0 and determine the electron’s minimum kinetic energy.

16/11/2018 Jinniu Hu Exercise class 188 Chapter 5 Wave Properties of Matter and Quantum Mechanics I 1.Treat the hydrogen atom as a one-dimensional entity of

length 2a0 and determine the electron’s minimum kinetic EXAMPLE 5.9 energy.

Solution: Treat the hydrogen atom as a one-dimensional entity of Solution Equation (5.43) gives length 2a0 and determine the electron’s minimum kinetic 2 U 2 Uc energy. K ϭ ϭ min 2m 2 2mc 2 2 / 1 2/ 197 eV # nm 2 Strategy We will use the uncertainty principle to deter- ϭ 6 2 ϭ 3.4 eV 2 0.511 ϫ 10 eV 2 ϫ 0.0529 nm mine K min. Equation (5.43) gives us the minimum kinetic 1 2 energy for a particle known to be located within a distance A calculation1 21 considering three21 dimensions 2 would give a /. result about twice this value. This simple calculation gives a reasonable value for the kinetic energy of the ground state 16/11/2018electron of the hydrogenJinniu atom. Hu

Energy-Time Uncertainty Principle Equation (5.40) is not the only form of the uncertainty principle. We can find another form by using Equation (5.23) from our study of wave motion. When we superimposed two waves to form a wave packet we found ⌬v ⌬t ϭ 2p. If we evaluate this same product using Gaussian packets, we will find 1 ¢v ¢t ϭ (5.44) 2 just as we did for the product ⌬k ⌬x. A relationship like this is easy to understand. If we are to localize a wave packet in a small time ⌬t (instead of over an infinite time as for a single wave), we must include the frequencies of many waves to have them cancel everywhere but over the time interval ⌬t. Because E ϭ hf, we have for each wave ¢v ¢E ϭ h ¢f ϭ h ϭ U ¢v 2p Therefore ¢E ¢E 1 ¢v ϭ and ¢v ¢t ϭ ¢t ϭ U U 2 We can therefore obtain another form of Heisenberg’s uncertainty principle: Heisenberg uncertainty U principle for energy ¢E ¢t Ն (5.45) and time 2

Other conjugate variables similar to px and x in Equation (5.40) also form uncertainty principle relations. The product of conjugate variables (such as px and x or E and t) must have the same dimensions as Planck’s constant. Conjugate variable pairs include the angular momentum L and angle u, as well as the rota- tional inertia I and angular velocity v. Similar uncertainty relations can be written for them. We once again must emphasize that the uncertainties expressed in Equations (5.40) and (5.45) are intrinsic. They are not due to our inability to construct better measuring equipment. No matter how well we can measure, no matter how accu-

03721_ch05_162-200.indd 188 9/29/11 9:48 AM Exercise class

2.An atom in an excited state normally remains in that state for a very short time ( 10-8s) before emitting a photon and returning to a lower energy state. The “lifetime” of the excited state can be regarded as an uncertainty in the time t associated with a measurement of the energy of the state. This, in turn, implies an “energy width,” namely, the corresponding energy uncertainty #E. Calculate (a) the characteristic “energy width” of such a state and (b) the uncertainty ratio of the frequency #f/f if the wavelength of the emitted photon is 300 nm.

16/11/2018 Jinniu Hu 5.7 Probability, Wave Functions, and the Copenhagen Interpretation 191

Strategy (a) We use the uncertainty principle, Equation (b) The frequency is found to be (5.45), to determine ⌬E because we know ⌬t. c 3 ϫ 108 m/s (b) We can determine ⌬f from the energy uncertainty f ϭ ϭ ϭ 1015 Hz (5.46) l ϫ Ϫ9 ⌬E by using E ϭ hf : ⌬E ϭ h ⌬f. We can determine the fre- 300 10 m Exercisequency class by f ϭ c/l. The uncertainty ⌬f is ¢E 3.3 ϫ 10Ϫ8 eV ¢f ϭ ϭ ϭ 8 ϫ 106 Hz (5.47) h ϫ Ϫ15 Solution: Solution(a) (a) Equation (5.45) gives 4.136 10 eV # s

Ϫ16 The uncertainty ratio of the frequency ⌬f/f is U 6.58 ϫ 10 eV # s Ϫ8 ¢E Ն ϭ Ϫ8 ϭ 3.3 ϫ 10 eV 6 f ϫ 5.7 Probability, Wave Functions,2 ¢t and the 2Copenhagen10 s Interpretation 191 ¢ 8 10 Hz Ϫ9 ϭ 15 ϭ 8 ϫ 10 This is a small energy, but many excited energy states have f 10 Hz Solution: (b) 1 21 2 Strategy (a) We use the uncertainty principle, Equation such energy(b) The widths. frequency For is stable found groundto be states, t ϭ q, and Modern instruments are capable of measuring ratios ap- (5.45), to determine ⌬E because we know ⌬t. ⌬E ϭ 0. For excited states in the nucleus, the lifetimes proaching 10Ϫ17, or 1 Hz in a frequency of 1017 Hz! Experi- c 3 ϫ 108 m/s (b) We can determine ⌬f from the energy uncertainty can be as shortf ϭas 10ϭϪ20 s (or shorter)ϭ 10with15 Hzenergy widths (5.46) of mental physicists have managed to improve this ratio by an l 300 ϫ 10Ϫ9 m ⌬E by using E ϭ hf : ⌬E ϭ h ⌬f. We can determine the fre- 100 keV (or more). irregular factor of 100 every three years over the past two quency by f ϭ c/l. The uncertainty ⌬f is decades. The experimental limitations are considerably bet- Ϫ8 ter than needed to measure the energy widths. ¢E 3.3 ϫ 10 eV ¢f ϭ ϭ ϭ 8 ϫ 106 Hz (5.47) h ϫ Ϫ15 Solution (a) Equation (5.45) gives 4.136 10 eV # s

Ϫ16 The uncertainty ratio of the frequency ⌬f/f is U 6.58 ϫ 10 eV # s Ϫ8 ¢E Ն ϭ Ϫ8 ϭ 3.3 ϫ 10 eV 6 2 ¢t 2 10 s ¢f 8 ϫ 10 Hz Ϫ ϭ ϭ 8 ϫ 10 9 f 1015 Hz This is a small energy, 1but21 many2 excited energy states have such energy widths. For stable ground states, t ϭ16/11/2018 q, and 5.7Modern instrumentsProbability, Jinniu are capableHu ofWave measuring Functions, ratios ap- and the ⌬E ϭ 0. For excited states in the nucleus, the lifetimes proaching 10Ϫ17, or 1 Hz in a frequency of 1017 Hz! Experi- can be as short as 10Ϫ20 s (or shorter) with energy widths of mental physicistsCopenhagen have managed to improve Interpretation this ratio by an 100 keV (or more). irregular factor of 100 every three years over the past two Wedecades. learned The in experimental elementary limitations physics arethat considerably the instantaneous bet- wave intensity of elec- 2 tromagneticter than needed radiation to measure (light) the energy is P0cE widths. where E is the electric field. Thus the probability of observing light is proportional to the square of the electric field. In the double-slit light experiment we can be assured that the electric field of the light wave is relatively large at the bright spots on the screen and small in the region of the dark places. If Young’s double-slit experiment is performed with very low intensity levels 5.7 Probability, Wave Functions,of light, and individual the flashes can be seen on the observing screen. We show a simula- tion of the experiment in Figure 5.19. After only 20 flashes (Figure 5.19a) we Copenhagen Interpretationcannot make any prediction as to the eventual pattern, but we still know that the We learned in elementary physics that the instantaneousprobability wave intensity of observing of elec- a flash is proportional to the square of the electric field. 2 tromagnetic radiation (light) is P0cE where E is the electricWe field. now Thus briefly the review prob- this calculation that is normally given in introductory ability of observing light is proportional to the square ofphysics the electric courses. field. If In the the distance from the central ray along the screen we are ob- double-slit light experiment we can be assured that the servingelectric infield an of experiment the light like that depicted in Figure 5.18a is denoted by y, the wave is relatively large at the bright spots on the screen and small in the region probability for the photon to be found between y and y ϩ dy is proportional to of the dark places. 2 If Young’s double-slit experiment is performed withthe very intensity low intensity of the levels wave (E ) times dy. For Young’s double-slit experiment, the of light, individual flashes can be seen on the observing screen.value of We the show electric a simula- field E produced by the two interfering waves is large where tion of the experiment in Figure 5.19. After only 20 flashesthe flash (Figure is likely 5.19a) to bewe observed and small where it is not likely to be seen. By cannot make any prediction as to the eventual pattern, butcounting we still the know number that the of flashes we relate the energy flux I (called the intensity) probability of observing a flash is proportional to the squareof the of light the electric to the numberfield. flux, N per unit area per unit time, of photons having 2 We now briefly review this calculation that is normallyenergy given hf in. In introductory the wave description, we have I ϭ P0c E , and in what appears to physics courses. If the distance from the central ray alongbe thethe screenparticle we description, are ob- I ϭ Nhf. The flux of photons N, or the probability P serving in an experiment like that depicted in Figure 5.18aof observing is denoted the by photons, y, the is proportional to the average8 9 value of the square of probability for the photon to be found between y and ythe ϩ dyelectric is proportional field E 2 .to the intensity of the wave (E 2) times dy. For Young’s double-slitHow experiment, can we interpret the the probability of finding the electron in the wave value of the electric field E produced by the two interferingdescription? waves is large 8where9 the flash is likely to be observed and small where it is not likely to be seen. By counting the number of flashes we relate the energy flux I (called the intensity) of the light to the number flux, N per unit area per unit time, of photons having 2 energy hf. In the wave description, we have I ϭ P0c E , and in what appears to be the particle description, I ϭ Nhf. The flux of photons N, or the probability P 03721_ch05_162-200.indd 191 9/29/11 9:48 AM of observing the photons, is proportional to the average8 9 value of the square of the electric field E 2 . How can we interpret the probability of finding the electron in the wave description? 8 9

03721_ch05_162-200.indd 191 9/29/11 9:48 AM Exercise class

3.Find the quantized energy levels of an electron constrained to move in a one-dimensional atom of size 0.1 nm.

16/11/2018 Jinniu Hu 5.8 Particle in a Box 195

2! c4 c4

2 c3 c3

2 c2 c2 Figure 5.23 Possible ways of ﬁtting waves into a one- dimensional box of length /. The left side shows the wave 2 c1 c1 functions for the four lowest energy values. The right side shows ᐉ ᐉ 0 0 the corresponding probability x x distributions.

2 which differs by a factor 1/p from the value of E1. Why the difference? Equa- tion (5.51) is based on the wave theory of physics, and as we shall see in the next chapter, it is a better calculation than the result in Equation (5.43), which used approximations. The probability of observing the particle between x and x ϩ dx in each state 2 is Pn dx r cn x dx. Notice that E0 ϭ 0 is not a possible state, because n ϭ 0 corresponds to c0 ϭ 0. The lowest energy level is therefore E1, with a probability 0 1 20 2 density P1 r c1 x , shown in Figure 5.23. The most probable location for the particle in the lowest energy state is in the middle of the box. This particle-in-a-box0 1 20 model is more important than it might seem. It is our first application of what we call “quantum theory” or “quantum mechanics.” No- tice how the quantization of energy arises from the need to fit a whole number of half-waves into the box and how we obtained the corresponding probability densi- ties of each of the states. The concept of energy levels, as first discussed in the Bohr model, has surfaced in a natural way by using waves. TheExercise procedure class followed is the same as finding the allowed modes of standing waves inside the box. We can use all the results that we learned about waves in elementary3.Find physics. the quantized energy levels of an electron constrained to move in a one-dimensional atom of size 0.1 nm. EXAMPLE 5.13

Solution:

2 2 2 Find the quantized energy levels of an electron constrained 2 h 2 h c En ϭ n ϭ n to move in a one-dimensional atom of size 0.1 nm. 8m/2 8mc 2/2 1239.8 eV nm 2 Strategy We previously found the minimum kinetic en- 2 # ϭ n 8 0.511 ϫ 106 eV 0.1 nm 2 ergy of an electron in a similar situation in Example 5.9. In 1 2 the present case we want to use quantum theory, so we use ϭ n 2 38 eV Equation (5.51) for the energy levels. 1 21 21 2 The ﬁrst three energy1 levels2 are E1 ϭ 38 eV, E2 ϭ 152 eV, Solution We use Equation (5.51) and insert the appropri- and E3 ϭ 342 eV. ate values for m and /. 16/11/2018 Jinniu Hu

03721_ch05_162-200.indd 195 9/29/11 9:48 AM Exercise class

4.Determine the expectation values for x, x2, p, and p2of a particle in an inﬁnite square well for the ﬁrst excited state.

16/11/2018 Jinniu Hu 6.3 Inﬁnite Square-Well Potential 215

EXAMPLE 6.8

2 2 Determine the expectation values for x, x , p, and p of a The expectation value p nϭ2 is determined by using particle in an infinite square well for the first excited state. Equation (6.23). 6.3 Infinite Square-Well8 9 Potential 215 2 L 2px d 2px Strategy The first excited state corresponds to n ϭ 2, be- p nϭ2 ϭ ϪiU sin sin dx L Ύ L dx L cause n ϭ 1 corresponds to the lowest energy state or the 0 ground state.EXAMPLE Equation (6.34) 6.8 gives us the wave function that which8 reduces9 1 to 2 a bc a bd we need to find the expectation values given in Section 6.2. L 4iU 2px 2px p ϭϪ sin cos dx ϭ 0 Determine the expectation values for x, x 2, p, and p2 of a The expectationnϭ2 2 valueΎ p is determined by using Solution The wave function for this case, according to L 0 nϭL2 L particleEquation in an (6.34), infinite is square well for the first excited state. Equation 8(6.23).9 a b a b Because the particle is moving8 9 left as often as right in the box, the average momentum2 L 2p isx zero.d 2px Strategy 2 2px ϭ The first excitedc ϭ state corresponds to n 2, be- p nϭ2 ϭ ϪiU sin 2 sin dx 2 x sin The expectationL Ύ value Lp nϭ2 dxis given byL cause n ϭ 1 corresponds to theB L lowest Lenergy state or the 0 ground state. Equation (6.34)1 2 gives us athe waveb function that which8 reduces9 21 toL 2 2px a 8 9 bcd ad bd 2px The expectation value x nϭ2 is 2 ϭ Ϫ Ϫ we need to find the expectation values given in Section 6.2. p nϭ2 Ύ sin iU iU sin dx L L 0 LL dx dx L 2 8 9 2px L 4iU 2px 2px 2 8 9 p ϭ ϭϪ a L sinba cosba bdx ϭa0 b x nϭ2 ϭ6.3 Infinitex sin Square-Well dx ϭPotential 215 n 2 22 Ύ 2px 2p d 2px Solution The wave functionL Ύ for thisL case, according2 to L2 0 L L 0 ϭ ϪiU sin cos dx ExerciseEquation class(6.34), is 8 9 L Ύ a L b La dx b L We evaluate all8 9 these integrationsa by lookingb up the integral Because the particle is0 moving left as often as right in the box, the average1 momentum2 2 La is zero.ba b a b EXAMPLE 6.8 in Appendix 3. As we expect,2 the2p x average2 position of2 the 2 8p 2px 2px 4.Determine the expectationc ϭ values for x, x , p, and p of a ϭϪϪU sin2 sin dx particle is in the2 middlex of thesin box (x ϭ L/2), even though The expectation Lvalue3 Ύ p nϭ2L is given byL particle in an infinite squareB L well forL the first excited 0 the actual probability of the particle being there is zero (see L1 2 2 2 a b a b 2 2state. 2 1 2 a b 2 4p U 2px 8 9 d d 2px Determine the expectation values for x, x , p, and p The of a cexpectation in FigureThe expectation value6.3). x nϭ 2 value is p nϭ2 is determined by using 2 ϭ Ϫ Ϫ p nϭ2 ϭ sin2 iU iU sin dx particle in an infinite square well for the first excited state. Equation (6.23). 2 L Ύ L L dx dx L The expectation2 Lvalue x 82np9ϭ2x of the squareL of the posi- 0 Solution:0 0 8 9 L 2 ϭ ϭ 8 9 a L ba ba b a b tion is givenx n ϭby2 2x sin 2px dxd 2px This value can be comparedp withp E2 [Equationp (6.35)]: Strategy The first excited state corresponds to n ϭ 2, be- p ϭ ϪLi Ύ sin L sin 2 dx 2 2 2 x 2 d 2 x nϭ2 U0 8 9 ϭ Ϫi sin cos dx LLΎ L dx L U cause n ϭ 1 corresponds to the lowest energy state or the 0 L Ύ L 2 2 L dx2 L 8 9 2 a 2pxb 0 p p ϭ We evaluate8 all9x 2these1ϭ integrations2 x 2 sina2 by lookingbc dx ϭ upa 0.32 thebdL integral2 4 U n 2 ground state. Equation (6.34) gives us the wave function that which reducesnϭ2 to E2 ϭ ϭ L Ύ L 1 2 2 La ba2 2mb a b we need to find the expectation values given in Section 6.2.in Appendix 3. As we expect,0 the average position of the 2 8p 2mL2px 2px L ϭϪϪU sin sin8 9 dx particle is in the8 9 middle24i ofU the box2ap x(x ϭb L/2),2px even though 3 Ύ 2 ϭ L L L ϭ The value p ofnϭ 2 ϭϪx nϭ2 2 issin 0.57L, largercos than dxx ϭnϭ02 0.5L. which is correct, because0 nonrelativistically we have E Solution The wave function for this case, accordingthe to actual probability ofL theΎ particleL being thereL is zero (see 2 Does this seem reasonable?0 (Hint: look again at the shape of p /2m ϩ4 pV1 2 and2 2 V ϭ 0. a b a b Equation (6.34), is c 2 in Figure8 6.3).9 8 9 a b a b 8 9 U the Becausewave function the particle in Figure is moving 6.3.) left as often as right in the ϭ 2 The expectation value x 2 of the square of the posi- L 2 2px box, the average momentumnϭ2 is zero. c ϭ 0 0 2 2 x sin tion is givenThe by expectation value p nϭ2 is given by This value can be compared with E [Equation (6.35)]: B L L 8 9 2 L L 1 2 a b 2 2px 8 p9 d d 2px 2 2 2 The expectation value x nϭ2 is 2 2 2 2 2 2 x 2 4p p nϭ2 16/11/2018 p xnϭ2 ϭEXAMPLEϭ sin x Jinniu sin ϪHuiU 6.9 dxϪϭiU0.32sinL dx U nϭ2 Ύ E2 ϭ ϭ L L 0 L Ύ L L dx dx L 2 2 8 9 2px L 0 2mL 2m 2 8 9 a L ba ba b a b 8 9 x nϭ2 ϭ x sin dx ϭ 8 9 ap b p p L Ύ L 2 2 2 2 2 2 x 2 d 2 x 0 The value of ϭ ϪxiUnϭ 2 is sin 0.57L, larger thancos xϪn14ϭ2 ϭ 0.5dx L. which is correct, because2 2 2 nonrelativistically2 we have2 E ϭ A typical diameter L ofΎ a nucleusL is L aboutdx 10 L m. Use the 2 p U c 1 p 197.3 eV # nm We evaluate all8 9 these integrationsa by lookingb up the integralDoes this seem reasonable?0 (Hint: look again at the shape of p /2m ϩ V and ϭV ϭ 0. ϭ infinite square-well potential to calculate the transition en- E1 2 2 2 Ϫ5 2 1 8 9 2 2 La ba b 8 a9 b 2mc L mc 2 10 nm in Appendix 3. As we expect, the average position ofthe the wave function in Figure2 8p 6.3.) 2px 2px 1 2 ergy from ϭϪ the Ϫ firstU excitedsin state tosin the ground dx state for a particle is in the middle of the box (x ϭ L/2), even though L3 Ύ L L proton confined to the 0 nucleus. Of course, this is only a 1 1 15 2 2 the actual probability of the particle being there is zero (see ϭ 1.92 ϫ 10 eV rough calculation4p1 2 2 for2 a protona in ab nucleus.a b mc 2 c 2 in Figure 6.3). U ϭ 2 2 L 1 2 2 The expectation value x nϭ2 of the square of the posi- EXAMPLE 6.9 The mass of the proton is 938.3 MeV/c , which gives 0 0 tion is given by This value can be compared with E2 [Equation (6.35)]: 15 2 8 9 Strategy To find the transition energy between the 1.92 ϫ 10 eV L 2 ϭ ϭ 2 2px ground and first excited energy2 2 states,p we use Equation E1 6 2.0 MeV 2 2 2 2 4p U nϭ2 Ϫ14 938.3 ϫ 102 eV 2 x nϭ2 ϭ x sin dx ϭ 0.32L ϭ ϭ 2 2 2 p Ύ A typical diameter of a E nucleus2 is2 about 10 m. Use the p U c 1 197.3 eV # nm L 0 L (6.35) to find E1 and E2. 2mL 2m 8 9 E1 ϭ ϭ Ϫ 8 9 a b infinite square-well potential to calculate the transition en- The first excited2mc state2L2 energymc 2 is found2 10 5[again nm 2 from Equation The value of 2 x 2 is 0.57L, larger than x ϭ 0.5L. which is correct, because nonrelativistically we have E ϭ 1 2 nϭ2 nϭ2 ergy from the first excited state to the ground state for a (6.35)] to be E2 ϭ 4E1 ϭ 8 MeV, and the transition energy 2 ϩ ϭ Does this seem reasonable? (Hint: look again at the shapeproton ofSolution p confined/2m V and to theV nucleus.0. Of course, this is only a is ⌬E ϭ E Ϫ E ϭ 6 MeV.1 This1 is a reasonable15 2 2 value for The energy of the ground state, from Equation 2 1 ϭ ϫ the wave function8 in9 Figure 6.3.) 8 9 2 1.92 10 eV rough(6.35), calculation is for a proton in a nucleus. protons in the nucleus. mc The mass of the proton is 938.31 MeV/c 2, which2 gives

15 2 Strategy To find the transition energy between the 1.92 ϫ 10 eV EXAMPLE 6.9 E ϭ ϭ 2.0 MeV ground and first excited energy states, we use Equation 1 938.3 ϫ 106 eV (6.35) to find E1 and E2. Ϫ14 2 2 2 2 2 The first excited state energy is found [again from Equation A typical diameter of a nucleus is about 10 m. Use the p U c 1 p 197.3 eV # nm E1 ϭ ϭ (6.35)] to be E ϭ 4E ϭ 8 MeV, and the transition energy infinite square-well potential to calculate the transition en- 2mc 2L2 mc 2 2 10Ϫ5 nm 2 2 1 1 2 ergy from the first excited state to the ground03721_ch06_201-240.indd state forSolution a 215 The energy of the ground state, from Equation is ⌬E ϭ E2 Ϫ E1 ϭ 6 MeV. This is a reasonable value for9/29/11 2:40 PM proton confined to the nucleus. Of course, this is only(6.35), a is 1 1 15 2 2 protons in the nucleus. ϭ 1.92 ϫ 10 eV rough calculation for a proton in a nucleus. mc 2 The mass of the proton is 938.31 MeV/c 2, which2 gives

15 2 Strategy To find the transition energy between the 1.92 ϫ 10 eV E ϭ ϭ 2.0 MeV ground and first excited energy states, we use Equation 1 938.3 ϫ 106 eV (6.35) to find E1 and E2. The first excited state energy is found [again from Equation (6.35)] to be E2 ϭ 4E1 ϭ 8 MeV, and the transition energy Solution The energy of the ground state,03721_ch06_201-240.indd from Equation 215is ⌬E ϭ E2 Ϫ E1 ϭ 6 MeV. This is a reasonable value for 9/29/11 2:40 PM (6.35), is protons in the nucleus.

03721_ch06_201-240.indd 215 9/29/11 2:40 PM 6.3 Inﬁnite Square-Well Potential 215

EXAMPLE 6.8

EXAMPLE 6.9

Ϫ14 2 2 A typical diameter of a nucleus is about 10 m. Use the p2 U 2c 2 1 p 197.3 eV # nm E ϭ ϭ infinite square-well potential to calculate the transition en- 1 2mc 2L2 mc 2 2 10Ϫ5 nm 2 ergy from the first excited state to the ground state for a 1 2 proton confined to the nucleus. Of course, this is only a 1 1 15 2 2 ϭ 1.92 ϫ 10 eV rough calculation for a proton in a nucleus. mc 2 The mass of the proton is 938.31 MeV/c 2, which2 gives

15 2 Strategy To find the transition energy between the 1.92 ϫ 10 eV E ϭ ϭ 2.0 MeV ground and first excited energy states, we use Equation 1 938.3 ϫ 106 eV (6.35) to find E1 and E2. The first excited state energy is found [again from Equation (6.35)] to be E2 ϭ 4E1 ϭ 8 MeV, and the transition energy Solution The energy of the ground state, from Equation is ⌬E ϭ E2 Ϫ E1 ϭ 6 MeV. This is a reasonable value for (6.35), is protons in the nucleus.

03721_ch06_201-240.indd 215 9/29/11 2:40 PM Exercise class

5.What are the possible quantum numbers for the state n=4 in atomic hydrogen?

16/11/2018 Jinniu Hu 7.3 Quantum Numbers 249 the wave functions have acceptable properties, including being single valued and ﬁnite. The restrictions imposed by the boundary conditions are n ϭ 1, 2, 3, 4, . . . Integer / ϭ 0, 1, 2, 3, . . . , n Ϫ 1 Integer (7.19)

m/ ϭϪ/, Ϫ/ ϩ 1, . . . , 0, 1, . . . , / Ϫ 1, / Integer These three quantum numbers must be integers. The orbital angular momentum quantum number must be less than the principal quantum number, / Ͻ n, and the magnitude of the magnetic quantum number (which may be positive or negative) must be less than or equal to the orbital angular momentum quantum number, m/ Յ /. We can summarize these conditions as 0 0 n Ͼ 0 / Ͻ n (7.20)

m/ Յ /

The lowest value of n is 1, and for n ϭ0 1,0 we must have / ϭ 0, m/ ϭ 0. For n ϭ 2, we may have / ϭ 0, m/ ϭ 0 as well as / ϭ 1, m/ ϭ Ϫ1, 0,Exercise ϩ1. class

5.What are the possible quantum numbers for the state n=4 in atomic hydrogen? CONCEPTUAL EXAMPLE 7.3 Solution:

What are the possible quantum numbers for the state n ϭ 4 n O mO in atomic hydrogen? 4 0 0 4 1 Ϫ1, 0, 1 Solution We want to apply the restrictions for the quan- 4 2 Ϫ2, Ϫ1, 0, 1, 2 tum numbers given in Equations (7.19) and (7.20). If n ϭ 4, 4 3 Ϫ3, Ϫ2, Ϫ1, 0, 1, 2, 3 then the possible values of / are / ϭ 0, 1, 2, 3, because

/max ϭ n Ϫ 1. For each value of /, m/ goes from Ϫ/ to ϩ/. We show the results in tabular form.

16/11/2018 Jinniu Hu

As yet these quantum numbers may seem to have little physical meaning. Let us examine each of them more carefully and try to ﬁnd classical analogies where possible.

Principal Quantum Number n The principal quantum number n results from the solution of the radial wave function R(r) in Equation (7.4). Because the radial equation includes the potential energy V(r), it is not surprising to ﬁnd that the boundary conditions on R(r) quantize the energy E. The result for this quantized energy is

2 2 Ϫm e 1 E0 En ϭ 2 ϭϪ 2 (7.21) 2 4pP0 U n n which is precisely the value found ina Chapterb 4 from the Bohr theory [Equations (4.25) and (4.26)]. So far, the energy levels of the hydrogen atom depend only

03721_ch07_241-271.indd 249 9/29/11 4:44 PM Exercise class

6. What is the value of the Bohr magneton? Use that value to calculate the energy difference between the ml=0 and ml=1 components in the 2p state of atomic hydrogen placed in an external ﬁeld of 2.00 T ?

16/11/2018 Jinniu Hu 7.4 Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect 255

ᐉ ϭ 1 mᐉ mᐉ 1 1 ᐉ ϭ 1 ⌬E ϭ m B ᐉ ϭ 1 n ϭ 2 B 0 2p 0 ⌬E Ϫ1 Ϫ1 B ϭ 0 B ϭ B0ˆk Energy

0 E0

Ϫ1 E0 Ϫ mBB

CONCEPTUAL EXAMPLE 7.6

6. What is the value of the Bohr magneton? Use that value

to calculate the energy difference between the ml=0 and EXAMPLE 7.7 ml=1 components in the 2p state of atomic hydrogen placed in an external field of 2.00 T ? What is the value of the Bohr magneton? Use that value to Solution The Bohr magneton is determined to be Solution: The Bohr magneton is determined to be calculate the energy difference between the m/ ϭ 0 and m/ e U ϭ ϩ1 components in the 2p state of atomic hydrogen placed m B ϭ in an external field of 2.00 T. 2m 1.602 ϫ 10Ϫ19 C 1.055 ϫ 10Ϫ34 J # s Strategy To find the Bohr magneton we insert the known ϭ Ϫ31 256 Chapter 7 The Hydrogen Atom 2 9.11 ϫ 10 kg values of e, U, and m into the equation for mB [see text after 1 21 2 Ϫ24 Equation (7.28)]. The energy difference is determined m B ϭ 9.27 ϫ 10 1 J/T 2 (7.32) from EquationThe international (7.31). system of units has been used (T ϭ tesla where ⌬m/ ϭ 1 Ϫ 0 ϭ 1. Hence, we have for magnetic field). The energy splitting is determined from E ϭ 9.27 ϫ 10Ϫ24 J/T 2.00 T ϭ 1.85 ϫ 10Ϫ23 J Equation (7.31) (see also Figure 7.6a): ¢ Ϫ4 1 21 2 ϭ 1.16 ϫ 10 eV ¢E ϭ mBB ¢m/ (7.33) Ϫ4 16/11/2018 An energy differenceJinniu of 10 Hu eV is easily observed by optical means. 03721_ch07_241-271.indd 255 9/29/11 4:44 PM

The splitting of spectral lines, called the normal Zeeman effect, can be partially explained by the application of external magnetic fields (see Figure 7.6). When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states (Figure 7.6a) with the energy difference given by Equation (7.33). A transition for an electron in the excited 2p level to the 1s ground state results in three different energy transitions as shown (greatly exaggerated) in Figure 7.6b. The energy differences between the three spectral lines shown in Figure 7.6b are quite small and were first observed by Pieter Zeeman in 1896. The application of external magnetic fields eliminates much of the energy degeneracy, because quantized states that previously had the same energy now have slight differences. When electrons make the transition between these states, the photons absorbed or produced have more widely varying energies. We will see in Section 7.6 that the selection rule for m/ does not allow more than AIP/Emilio Segrè Visual Archives, Segrè Collection. three different spectral lines in the normal Zeeman effect (see Problem 33). Efforts were begun in the 1920s to detect the effects of space quantization (m ) Otto Stern (1888– 1969) was / born in a part of Germany (now by measuring the energy difference ⌬E as in Example 7.7. In 1922 O. Stern and W. in Poland), where he was edu- Gerlach reported the results of an experiment that clearly showed evidence for cated. He worked in several uni- space quantization. If an external magnetic field is inhomogeneous—for example, versities until he left Germany in if it is stronger at the south pole than at the north pole—then there will be a net 1933 to avoid Nazi persecution force on a magnet placed in the field as well as a torque. This force is represented and emigrated to the United in Figure 7.7, where the net force on m (direction of S to N in bar magnet) is dif- States. He was educated and ferent for different orientations of m in the inhomogeneous magnetic field B. trained as a theorist but changed to experimentation when he be- gan his molecular beam experiments in 1920 at the University of Frankfurt with Walter Gerlach. He continued his distinguished ca- reer in Hamburg and later at South Carnegie Institute of Technology B in Pittsburgh. He received the No- greater bel Prize in 1943.

Net! N S force Figure 7.7 An inhomogeneous S N magnetic ﬁeld is created by the Net! smaller south pole. Two bar mag- force B nets representing atomic mag- lower netic moments have m in opposite North directions. Because the force on the top of the bar magnets is greater than that on the bottom, there will be a net translational force on the bar magnets (atoms).

03721_ch07_241-271.indd 256 9/29/11 4:44 PM Exercise class

7. Show that an energy difference of 2*10-3eV for the 3p subshell of sodium accounts for the 0.6nm splitting of a spectral line at 589.3 nm.

16/11/2018 Jinniu Hu 8.2 Total Angular Momentum 285

EXAMPLE 8.3 8.2 Total Angular Momentum 285 8.2 Total Angular Momentum 285 Exercise class Show that an energy difference of 2 ϫ 10Ϫ3 eV for the 3p Then, letting ⌬E ϭ dE and ⌬l ϭ dl and taking absolute EXAMPLE 8.3 EXAMPLE 8.3 subshell of7. sodiumShow that accounts an energy for difference the 0.6-nm of 2*10 splitting-3eV for of the a 3pvalues yields spectral line at 589.3 nm. subshell of sodium accounts for the 0.6nm splitting of a hc l2 E ϭ l or l ϭ E Show that an energy difference of 2 ϫ 10Ϫ3 eV for the 3p spectralThen, lettingline at ⌬589.3E ϭ nm.dE and ⌬l ϭ dl and taking absolute ¢ 2 ¢ ¢ ¢ Ϫ l hc Showsubshell that an of energy sodium difference accounts of for 2 theϫ 10 0.6-nm3 eV for splitting theStrategy 3 p of aThen, values The letting wavelength yields ⌬E ϭ dEl of and a photon ⌬l ϭ d isl andrelated taking to the absolute Solution: since 0 0 0 0 0 0 0 0 subshellspectral of line sodium at 589.3 accounts nm. for the 0.6-nm splittingenergy of a ofvalues a transition yields by Solution We insert the values of l, ⌬E, and hc to obtain hc l2 spectral line at 589.3 nm. E ϭ l or l ϭ2 E 2 Ϫ3 ¢ hc 2hc¢ ¢ l ¢ 589.3 nm 2 ϫ 10 eV E ϭ l l or l ϭ hcE Strategy The wavelength l of a photon is related to the ¢ E2ϭ¢ ¢ ¢ ¢l ϭ ϭ 0.6 nm l l hc ϫ 3 Strategyenergy of The a transition wavelength by l of a photon is related to the Solution We0 insert0 the0 values0 of 0 l, ⌬0 E, and0 hc to0 obtain 1.240 10 eV # nm 0 0 0 0 0 0 0 0 1 2 1 2 energy of a transition by For a smallSolution splitting, We insert we the approximate values of l , ⌬⌬EE, and by hc using to obtain a 0 0 hc 589.3 nm 2 2 ϫ 10Ϫ3 eV The value of 0.6 nm agrees with the experimental measure- E ϭ dif ferential: l ϭ 2 Ϫ3 ϭ 0.6 nm hc l ¢ 589.3 nm 2 ϫ 310 eV ment for sodium. E ϭ ¢l ϭ 1.240 ϫ 10 eV # nm ϭ 0.6 nm l 1.2401 Ϫhϫc 1023 eV1 nm 2 For a small splitting, we approximate ⌬E by using a 0 0 1 2 1 # 2 The value of dE0.6 ϭnm agrees2 dl with the experimental measure- Fordif a ferential: small splitting, we approximate ⌬E by using a The value0 of0 0.6 nm agreesl with the experimental measurement for sodium. Jinniu Hu dif ferential: ment16/11/2018 for sodium. Ϫhc dE ϭ dl Ϫhc l2 dE ϭ dl l2 Many-Electron Atoms The interaction of the various spins and angular momenta becomes formidable Many-Electron Atoms Many-Electron Atoms for more than two electrons outside an inert core. Various empirical rules help The interaction of the various spins and angularin applyingmomenta the becomes quantization formidable results to such atoms. The best-known rule set is Thefor interaction more than of two the electrons various spins outside and an angular inert called core.momenta Various Hund’s becomes empirical rules, formidable introduced rules help in 1925 by the German physicist Friedrich Hund for inmore applying than twothe electronsquantization outside results an toinert such core. atoms.(1896– Various The1997), empirical best-known who is rules known rule help set mainly is for his work on the electronic structure of in applyingcalled Hund’s the quantization rules, introduced results in to1925 such by atoms.theatoms German The and best-knownphysicist molecules. Friedrich rule We set Hund consideris here the case of two electrons outside a called(1896– Hund’s 1997), rules, who introduced is known mainlyin 1925 forby thehis Germanworkclosed on physicistshellthe electronic (for Friedrich example, structure Hund helium of and the alkaline earths).* (1896–atoms1997), and who molecules. is known We mainly consider for his here work the on case Thethe of electronic order two electrons in whichstructure outside a given of a subshell is filled is governed by Hund’s rules: atomsclosed and shell molecules. (for example, We consider helium and here the the alkaline case of earths).* two electrons outside a closed shellThe order(for example, in which helium a given andsubshell the alkaline is filled earths).*isRule governed 1. The by Hund’stotal spin rules: angular momentum S should be maximized to the The order in which a given subshell is filled is governed byextent Hund’s possible rules: without violating the Pauli exclusion principle. Rule 1. The total spin angular momentum S Ruleshould 2. be Insofar maximized as rule to 1the is not violated, L should also be maximized. Hund’s rules Rule 1. Theextent total possiblespin angular without momentum violating Sthe should Pauli beexclusion maximized principle. to the Rule 3. For atoms having subshells less than half full, J should be minimized. Rule extent2. Insofar possible as rule without 1 is not violating violated, the L shouldPauli exclusion also be maximized. principle. Hund’s rules RuleRule 2. 3. Insofar For atoms as rule having 1 is not subshells violated, less L than shouldFor half example,also full, be J should maximized. the befirst minimized. five electronsHund’s to rulesoccupy a d subshell should all have the Rule 3. For atoms having subshells less than half full, J should be minimized. For example, the first five electrons to occupysame a d subshellvalue of shouldms. This all requires have the that each one has a different m/ (because the al- For example, the first five electrons to occupy alowed d subshell m/ values should are all Ϫ have2, Ϫ 1,the 0, 1, 2). By rule 2 the first two electrons to occupy a same value of ms. This requires that each one has a different m/ (because the al- same value of m . This requires that each one hasd a subshell different should m (because have mthe ϭ al- 2 and m ϭ 1 or m ϭ Ϫ2 and m ϭ Ϫ1. lowed m/ valuess are Ϫ2, Ϫ1, 0, 1, 2). By rule 2 the first two electrons/ to occupy/ a / / / lowed m values are Ϫ2, Ϫ1, 0, 1, 2). By rule 2 the firstBesides two electrons the spin-orbit to occupy interaction a already discussed, there are now spin-spin d subshell/ should have m/ ϭ 2 and m/ ϭ 1 or m/ ϭ Ϫ2 and m/ ϭ Ϫ1. d subshellBesides should the have spin-orbit m/ ϭ 2interaction and m/ ϭ 1already or m/ andϭdiscussed, Ϫ 2 orbital-orbital and therem/ ϭ Ϫare1. now interactions. spin-spin There are also effects due to the spin of the andBesides orbital-orbital the spin-orbit interactions. interaction There already are discussed, alsonucleus effects there that due arelead to now the to spin-spinhyperfine spin of the structure (see Special Topic “Hydrogen and the andnucleus orbital-orbital that lead interactions. to hyperfine Therestructure are (see also Special 21-cm effects LineTopic due toTransition” “Hydrogen the spin of inand theChapter the 7), but the nuclear effects are much smaller nucleus21-cm that Line lead Transition” to hyperfine in Chapterstructure 7), (see but Special thethan nuclear Topic the oneseffects “Hydrogen we are are much currentlyand smaller the considering. For the two-electron atom, we label 21-cm Line Transition” in Chapter 7), but the nuclear effects are much smaller than the ones we are currently considering. Forthe the electrons two-electron 1 and atom, 2 so thatwe label we have L1, S1 and L2, S2. The total angular momen- thanthe the electrons ones we 1 are and currently 2 so that considering.we have L1, S For1 and thetum L2 ,two-electron SJ 2is. The the totalvector atom,angular sum we of momen- label the four angular momenta: thetum electrons J is the 1 vectorand 2 sosum that of we the have four L angular1, S1 and momenta: L2, S2. The total angular momentum J is the vector sum of the four angular momenta: J ϭ L1 ϩ L 2 ϩ S1 ϩ S 2 (8.9) J ϭ L1 ϩ L 2 ϩ S1 ϩ S 2 (8.9) J ϭ L1 ϩ L 2 ϩ S1 ϩThereS 2 are two schemes, called (8.9) LS coupling and jj coupling, for combining the There are two schemes, called LS coupling andfour jj angularcoupling ,momenta for combining to form the J. We shall discuss these next. The decision of Therefour are angular two schemes, momenta called to form LS couplingJ. We shall and discuss jj coupling these, next.for combining The decision the of four angular momenta to form J. We shall discuss these next. The decision of *See H. G. Kuhn, Atomic Spectra, 2nd ed., New York: Academic Press (1969), or H. E. White, Introduc- *See H. G. Kuhn, Atomic Spectra, 2nd ed., New York: Academic Press (1969), or H. E. White, Introduc- tion to Atomic Spectra, New York: McGraw-Hill (1934) for further study. *Seetion H. toG. Atomic Kuhn, Spectra Atomic, SpectraNew York:, 2nd McGraw-Hill ed., New York: (1934) Academic for further Press (1969),study. or H. E. White, Introduc- tion to Atomic Spectra, New York: McGraw-Hill (1934) for further study.

03721_ch08_272-297.indd 285 03721_ch08_272-297.indd 285 9/29/11 10:12 AM 9/29/11 10:12 AM 03721_ch08_272-297.indd 285 9/29/11 10:12 AM Exercise class

8. If the spin-orbit splitting of the 3P3/2 and 3P1/2 states of sodium is 0.0021 eV, what is the internal magnetic ﬁeld causing the splitting?

16/11/2018 Jinniu Hu 8.2 Total Angular Momentum 291

Solution The possibilities are Energy (eV) 1 1 Singlet Triplet 1s 2s L ϭ 0 0

If S ϭ 0, then J ϭ 0 Ϫ1 1 If ϭ 1, then ϭ 1 1s3s 3 S 0 3 S J Ϫ2 3 S 1 with S ϭ 1 being lowest in energy. The lowest excited state Ϫ3 3 1 1 is S1 and then comes S0. 1s2p 2 P 1 3 Ϫ4 1 2 P 0,1,2 1 1 1s2s 2 S 0 1s 2p L ϭ 1 3 Ϫ5 2 S 1 If S ϭ 0, then J ϭ 1 If S ϭ 1, then J ϭ 0, 1, 2

3 Ϫ23 The state P0 has the lowest energy of these states, followed 3 3 1 by P1, P2, and P1. The energy-level diagram for helium is Ϫ24 shown in Figure 8.13. 2 1 Ϫ25 (1s) 1 S 0 (Ground state)

Figure 8.13 The low-lying atomic states of helium are shown. 1 The ground state ( S0) is some 20 eV below the grouping of the 3 lowest excited states. The level indicated by P0,1,2 is actually three 3 3 3 states ( P0, P1, P2), but the separations are too small to be indicated.

EXAMPLE 8.8 Exercise class

If the spin-orbit splitting of the 3P3/2 and 3P1/2 states of so- 8. intoIf the Equation spin-orbit (8.17) splitting to find of B ,the because 3P3/2 and we are3P 1/2 given states the of dium is 0.0021 eV, what is the internal magnetic field caus- sodiumenergy is splitting 0.0021 ⌬eV,E. what is the internal magnetic field ing the splitting? causing the splitting? Solution The difference in spins between the 3P3/2 and Strategy The potential energy due to the spin magnetic Solution:3P1/2 states is U so that moment is e U U e U ¢E ϭ gs B ϭ B V ϭϪ ms # B (8.17) 2m U m By analogy with Equations (7.34), the z component of the Then a b total magnetic moment is Ϫ31 m ¢E 9.11 ϫ 10 kg 0.0021 eV B ϭ ϭ e Jz Ϫ19 Ϫ16 U e U 1.6 ϫ 10 C 6.58 ϫ 10 eV # s mz ϭϪgz (8.18) 1 2 1 2 2m U ϭ 181 T 2 1 2 where we have used the gyromagnetica b ratio g ϭ 2, because s This is a large magnetic field, as internal magnetic fi elds this splitting is actually due to spin. We associate the energy often are. splitting, denoted ⌬E, with the potential energy, ⌬E ϭ V. We 16/11/2018 Jinniu Hu determine the value of mz in Equation (8.18) and insert this

EXAMPLE 8.9

What are the possible energy states for atomic carbon? / ϭ 1, so we have L ϭ 0, 1, or 2 using the LS coupling scheme. The spin angular momentum is S ϭ 0 or 1. Strategy The element carbon has two 2p subshell electrons outside the closed 2s2 subshell. Both electrons have

03721_ch08_272-297.indd 291 9/29/11 10:12 AM