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Unformatted text preview: Chapter 22: Quantum States for ManyElectron Atoms and Atomic
Spectroscopy Problem numbers in italics indicate that the solution is included in the Student’s Solutions
Manual. Questions on Concepts (222.1) The statement that the Coulomb integral J deﬁned in Equation (22.20) is positive by explicitly formulating the integral that describes the interaction between two negative classical charge clouds. J,,= e2 ”[1s(1)]2[ 1 ][23(2)]2drldrz 87:80 r1 — r2 6 2
1 62 1 __/ 1 I”? g /’)
:_ __ flag 2___~_ ’2 ~“od d9
V32 87Z280 [a0] ”e e[171—172]{ a0} e II T“ The angular coordinates integrate to give a factor 161:2, leaving only the integration over r1 and r2, which 1
_... ’i‘rz can vary between 0 and 00. The terms , e‘ﬁ/a" and 642/20” are positive as r1 and r2 vary over their allowed ranges. The only other factor involving r1 or r2 appears as the square, and is therefore
always positive. Therefore, the integrand is always positive, and we can conclude that J12 must be
positive. Q22.2) Without invoking equations, explain why the energy of the triplet state is lower than that of the singlet state for He in the 151251 configuration. The probability density is different in the singlet and triplet states. For the triplet state, the electrons are
on average closer to the nucleus. This lowers the electrostatic energy arising from the attractive
interaction between the electrons and the nucleus. It also increases the energy arising from the electron
electron repulsion, but the net effect is to decrease the total energy of the atom in the triplet state. Q22.3) How can the width of a laser line be less than that determined by Doppler broadening? The laser line is narrowed further than what arises from Doppler broadening as a result of the action of
the optical resonator. Not all lasing frequencies are enhanced through the resonator, thereby leading to a
narrowing of the frequencies emitted. Q22.4) Why is an electronically excited atom more reactive than the same ground—state atom? It is more reactive because it contains excess energy in the form of electronic energy that can be used to overcome an activation barrier. 484 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy Q22.5) Why is atomic absorption spectroscopy more sensitive in many applications than atomic
emission spectroscopy? Absorption spectra rely on the population of lowlying states whereas emission spectroscopy relies on
the population of excited states. The lowlying states have a higher population, with few exceptions.
Therefore absorption spectroscopy generally has a better signal to noise ratio than emission
spectroscopy. Q22.6) Why does the Doppler effect lead to a shift in the wavelength of a star, but to a broadening of a
transition in a gas? The wavelength of a star is shifted because all the photon emitters are moving in the same direction with
respect to the observer. In a gas, the atoms move randomly and with equal probability in all directions
with respect to the observer. This leads to a broadening of the frequency rather than a shift. Q22.7) Why are n, l, m], and m not good quantum numbers for many electron atoms? They are only good quantum numbers for one electron atoms or ions. Once electron—electron repulsion
is taken into account, L, S, ML, and M5 are the quantums that are valid for many electron atoms. Q22.8) Write an equation giving the relationship between the Rydberg constant for H and for Li2+. . . R . .
Because the Rydberg constant varies as the reduced mass, y, the ratio H IS given by
R 12+
L1
H Li“
RH : mnuclcar melecrron + mmlclear
R m”+ m + mH L i2+ nuclear electron nuclear Q22.9) Can the individual states in Table 22.1 be distinguished experimentally? No. The energy level that corresponds to an individual term can be determined and in the limit of J—J
coupling in the presence of an external ﬁeld, the energy levels corresponding to individual states can be
determined. However, the states themselves cannot be determined. 022.10) How is it possible to determine the L and S value of a term knowing only the ML and M5 values
of the states? The fact that —S S M S S +S and — L S M ,1 S +L allows us to determine the L and S values of a term as shown in Section 22.2.
Q22.11) What is the origin of the chemical shift in XPS?
The origin of the chemical shift in XPS is shielding. If electrons are withdrawn from the atom of interest, the binding energy of the remaining electrons will increase because the orbital is less shielded. This leads to a decrease in the kinetic energy of the emitted electron. 485 Chapter 22/ Quantum States for Many—Electron Atoms and Atomic Spectroscopy Q22.12) Why are two medium—energy photons rather than one highenergy photon used in laser isotope separation? One high energy photon would ionize all atoms, regardless of their mass. Two photons are needed to
utilize the fact that the excited state energies are different for the various isotopes. Q22.l3) Why does one need to put a sample in a vacuum chamber to study it with XPS or AES? The electrons will contain the information on the atom from which they were ejected only if they do not
undergo inelastic collisions with gas phase atoms. Therefore, the sample needs to be in a vacuum.
Q22.l4) Why is XPS a surfacesensitive technique? XPS is a surfacesensitive technique because only electrons that do not undergo inelastic collisions
before they exit into the vacuum have an energy characteristic of the atom from which they were
ejected. Because the inelastic mean free path is on the order of a nanometer, the surface sensitivity is
very high. Q22.15) Explain the direction of the chemical shifts for Fe(0), Fe(ll), and F e(IlI) in Figure 22.20. As electrons are removed from Fe to form Fe2+ or Fey, the Fe 2p electrons are less well shielded from
the nuclear charge. Therefore, they will experience a higher binding energy. Because the shielding is less effective for F e3: it will exhibit the greatest binding energy. Problems P22.1) The principal line in the emission spectrum of potassium is violet. On close examination, the
line is seen to be a doublet with wavelengths of 393.366 and 396.847 nm. Explain the source of this
doublet. The lower lying state is the level and state 281/2, and the upper level is 2P, which contains the states 2P1/2
and 2P3/2. Both transitions are allowed, and the 589.0 nm wavelength corresponds to the transition
between the ground state and 2P3/2 because states with higher J lie lower in energy. The 589.6 nm
wavelength corresponds to the transition between the ground state and 2P1 /2. P222) The absorption spectrum of the hydrogen atom shows lines at 5334, 7804, 9145, 9953, and
10,478 cm‘l. There are no lower frequency lines in the spectrum. Use the graphical methods discussed
in Example Problem 22.6 to determine ”initial and the ionization energy of the hydrogen atom in this state. Assume values for 11mm] of l, 2, and 3. 486 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy ~ 1 . . .
The slope of a plot of v versus assumed values of , W111 have a slope of 7RH and an Intercept With
11,
flnal
. RH . .
the frequency ax1s of 2 . However, both ninmal and nﬁna/ are unknown, so that 1n plotting the data, we
n Imlial have to assign nﬁm] values to the observed frequencies. Because there are no lower frequency lines in the spectrum, we know that the lowest value for nﬁm, is nmm, + 1. We try different combinations of nﬁm; RH
2
Initial .In values and see if the slope and intercept are consistent with the expected values of —RH and
n this case, we assume that the sequence of spectral lines corresponds to nﬁnal = 2, 3, 4, 5, and 6 for an
assumed value of 11mm]: 1, nfma; = 3, 4, 5, 6, and 7 for an assumed value of nimial : 2, and ”ﬁnal 2 4, 5, 6, 7, and 8 for an assumed value of nimm; = 3. The plots are shown below. Frequency/(cm'l ninitiaFZ 0.05 0.1 0.15 0.2 0.25 2
i/“final The slopes and intercepts calculated for these assumed values of I’llmm] are: Assumed n,,,,,,a/ Slope (cm1) Intercept (cm‘l)
1 —2.24 x 104 1.07 x 104
2 —5.63><104 l.15><104
3 —1.09x105 1.22x104 By examining the consistency of these values with the expected values, we conclude that ”initial = 3. The thH
32 , corresponding to n ﬁnal ionization energy of the hydrogen atom in this state is —>00, or 2.42x10719J. 487 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy P223 Using Table 22.3, which lists the possible terms that arise from a given conﬁguration, and Hund’s rules, write the term symbols for the ground state of the atoms H through F in the form (25 HE J. We use Hund’s rule that the term with the highest multiplicity is the lowest in energy to get the left
superscript. For the right subscript, if there are several choices for J, the lowest J value gives the lowest
energy if the subshell is less than half full, and the highest J value gives the lowest energy if the subshell is exactly or more than half full. Applying these rules gives rise to the following term symbols. Atom Conﬁgurationl Ground state term symbol
H J lsl 281/2 He 152 1So Li ls22sI [281/2 _
Be 1522s2 —l 1So ‘—‘
B i ls22522pl 2PM C 1522522p2 3P0 N 1522522193 433/2 0 1322s22p4 3P2 F i1392522105 2P3/2 j P22.4) In this problem you will supply the missing steps in the derivation of the formula
E sin g! e! : E13, + E25 + J + K for the singlet level of the 1312s1 conﬁguration of He.
a. Expand Equation (22.17) to obtain i”[1s(1)2s(2)+25(1)1s(2)](H1)
2 [ls(l)2s (2)+2s(l)ls (2)]drld12 E ﬂing/cl : + _1_ ”[15 (1)28 (2) + ZS (1)15 (2)](192)
2 [ls (1) 2s (2) + ZS (l) ls (2)] drldrz 1% j j [ls(1) 251(2) + 23(1) 1s (2)] [—4—] 47:50 rl —r2 [ls(l)25(2)+2s(l)ls(2)]drldrz 488 Chapter 22/ Quantum States for Many—Electron Atoms and Atomic Spectroscopy The integral n+=*11115() 2(2 ><+2s >(2>1121+12+ [13(1)2S2)(1+2S(”2)](1da2 4m:O 111“ r 1 can be expressed as the sum of three integrals, each containing one of the three operators. This gives the desired result.
b. Starting from the equations H11S(i) = El_\_lS(i) and H [.2S(i) = E232S(i), show that E Sing/cl : E  .s + E2S +é ”[1S (1) 2s (2) + 2s (1)1s (2)](—43;] 47r601rl —r2 [1s(1)2S(2)+2S(l)1S(2)]drldz'2 [lSWS( 2)(+2S(1)ls(2 )]a’rla’r2 EmgltZ—UDSU) 2S(2 )(+2S )1(1)1S(2)](1€{1+:€;2 +
:311[ls(1)2s(2))(1+283(2)](H‘)DS(1)2 )l+23()(1S 2 )]dmzr2 +lﬂ[ls(l)2s(2)+2S(l 52()H](2)[1S(1)2 2+(2Sl (2)_]drldr +—11[1s(1)1)22(s )(+2s1(1)1:(2)]]2[ Letting the operators act on the following wave function gives 47rgo 1:1— r 21 1( 2(11
47rsolrl— r21 S]:2((1)2ss )+ S( )S(2)]a’1,oilr2 =%]][1s(1)2s(2) )(+2s1 (2)L][EY 15(1 2)+E 2s (1)1s(2) ]drldr singlet +12”[1s(1)2s(2)+ 2S(1)1S(2)]1:E2X1S(1)2S(2)+ E 2s(1 (2)]dr drz +— 211M )(223 )1+(2s (1)1 S)(2)]( 92 ][1S(1)2S()ls(2)(1)+2s (2)]517121222 47I6‘0 Ir1 — r21
Expanding the previous expression gives
1
Esmgm 2 5E” 11[1s(1)2s(2)+ 2s(1(2)(1)][1s2s(2)]c1212172 +5151.) H[1S(1)2S()+2S(1)1 (2)][2S(l) 1S (2)]alrldr2 +1E2S ]][1s(1)2s)(+2s(1)1s(2) )(1][1s) 2s (2)]2126122 +%E25H1:13(1)2S)1+2(s(2)(1)][25 1s (2)]drdr2 [1s()( 2)+2S( (2)]drla’r2 é1111s<1>2s(2>+2s<1>12<2111 472'8011'1— r 1 489 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy Each ofthe ﬁrst four integrals of the previous expression can be expanded into the sum oftwo integrals. For example, lEHI III 1s(1)2s(2)+2.1(1)15(2)][11(1)2s(2)]arr,a/r2 :—E1(IJ[ls(l)2s(2)[ls(l)(2)]clrdTZ+;EHII[25(1)1S(2)][15(1)2S(2)]d2'ldr2 Because the orbitals are orthogonal and normalized, the ﬁrst integral has the value one, and the second has the value zero. Therefore, the previous expression can be simpliﬁed to
l 1 l 1 E11111 : +—E,s +—Ely +—E23 +—E2X
. 2 2 2 2
1L; III:15(1)23(2)+23(l)ls(2)][ZEOL—ZJDS(1)23(2)+25(l)ls(2)]drldrz
Ewe, 2E“. +E2,1%II[1s(1)2s(2)+2s(1)11(2)]I47m——0———‘IIr _r ):ls(12)( ).+23' (1)1 3(2)]d21d2, This is the desired result 0. Expand the previous equation using the deﬁnitions 28:50 le[15(1)]2 [If] —r2I][2S(2)] dz‘ldfz and =8:802HI18(1) 25 (2)][1 ][15 (2)23 (1)]drldrz =ELY+E23+J+K. lr1—r2l to obtain the desired result, E xmg/cl 7
. . e‘
We expand the term 1nvolv1ng the operator ——
47:50 rl — r I Because the operator involves only multiplication the order ofthe terms is unimportant, giving
47750Ir17 r7 éIII1.1(1)21(2)+21(1)1.1(2)__]I
:_;II.(1)[11)2S(2)]2[R_2__]1121dr2 +2 III2s(1)1s(2)]2 [mjdrldfz r2 + II[1S(1)23(2)] [4ﬂ80l:1“ —l‘ 2"]l280) 15(2)]dZ’1dT2 Because the labels in the ﬁrst two terms are irrelevant to the value for the integral, ——.(1)]l:lr25( 2)(l+2s ()ls(2)]drldr2 We can switch the labels 1 and 2 in the second integral. 1111 1—111 er II21(1)11(2)]dr,dr, 47:50 Irl r, :J+K
490 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy Adding E“. + E23, to this result gives the desired answer. P22.5) What .1 values are possible for a 6H term? Calculate the number of states associated with each level and show that the total number of states is the same as that calculated from (2S + 1)(2L + 1).
S = 5/2, L = 5.
J lies between L + S and lL — Sl and can have the values 15/2, 13/2, 11/2, 9/2, 7/2, and 5/2. The number of states is 2.1+ 1 or 16, 14, 12, 10, 8, and 6, respectively. This gives a total number of states of 16 + 14 + 12 + 10 + 8 + 6 = 66. (2S+1)(2L + 1) = 6x11 = 66 also. P22.6) Using Table 22.3, which lists the possible terms that arise from a given conﬁguration, and
Hund’s rules, write the conﬁgurations and term symbols for the ground state of the ions F and Ca2 + in the form (254.1)”. We use Hund’s rule that the term with the highest multiplicity is the lowest in energy to get the left
superscript. For the right subscript, if there are several choices for J, the lowest J value gives the lowest
energy if the subshell is less than half full, and the highest J value gives the lowest energy if the subshell
is exactly or more than half full. Applying these rules gives rise to the following term symbols. F': 1522522p6, 180 Ca2+: ls22s22p63sz3p6, 180 P22.7) The Doppler broadening in a gas can be expressed as A v = 2VO /CW , where M is the molar mass. For the sodium 3p 2PM ——> 33 231/2 transition, v0 25.0933x1014 s". Calculate Av and Av/V0 at 500. K. 14 71 e23 *1
AV:_2_t: 2kT1n222x5.0933x10 s x 21n2x1.381><10 JK XSOO'K:I.7O><109 3,1 8 ~l 727
c m 2.998x10 ms 22.99 amux1.661><10 kg am U 9 ,1
ﬂ_ 1.70><10 s :333X10, v _ 5.0933x10'4 s"1 P22.8) Calculate the transition dipole moment, pf” = 11/1; (T) ,uz w” (2') d1 where ,u: : —er cos/9 for a transition from the ls level to the 2pz level in H. Show that this transition is allowed. The integration is over r, (9, and (Ii Use 3/2
1 l r _, a
W210<r9 9: ¢) : m [a—] :6 /2 0 C059
0 0 for the 2192 wave function. 491 Chapter 22/ Quantum States for Many—Electron Atoms and Atomic Spectroscopy
,uf‘ : —e Jwgyzwldr = A5327! J‘sin 9d6 {ﬂy/gm (r)r cos 91/45, (r)dr
0 0 3/2 3/2
I 1 1 1 ” °° r .
:‘327I — ~—— —— sinﬂcoslﬁdﬂ r2 ——e“’/2”“r0050] e"/"0 dr
Wield rial l J i l l 3 I7 00
= —e 2:5[1—J Jsin Hcoslﬁdﬂ Jrz [fer/2“” ]r0036(e’/"" )dr 0 3 00 r
—1 003319” r4 e—%’“ dr
tr iM ]
0 :—e
2J2ia0]
__el ilgx 4! _ e >(2X41? 1 4a5i_l28\/2ea
zﬁa03352ﬁ335a0° 2430
270 Because the transition dipole moment is different from zero, the transition is allowed.
P22.9) Consider the Is np 3 P —> Is nd 3 D transition in He. Draw an energylevel diagram, taking the spinorbit coupling that splits terms into levels into account. Into how many levels does each term split? The selection rule for transitions in this case is AJ = 0,i1 . How many transitions will be observed in an absorption spectrum? Show the allowed transitions in your energy diagram. For the 3 P term, L = l, S = 1, and J can have the values 2, 1, and 0. There will be three levels. For the 3 D
term, L : 2, S : l, and J can have the values 3, 2, and 1. There will be three levels. For the given selection rules, there will be 6 transitions as indicated in the ﬁgure below. Energy 492 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy P22.10) Atomic emission experiments of a mixture show a calcium line at 422.673 nm corresponding to
a lP1 —> ISO transition and a doublet due to potassium 2P30 —> 281/2 and ZPl/z —> 281/2 transitions at 764.494 and 769.901 nm, respectively. a. Calculate the ratio gamer /g,0W for each of these transitions. b. Calculate nupper /n,0wer for a temperature of 1600. °C for each transition. a) Calculate the ratio gupp” for each of these transitions.
glower
gupper ‘ 2Jupper +1
g lower 2Jlowur + 1
ll) er 3 u) 8" 4
‘Pl—>' 50: g” =— 2P —>ZS g'” =—=2 glowcr 1 / 5% glower 2 7 g 2
.. P1—)ZS upper 1
/—) SA gluwcr 2 n .
b) Calculate W” for each transition at a temperature of 1600. °C.
”lower ”upper __ guppcr e—N%T g g upper eihc%'li ”lower g/(m'er glower
1P _) IS I ”upper :g__uppere h/l_ _3€Xp 6.626X10_34 J SX2.998X108 m SA]
1 0 ”7...... game 1.381x10‘23 JK“ x1873 Kx422.673><10"9 m
= 3.86 ><10”8
2 P _> 25 . ”1W g__,.,,,..re J... /,_ _26Xp_ 6.626x10'34 J sx2.998x108 m s"
% %’ nW gwee, 1.381x10‘23J K“ x1873 K x 764.494x10"9m
: 8.67 ><10'5
11...... 8...... m ,, 6.626x10‘34 J s><2.998><108m s“‘
2% A 25%: n“ 2 H e A] :eXp[ 1381x10‘” JK“><1873 K><769 901x10‘9
lower glower ‘ ’ 1n
= 4.65 ><10‘5 P22.11) How many ways are there to place three electrons into an f subshell? What is the ground—state term for the f conﬁguration, and how many states are associated with this term? See Problem 21.5. The ﬁrst electrons can have any combination of 7 m1 and 2 ms values so that n = 14 and m = 3. The I
number of states is ———1£—— = 364.
31(14 — 3)! Using the method discussed in Example Problem 21.7, MLmax = 6 and Mgmax = 3/2. Therefore the ground
state term is 4I, which has (2L + 1)(2S + 1) = (2 x 6 + 1)(4) = 52 states. 493 Chapter 22/ Quantum States for ManyElectron Atoms and Atomic Spectroscopy P22.12) Calculate the wavelengths of the ﬁrst three lines of the Lyman, Balmer. and Paschen series, and the series limit (the shortest wavelength) for each series. Lyman Series. E 2HR [—1 ~ 17—] 1 n" n22 E,[email protected]—l)=§RH=8zm7scm* 4=LM569mn
4 4 9 n=3 E,=RH[—1J=§RH=9wwo7an* 4:1mMW4nm
[ 16 J 2£R 2102822 cm 4 2 97.2553 nm —]: RH2109678cn1 4==9i17681nn . 1 1
Balmer Series: E" 2 RH [—27 — 71—2]
1 1 5
1123 E22RH ——— 2—6RH 2152329cm 22656473 nm
n2 3 E2 2RH [77;]1 2 ——RH — 20564.4 cm‘I xi 2 486.276 nm RH 223032. 2 cm /1 2 434.175 nm RH 2 27419.3 cm" A 2 364.707 nm 4 stlﬂ 100 Paschen Series: En 2 RH [517— _ 17] n 1 1 7 7224 E22RH ———— 2————RH2533152cm 421875.64nm
9 16 144
1 1 16 n2 5 E_ 2RH ——— 2—RH2779925cm 221282.17nm
9 25 225 n2 6 E2 2RH l~—1— 2LRH29139.75cm] 121094.12nm
9 36 12
1 1 1 _l n2 00 E2 2RH ——— 2—RH212186cm 42820.591nm
9 oo 9 P22.13) The Lyman series in the hydrogen atom corresponds to transitions that originate from the n 2 1 level in absorption or that terminate in the n = 1 level for emission. Calculate the energy, frequency (in 494 Chapter 22/ Quantum States for Many—Electron Atoms and Atomic Spectroscopy inverse seconds and inverse centimeters), and wavelength of the least and most energetic transition in this series. ~ 1 l
v = R H 2 — 2
ninilml ”ﬁnal 17max 2 RH [1—i]= 109678 cm‘1 1 oo E a, =th =6.626x10‘34 J sx2.998x10‘° ems—‘x109678 cm" =2.17872><10~18 1 [TI IHBX v =c17 =2.998><1010 cm s“><109678 cm’1 = 3.28813x10'5 5‘1 AW =;=9117.63 nm Vmax —1
ﬁlnin : RH (1—1;): 3XIO9678 cm : 82258.2 cmTl
1 2 4 Em = hcﬁmm = 6.626><10‘34 J sx2.998x10'° cms‘l x82302.8 cm" = 1.63404x10“18 1
vmin =617min =2.998><l()10 cm s‘lx82302.8 cmTl = 2.4661><10l5 s“l
’lmm 271—: 121.568 nm vmin P2214) The inelastic mean free path of electrons in a solid, 1, governs the surface sensitivity of
techniques such as ABS and XPS. The electrons generated below the surface must make their way to the
surface without losing energy in order to give elemental and chemical shift information. An empirical expression ...
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