Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
,Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
Chapter 1
The Schrödinger Equation
1.1 (a) F; (b) T; (c) T.
1.2 (a) Ephoton = h = hc/ = (6.626 × 10–34 J s)(2.998 × 108 m/s)/(1064 × 10–9 m) =
1.867 × 10–19 J.
(b) E = (5 × 106 J/s)(2 × 10–8 s) = 0.1 J = n(1.867 × 10–19 J) and n = 5 × 1017.
1.3 Use of Ephoton = hc/ gives
(6.022 1023)(6.626 10−34 J s)(2.998 108 m/s)
E= = 399 kJ
300 10−9 m
1.4 (a) Tmax = h − =
(6.626 × 10–34 J s)(2.998 × 108 m/s)/(200 × 10–9 m) – (2.75 eV)(1.602 × 10–19 J/eV) =
5.53 × 10–19 J = 3.45 eV.
(b) The minimum photon energy needed to produce the photoelectric effect is
(2.75 eV)(1.602 × 10–19 J/eV) = hν =hc/λ = (6.626 × 10–34 J s)(2.998 × 108 m/s)/λ
and λ = 4.51 × 10–7 m = 451 nm.
(c) Since the impure metal has a smaller work function, there will be more energy left
over after the electron escapes and the maximum T is larger for impure Na.
1.5 (a) At high frequencies, we have eb /T 1 and the −1 in the denominator of Planck’s
formula can be neglected to give Wien’s formula.
(b) The Taylor series for the exponential function is ex = 1 + x + x2 /2! + ⋯. For x 1,
we can neglect x2 and higher powers to give ex − 1 x. Taking x h /kT , we have for
Planck’s formula at low frequencies
a 3 2 h 3 2 h 3 2 2kT
= =
eb /T − 1 c2 (eh /kT − 1) c2 (h /kT ) c2
1.6 = h/ mv = 137h/ mc = 137(6.626 × 10–34 J s)/(9.109 × 10–31 kg)(2.998 × 108 m/s) =
3.32 × 10–10 m = 0.332 nm.
1-1
Copyright © 2014 Pearson Education, Inc.
,Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
1.7 Integration gives x = − 21 gt 2 + (gt0 + v0 )t + c2. If we know that the particle had position
x0 at time t0, then x0 = − 1 gt 2 + (gt0 + v0 )t0 + c2 and c2 = x0 − 1 gt 2 − v0t0. Substitution
2 0 2 0
of the expression for c2 into the equation for x gives x = x0 − 21 g(t − t0 )2 + v0 (t − t0 ).
2
/h
1.8 −(h / i)( / t) = −(h2 /2m)(2/x2 ) + V . For = ae−ibte−bmx , we find
/t = −ib , / x = −2bmh−1x, and 2 / x2 = −2bmh−1 − 2bmh−1x(/x)
= −2bmh−1 − 2bmh−1x(−2bmh−1x) = −2bmh−1 + 4b2m2h−2x2 . Substituting into the
time-dependent Schrödinger equation and then dividing by Ψ, we get
−(h / i)(−ib) = −(hm)(−2bmh−1 + 4b2m2h−2 x2 ) + V and V = 2b2mx2 .
1.9 (a) F; (b) F. (These statements are valid only for stationary states.)
2 2
1.10 ψ satisfies the time-independent Schrödinger (1.19). / x = be−cx −2bcx2e−cx ;
2 2 2 2 2
2 / x2 = −2bcxe−cx − 4bcxe−cx + 4bc2x3e−cx = −6bcxe−cx + 4bc2x3e−cx . Equation
2 2 2 2
(1.19) becomes (−h2 /2m)(−6bcxe−cx + 4bc2x3e−cx ) + (2c2h2x2 /m)bxe−cx = Ebxe−cx .
The x3 terms cancel and E = 3h2c/m =
3(6.626 × 10–34 J s)22.00(10–9 m)–2/4π2(1.00 × 10–30 kg) = 6.67 × 10–20 J.
1.11 Only the time-dependent equation.
1.12 (a) | |2 dx = (2/b3)x2e−2|x|/b dx =
2(3.0 10−9 m)−3(0.90 10−9 m)2e−2(0.90 nm)/(3.0 nm) (0.0001 10−9 m) = 3.29 × 10–6.
(b) For x2 0, we have | = x2 −and
| 2xnm the probability is given by (1.23) and (A.7) as
2 nm
|| dx = (2 / b3) x e 2x / b dx = (2 / b3)e−2x /b (−bx2 /2 − xb2 /2 − b3 /4) |2 nm =
0 0 0
−e−2x/b (x2 /b2 + x/b + 1/2) |20 nm = −e−4/3(4/9 + 2/3 + 1/2) + 1/2 = 0.0753.
(c) Ψ is zero at x =0, and this is the minimum possible probability density.
0
− | | dx = (2/b3)
dx + (2/b3) x2e−2x/ b dx. Let w = –x in the first
− x e 0
2 2 2x/ b
(d)
w2e−2w /b (−dw) = w2e−2w /b dw, which
0
integral on the right. This integral becomes 0
equals the second integral on the right [see Eq. (4.10)]. Hence
|| 2 dx = (4 / b3) x2e−2x/b dx = (4 / b3)[2!/ (b / 2)3] = 1, where (A.8) in the
− 0
Appendix was used.
1-2
Copyright © 2014 Pearson Education, Inc.
, Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
1.13 The interval is small enough to be considered infinitesimal (since changes negligibly
2
/ c2
within this interval). At t = 0, we have | |2 dx = (32 / c6 )1/2 x2e−2x dx =
6 1/2 2 –2
[32/π(2.00 Å) ] (2.00 Å) e (0.001 Å) = 0.000216.
a || dx =
b 1.5001 nm
1.14 a−1e−2 x / a dx = −e−2 x / a /2 | 1.5000 nm= (–e–3.0002 + e–3.0000)/2 =
2 1.5001 nm
1.5000 nm
–6
4.978 × 10 .
1.15 (a) This function is not real and cannot be a probability density.
(b) This function is negative when x < 0 and cannot be a probability density.
(c) This function is not normalized (unless b = ) and can’t be a probability density.
1.16 (a) There are four equally probable cases for two children: BB, BG, GB, GG, where the
first letter gives the gender of the older child. The BB possibility is eliminated by the
given information. Of the remaining three possibilities BG, GB, GG, only one has two
girls, so the probability that they have two girls is 1/3.
(b) The fact that the older child is a girl eliminates the BB and BG cases, leaving GB and
GG, so the probability is 1/2 that the younger child is a girl.
1.17 The 138 peak arises from the case 12C12CF6, whose probability is (0.9889)2 = 0.9779.
The 139 peak arises from the cases 12C13CF6 and 13C12CF6, whose probability is
(0.9889)(0.0111) + (0.0111)(0.9889) = 0.02195. The 140 peak arises from 13C13CF6,
whose probability is (0.0111)2 = 0.000123. (As a check, these add to 1.) The 139 peak
height is (0.02195/0.9779)100 = 2.24. The 140 peak height is (0.000123/0.9779)100 =
0.0126.
1.18 There are 26 cards, 2 spades and 24 nonspades, to be distributed between B and D.
Imagine that 13 cards, picked at random from the 26, are dealt to B. The probability that
every card dealt to B is a nonspade is 24 23 22 21 ⋯ 14 13 12 = 13(12) = 6 . Likewise, the
26 25 24 23 16 15 14 26(25) 25
6
probability that D gets 13 nonspades is 25
. If B does not get all nonspades and D does not
get all nonspades, then each must get one of the two spades and the probability that each
gets one spade is 1 − 256 − 256 = 13 /25 . (A commonly given answer is: There are four
possible outcomes, namely, both spades to B, both spades to D, spade 1 to B and spade 2
to D, spade 2 to B and spade 1 to D, so the probability that each gets one spade is 2/4 =
1/2. This answer is wrong, because the four outcomes are not all equally likely.)
1-3
Copyright © 2014 Pearson Education, Inc.
Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
,Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
Chapter 1
The Schrödinger Equation
1.1 (a) F; (b) T; (c) T.
1.2 (a) Ephoton = h = hc/ = (6.626 × 10–34 J s)(2.998 × 108 m/s)/(1064 × 10–9 m) =
1.867 × 10–19 J.
(b) E = (5 × 106 J/s)(2 × 10–8 s) = 0.1 J = n(1.867 × 10–19 J) and n = 5 × 1017.
1.3 Use of Ephoton = hc/ gives
(6.022 1023)(6.626 10−34 J s)(2.998 108 m/s)
E= = 399 kJ
300 10−9 m
1.4 (a) Tmax = h − =
(6.626 × 10–34 J s)(2.998 × 108 m/s)/(200 × 10–9 m) – (2.75 eV)(1.602 × 10–19 J/eV) =
5.53 × 10–19 J = 3.45 eV.
(b) The minimum photon energy needed to produce the photoelectric effect is
(2.75 eV)(1.602 × 10–19 J/eV) = hν =hc/λ = (6.626 × 10–34 J s)(2.998 × 108 m/s)/λ
and λ = 4.51 × 10–7 m = 451 nm.
(c) Since the impure metal has a smaller work function, there will be more energy left
over after the electron escapes and the maximum T is larger for impure Na.
1.5 (a) At high frequencies, we have eb /T 1 and the −1 in the denominator of Planck’s
formula can be neglected to give Wien’s formula.
(b) The Taylor series for the exponential function is ex = 1 + x + x2 /2! + ⋯. For x 1,
we can neglect x2 and higher powers to give ex − 1 x. Taking x h /kT , we have for
Planck’s formula at low frequencies
a 3 2 h 3 2 h 3 2 2kT
= =
eb /T − 1 c2 (eh /kT − 1) c2 (h /kT ) c2
1.6 = h/ mv = 137h/ mc = 137(6.626 × 10–34 J s)/(9.109 × 10–31 kg)(2.998 × 108 m/s) =
3.32 × 10–10 m = 0.332 nm.
1-1
Copyright © 2014 Pearson Education, Inc.
,Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
1.7 Integration gives x = − 21 gt 2 + (gt0 + v0 )t + c2. If we know that the particle had position
x0 at time t0, then x0 = − 1 gt 2 + (gt0 + v0 )t0 + c2 and c2 = x0 − 1 gt 2 − v0t0. Substitution
2 0 2 0
of the expression for c2 into the equation for x gives x = x0 − 21 g(t − t0 )2 + v0 (t − t0 ).
2
/h
1.8 −(h / i)( / t) = −(h2 /2m)(2/x2 ) + V . For = ae−ibte−bmx , we find
/t = −ib , / x = −2bmh−1x, and 2 / x2 = −2bmh−1 − 2bmh−1x(/x)
= −2bmh−1 − 2bmh−1x(−2bmh−1x) = −2bmh−1 + 4b2m2h−2x2 . Substituting into the
time-dependent Schrödinger equation and then dividing by Ψ, we get
−(h / i)(−ib) = −(hm)(−2bmh−1 + 4b2m2h−2 x2 ) + V and V = 2b2mx2 .
1.9 (a) F; (b) F. (These statements are valid only for stationary states.)
2 2
1.10 ψ satisfies the time-independent Schrödinger (1.19). / x = be−cx −2bcx2e−cx ;
2 2 2 2 2
2 / x2 = −2bcxe−cx − 4bcxe−cx + 4bc2x3e−cx = −6bcxe−cx + 4bc2x3e−cx . Equation
2 2 2 2
(1.19) becomes (−h2 /2m)(−6bcxe−cx + 4bc2x3e−cx ) + (2c2h2x2 /m)bxe−cx = Ebxe−cx .
The x3 terms cancel and E = 3h2c/m =
3(6.626 × 10–34 J s)22.00(10–9 m)–2/4π2(1.00 × 10–30 kg) = 6.67 × 10–20 J.
1.11 Only the time-dependent equation.
1.12 (a) | |2 dx = (2/b3)x2e−2|x|/b dx =
2(3.0 10−9 m)−3(0.90 10−9 m)2e−2(0.90 nm)/(3.0 nm) (0.0001 10−9 m) = 3.29 × 10–6.
(b) For x2 0, we have | = x2 −and
| 2xnm the probability is given by (1.23) and (A.7) as
2 nm
|| dx = (2 / b3) x e 2x / b dx = (2 / b3)e−2x /b (−bx2 /2 − xb2 /2 − b3 /4) |2 nm =
0 0 0
−e−2x/b (x2 /b2 + x/b + 1/2) |20 nm = −e−4/3(4/9 + 2/3 + 1/2) + 1/2 = 0.0753.
(c) Ψ is zero at x =0, and this is the minimum possible probability density.
0
− | | dx = (2/b3)
dx + (2/b3) x2e−2x/ b dx. Let w = –x in the first
− x e 0
2 2 2x/ b
(d)
w2e−2w /b (−dw) = w2e−2w /b dw, which
0
integral on the right. This integral becomes 0
equals the second integral on the right [see Eq. (4.10)]. Hence
|| 2 dx = (4 / b3) x2e−2x/b dx = (4 / b3)[2!/ (b / 2)3] = 1, where (A.8) in the
− 0
Appendix was used.
1-2
Copyright © 2014 Pearson Education, Inc.
, Solutions Manual for Quantum Chemistry 7th Edition by Ira N. Levin
1.13 The interval is small enough to be considered infinitesimal (since changes negligibly
2
/ c2
within this interval). At t = 0, we have | |2 dx = (32 / c6 )1/2 x2e−2x dx =
6 1/2 2 –2
[32/π(2.00 Å) ] (2.00 Å) e (0.001 Å) = 0.000216.
a || dx =
b 1.5001 nm
1.14 a−1e−2 x / a dx = −e−2 x / a /2 | 1.5000 nm= (–e–3.0002 + e–3.0000)/2 =
2 1.5001 nm
1.5000 nm
–6
4.978 × 10 .
1.15 (a) This function is not real and cannot be a probability density.
(b) This function is negative when x < 0 and cannot be a probability density.
(c) This function is not normalized (unless b = ) and can’t be a probability density.
1.16 (a) There are four equally probable cases for two children: BB, BG, GB, GG, where the
first letter gives the gender of the older child. The BB possibility is eliminated by the
given information. Of the remaining three possibilities BG, GB, GG, only one has two
girls, so the probability that they have two girls is 1/3.
(b) The fact that the older child is a girl eliminates the BB and BG cases, leaving GB and
GG, so the probability is 1/2 that the younger child is a girl.
1.17 The 138 peak arises from the case 12C12CF6, whose probability is (0.9889)2 = 0.9779.
The 139 peak arises from the cases 12C13CF6 and 13C12CF6, whose probability is
(0.9889)(0.0111) + (0.0111)(0.9889) = 0.02195. The 140 peak arises from 13C13CF6,
whose probability is (0.0111)2 = 0.000123. (As a check, these add to 1.) The 139 peak
height is (0.02195/0.9779)100 = 2.24. The 140 peak height is (0.000123/0.9779)100 =
0.0126.
1.18 There are 26 cards, 2 spades and 24 nonspades, to be distributed between B and D.
Imagine that 13 cards, picked at random from the 26, are dealt to B. The probability that
every card dealt to B is a nonspade is 24 23 22 21 ⋯ 14 13 12 = 13(12) = 6 . Likewise, the
26 25 24 23 16 15 14 26(25) 25
6
probability that D gets 13 nonspades is 25
. If B does not get all nonspades and D does not
get all nonspades, then each must get one of the two spades and the probability that each
gets one spade is 1 − 256 − 256 = 13 /25 . (A commonly given answer is: There are four
possible outcomes, namely, both spades to B, both spades to D, spade 1 to B and spade 2
to D, spade 2 to B and spade 1 to D, so the probability that each gets one spade is 2/4 =
1/2. This answer is wrong, because the four outcomes are not all equally likely.)
1-3
Copyright © 2014 Pearson Education, Inc.
