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40.1: a)
2
234
2
2
1
2
22
m)kg)(1.520.0(8
s)J1063.6(
88
mL
h
E
mL
hn
E
n
J.1022.1
67
1
E
b)
sm101.1
kg0.20
J)102.1(22
2
1
33
67
2
m
E
vmvE
s.104.1
sm101.1
m5.1
33
33
v
d
t
c)
J.103.7J)1022.1(3
8
3
)14(
8
6767
2
2
2
2
12
mL
h
mL
h
EE
d) No. The spacing between energy levels is so small that the energy appears
continuous and the balls particle-like (as opposed to wave-like).
40.2:
1
8mE
h
L
m.104.6
)eVJ10eV)(1.60210kg)(5.0108(1.673
s)J10626.6(
15
19627
34
40.3:
)(8
3
8
3
)14(
8
12
2
2
2
2
12
EEm
hL
mL
h
mL
h
EE
)eVJ10eV)(1.60kg)(3.0108(9.11
3
s)J1063.6(
1931
34
nm.0.61m101.6
10
L
40.4:
a) The energy of the given photon is
J.1063.1
m)10(122
m/s)10(3.00
s)J1063.6(
λ
18
9
3
34
c
hhfE
The energy levels of a particle in a box are given by Eq. 40.9
J)10kg)(1.631011.9(8
)12(s)J1063.6(
8
)(
)(
8
2031
22234222
22
2
2
Em
mnh
Lmn
mL
h
E
m.1033.3
10
b) The ground state energy for an electron in a box of the calculated dimensions is
eV3.40J1043.5
19
m)10kg)(3.331011.9(8
)sJ1063.6(
8
21031
234
2
2
mL
h
E
(one-third of the original
photon energy), which does not correspond to the
eV6.13
ground state energy of the
hydrogen atom. Note that the energy levels for a particle in a box are proportional to
2
n
,
whereas the energy levels for the hydrogen atom are proportional to
2
1
n
.
40.5:
J105.2)smkg)(0.0101000.5(
723
2
1
2
2
1
mvE
b)
1
2
2
1
8
so
8
mE
h
L
mL
h
E
m106.6givesJ105.2
307
1
LE
40.6: a) The wave function for
1
n
vanishes only at
0
x
and
Lx
in the range
.0 Lx
b) In the range for
,
x
the sine term is a maximum only at the middle of the
box,
.2/Lx
c) The answers to parts (a) and (b) are consistent with the figure.
40.7: The first excited state or
)2(
n
wave function is
.
2
sin
2
)(
2
L
πx
L
x
ψ
So
.
2
sin
2
)(
2
2
2
L
πx
L
x
ψ
a) If the probability amplitude is zero, then
0
2
sin
2
L
πx
2,1,0,
2
2
m
Lm
xm
π
L
πx
So probability is zero for
.,
2
,0 L
L
x
b) The probability is largest if
.1
2
sin
L
πx
.
4
)12(
2
)12(
2
L
mx
π
m
L
πx
So probability is largest for
.
4
3
and
4
LL
x
c) These answers are consistent with the zeros and maxima of Fig. 40.5.
40.8: a) The third excited state is
,4
n
so
2
2
2
8
)14(
mL
h
E
eV.361J1078.5
m)10kg)(0.1251011.9(8
s)J10626.6(15
17
2931
234
b)
J105.78
m/s)10s)(3.0J1063.6(
λ
17
834
E
hc
nm.44.3λ
40.9: Recall
.
2
λ
mE
h
p
h
a)
m106.0m)100.3(22
8/2
λ
8
1010
22
1
2
2
1
L
mLmh
h
mL
h
E
m/skg101.1
m106.0
s)J1063.6(
λ
24
10
34
1
1
h
p
b)
m100.3λ
8
4
10
2
2
2
2
L
mL
h
E
m/s.kg102.22
λ
24
1
2
2
p
h
p
c)
m100.2
3
2
λ
8
9
10
3
2
2
3
L
mL
h
E
m/s.kg103.33
24
13
pp
40.10:
,
2
2
2
ψk
dx
ψd
and for
ψ
to be a solution of Eq.
.
28
),3.40(
22
2
2
m
E
h
mπ
Ek
b) The wave function must vanish at the rigid walls; the given function will vanish at
0
x
for any
,k
but to vanish at
nπkLLx
,
for integer
.
n
40.11: a)
.
8
:Eq.(40.3)
2
2
2
2
Eψ
dx
ψd
.
m
π
h
kxAkkxAk
dx
d
kxA
dx
d
ψ
dx
d
cos)sin()cos(
2
2
2
2
2
.
28
8
coscos
8
2
2
2
22
2
22
mE
h
mE
π
k
m
π
hk
E
kxEAkx
m
π
hAk
b) This is not an acceptable wave function for a box with rigid walls since we need
,0)()0(
Lψψ
but this
)(xψ
has maxima there. It doesn’t satisfy the boundary
condition.
40.12:
UCψ
dx
ψd
C
m
ψU
dx
ψd
m
2
22
2
22
2
2
,
2
2
22
ψEECψCEψ
Uψ
dx
ψd
m
C
and so
ψ
is a solution to
(40.1)Eq.
with the same energy.
40.13: a)
EψUψ
dx
ψd
m
2
22
2
:Eq.(40.1)
Left-hand side:
kxAUkxA
dx
d
m
sin)sin(
2
0
2
22
kxAUkxA
m
k
sinsin
2
0
22
.
2
0
22
ψU
m
k
But
EUU
m
k
00
22
2
for constant
.k
But
0
22
2
U
m
k
should equal
E
no solution.
b) If
,
0
UE
then
EU
m
k
0
22
2
is consistent and so
kxAψ sin
is a solution of
)1.40.(Eq
for this case.
40.14:
According to Eq.40.17, the wavelength of the electron inside of the square well is
given by
.
)3(2
λ
2
0
in
Um
hmE
k
By an analysis similar to that used to derive Eq.40.17, we can show that outside the box
.
)2(2)(2
λ
00
out
Um
h
UEm
h
Thus, the ratio of the wavelengths is
.
2
3
)2(2
)3(2
λ
λ
0
0
in
out
Um
Um
40.15:
J103.20eV00.2;
2
625.0625.0
19
1
2
22
1
E
mL
π
EE
m1043.3
J)10kg)(3.2010109.9(2
625.0
10
2/1
1931
πL
40.16: Since
EU 6
0
we can use the result
EE 625.0
1
from Section 40.3, so
EEU 375.5
10
and the maximum wavelength of the photon would be
h
cmL
mLh
hc
EU
hc
)375.5(
8
)8)(375.5(
λ
2
22
10
m.1038.1
s)J1063(5.375)(6.
m/s)1000.3(m)10kg)(1.501011.9(8
6
34
82931
40.17: Since
,6
0
EU
we can use the results from Section 40.3,
09
.5,625.0
31
EEE
and
21527
2342
2
22
m)10kg)(4.01067.1(2
s.)J10054.1(
2
π
mL
π
E
J.1005.2
12
E
The transition energy is
J.109.15J)1005.2)(625.009.5(
1212
13
EE
The wavelength of the photon absorbed is then
m.1017.2
J109.15
m/s)10s)(3.00J1063.6(
λ
14
12
834
13
EE
hc
40.18: Since
,6
0
EU
we can use the results from Section 40.3.
m1099.4
)sm10kg)(3.00108(9.11
m)10s)(4.55J1063.6)(805.1(
8
λ)805.1(
So
.
8
805.1
)625.043.2(
λ
.625.0and43.2
.
8
10
831
734
2
2
12
12
2
2
mc
h
L
mL
h
E
hc
EEE
EEEE
mL
h
E
40.19:
x
mE
Bx
m
A
ψ
2
cos
E2
sin:Eq.(40.16)
.Eq.(40.15))(
2
2
cos
22
sin
2
2
222
2
ψ
mE
x
mEmE
Bx
mEmE
A
dx
ψd
40.20:
),(
xx
DeCe
dx
dψ
ψDeCe
dx
ψd
xx 22
2
2
)(
for all constants
C
and
.D
Hence
ψ
is a solution to Eq. (40.1) for
,)](2[or,
2
2/1
00
2
2
EUmEU
m
and
is real for
.
0
UE
40.21:
L.e
U
E
U
E
T
EUm
)(22
00
0
116
eV11.0
eV0.6
0
U
E
and
J.108.0eV5
19
0
UE
a)
m1080.0
9
L
m)1080.0(
sJ10055.1
J)10kg)(8.01011.9(22
9
34
1931
eV11.0
ev0.6
1
eV11.0
eV0.6
16
eT
.104.4
8
b)
m1040.0
9
L
.102.4
4
T
40.22:
(Also see Problem 40.25). The transmission coefficient is
LEUm
e
U
E
U
E
T
)(22
00
0
116
with
andm,1060.0eV,0.5
9
LE
kg109.11m
31
a)
.105.5eV0.7
4
0
TU
b)
5
0
101.8eV0.9
TU
c)
.101.1eV0.13
7
0
TU
40.23:
KmKhph λso,2λ
is constant
12211
λ;λλ KK
and
1
K
are for
01
2where UKLx
and
2
λ
and
2
K
are for
002
where0 UUEKLx
2
1
2
λ
λ
0
0
1
2
2
1
U
U
K
K
40.24:
Using Eq. 40.21
56.2
eV15.0
eV0.12
1
eV15.0
eV0.12
16116
00
U
E
U
E
G
nm.26.0
0.025
2.56
ln
)m109.8(2
1
)ln(
2
1
m109.8
2s)J1063.6(
J/eV)10eV)(1.6012.0kg)(15.01011.9(2
)(2
18
2
19
34
1931
0
TGLGeT
π
EUm
L
40.25: a) Probability of tunneling is
L
GeT
2
where
74.2
41
32
1
41
32
16116
00
U
E
U
E
G
and
sJ10054.1
J/eV).10eV)(1.6032kg)(41eV1011.9(2
)(2
34
931
0
EUm
.102.174.274.2So
.m1054.1
37.7m)105.2)(m1054.1(2
110
10110
eeT
e
b) For a proton,
.
1011.9
1067.1
31
27
e
p
m
m
zero. toequalpurposespracticalallfor andisThe.10
74.274.2
m1059.6
143
330m)105.2)(m1059.6(2
111
10111
smallT
eeT
40.26:
,
)(2
and116with
0
00
2
EUm
U
E
U
E
GGeT
L
.116Giving
)(22
00
0
L
e
U
E
U
E
T
EUm
a) If
EUmLU
0
27156
0
andkg1064.6m,100.2eV,100.30
.090.0eV),100.29eV(100.1
66
TE
b) If
.014.0eV)100.20(eV100.10
66
0
TEEU
40.27: The ground state energy of a simple harmonic oscillator is, with
,0
n
eV.1038.1
J1022.22Also
eV106.91J1011.1
kg0.250
N/m110
2
s)J10055.1(
2
1
2
1
14
33
01
1533
0
34
0
EωEE
E
m
k
ωE
nn
Such tiny energies are unimportant for the motion of the block so quantum effects are not
important.
40.28: Let
aisand,24(and2soand,2
22
2
2
ψδ)ψδx
dx
ψd
ψx
dx
dψ
δ
km
ω./mkδ
m
E
2
1
2
1
ifEq.(40.21)ofsolution
2
40.29: The photon’s energy is
.
λ
hc
E
The transition energy is
m
k
ωωEEE
2
1
2
3
01
24
26282
2
22
m)10(5.25
kg)104.9()sm1000.3(4
λ
4
λ
2
πmcπ
k
m
kc
π
.mN2.1
k
40.30:
According to Eq.40.26, the energy released during the transition between two
adjacent levels is twice the ground state energy
eV.2.112
023
EωEE
For a photon of energy
E
nm.111
J/eV)10eV)(1.60(11.2
)sm10s)(3.00J.1063.6(
λ
19
834
E
hc
f
c
hfE
40.31: a)
ωnE
)(
2
1
For
J10544.1,1
21
mkωEn
m129)(λsoλ μEhchcE
b)
meV4.82J)10544.1(
21
2
1
2
1
0
ωE
40.32: a)
.368.0expexp
)0(
)(
12
2
2
e
k
ω
kmA
km
ψ
Aψ
This is more or less what is shown in Fig. (40.19).
b)
.1083.14exp)2(exp
)0(
)2(
242
2
2
e
k
ω
kmA
km
ψ
Aψ
This figure cannot be read this precisely, but the qualitative decrease in amplitude with
distance is clear.
40.33:
For an excited level of the harmonic oscillator
2
2
1
2
1
AkωnE
n
.
)12(
k
ωn
A
This is the uncertainty in the position.
Also
max
2
2
1
2
1
mvωnE
n
.)12()12(
)12(
)12(
)(
)12(
max
max
n
k
m
ωn
ωmn
k
ωn
mvApx
m
ωn
v
,)12(So
npx
which agrees for the ground state
.with)0(
pxn
The uncertainty is seen to increase with n.
40.34: a)
4/
0
4/
0
2
2
cos1
2
12
sin
2
LL
dx
L
πx
L
dx
L
πx
L
,
2
1
4
1
2
sin
2
1
4
0
π
L
πxL
x
L
L
about 0.0908.
b) Repeating with limits of
2and4 LL
gives
,
2
1
4
12
sin
2
1
2
4
L
L
L
πx
π
L
x
L
about 0.0409. c) The particle is much likely to be nearer the middle of the box than the
edge. d) The results sum to exactly
21
, which means that the particle is as likely to be
between
2and0 Lx
as it is to be between
.and2 LxLx
e) These results are
represented in Fig. (40.5b).
40.35: a)
4/3
4/
22
1
4/3
4/
sin
2
||
L
L
L
L
dx
L
πx
L
dx
ψP
.818.0
1
2
1
2sin
2
11
sin
2
Let
43
4
4/3
4/
2
π
zz
π
zdzP
L
π
dxdz
L
πx
z
π
π
b)
4/3
4/
2
4/3
4/
2
2
2
sin
2
||
L
L
L
L
dx
L
πx
L
dx
ψP
2
1
2sin
2
1
2
1
sin
1
23
2
2/3
2/
2
π
π
π
π
zz
π
dzz
π
P
c) This is consistent with Fig. 40.5(b) since more of
1
ψ
is between
than
4
3
and
4
ψ
LL
x
and the proportions appear correct.
40.36: Using the normalized wave function
,LπxLψ )sin(2
1
the probabilities
and2)2(sin)2(b),4sin)2()aare||
222
LdxdxπLLdx)dxπ(Ldxψ
.)43(sin)2(c)
2
LdxπL
40.37:
L
πx
L
x
ψ
2
sin
2
)(
2
dx
L
πx
L
dx
ψ
2
sin
2
||
22
2
a)
L
dx
dx
π
L
dx
ψ
L
x
2
2
sin
2
|:|
4
22
2
b)
0sin
2
|:|
2
22
2
dxπ
L
dxψ
L
x
c)
L
dx
dx
π
L
dx
ψ
L
x
2
2
3
sin
2
|:|
4
3
22
2
40.38: a)
.
1212)1(
222
22
n
n
n
n
n
nn
R
n
This is never larger than it is for
.3and,1
Rn
b) R approaches zero; in the classical
limit, there is no quantization, and the spacing of successive levels is vanishingly small
compared to the energy levels.
[...]... s) 40. 39: a) The transition energy E E2 E1 λ hc 8mcL2 8(9.11 10 31 E 3h λ 1.92 10 5 m 8(9.11 10 31 kg)(3.00 108 m s)(4.18 10 9 m) 2 b) λ 1.15 10 5 m 2 2 34 (3 2 )(6.63 10 J s) dψ π 2 dψ dψ cos(πx L) , and as x L, 0 b) 2π 2 L3 dx L L dx dx Clearly not In Eq (40. 1), any point where U (x) is singular necessitates a discontinuity in dψ dx 40. 40:... 2 m 2 2 b) Similar to the second graph in Fig 40. 18 40. 52: With u ( x) x 2 , a mω , | ψ |2 is of the form C *Cu(x)e -au(x) , which has a maximum mω 2 2 x 1, x m b) ψ and hence ψ vanish at x 0, and as x at au 1, or 2 ψ 0 the result of part (a) is A 3 , found from n 1 in Eq (40. 26), and this is consistent with Fig (40. 19) The figure also shows a minimum at x 0 and... detected 40. 50: a) E 1 2 mv (n (1 2))ω (n (1 2))hf , and solving for n, 2 1 2 mv 1 (1 2)(0.020 kg)(0.360 m s) 2 1 n 2 1.3 1030 34 2 (6.63 10 J s)(1.50 Hz) 2 hf b) The difference between energies is ω hf (6.63 10 34 J s)(1.50 Hz) 9.95 10 34 J This energy is too small to be detected with current technology 2 d 2ψ 1 2 k x ψ Eψ 40. 51: a) Eq (40. 21) : 2m... E 2 2mE k2 2 1 1 1 kL 2 1 since 2 d 2ψ k 2ψ , and using this in dx 2 Eq (40. 1) with U U 0 E gives k 2 2m( E U 0 ) 2 , or k i 2m(U 0 E ) i 40. 48: For a wave function ψ with wave number k , 40. 49: The angular frequency ω 2π 2π 12.6 rad s Ground state harmonic 0.500 s T oscillator energy is given by 1 1 E0 ω (1.054... state First excited state energy E100 E001 E010 40. 54: Let 1 k1 m , ω2 k 2 m , ψ nx ( x) be a solution of Eq (40. 21) with Enx 1 nx ω1 , ψ nx ( y ) be a similar solution, and ψ nz ( z ) be a solution of Eq.(42 - 25) but 2 1 with z as the independent variable instead of x, and energy Enz nz ω2 (a) As in 2 Problem 40. 53, look for a solution of the form ψ ( x, y, z... energy n2h2 En levels as given in Eq (40. 9), where 8mL2 d) Part (a)’s wave functions are odd, and part (b)’s are even En 40. 56: a) As with the particle in a box, ψ ( x) A sin kx, where A is a constant and k 2 2mE 2 Unlike the particle in a box, however, k and hence E do not have simple forms b) For x L, the wave function must have the form of Eq (40. 18) For the wave function to remain... U ( x)) 2 dx h Note that if E U (x) we are inside the well and A 0 From the above equation, this 40. 44: b) d 2ψ ( x ) implies has the opposite sign of ψ (x) dx 2 If E U (x), we are outside the well and A 0 From the above equation, this implies d 2ψ ( x ) has the same sign as ψ (x) dx 2 40. 45: a) We set the solutions for inside and outside the well equal to each other at the well boundaries,... L L3 dx 8π 2 8π 2 dψ 2 2πx cos for x L, since cos 2π 1 L3 dx L3 L e) The slope of the wave function is greatest for ψ 2 (n 2) close to the walls of the box, as shown in Fig 40. 5 d) 40. 42: p pfinal pinitial nπ hn p k L 2L hn ˆ i and the final momentum, At x 0 the initial momentum at the wall is pinitial 2L hn ˆ hn ˆ hn ˆ hn ˆ i So p i ... J s) dψ π 2 dψ dψ cos(πx L) , and as x L, 0 b) 2π 2 L3 dx L L dx dx Clearly not In Eq (40. 1), any point where U (x) is singular necessitates a discontinuity in dψ dx 40. 40: a) ψ 0, so 40. 41: a) ψ1 dψ 1 dx 2 πx sin L L 2π 2 1 πx 1 L3 2 L 2 2 π πx cos L L L dψ1 2π 2 for x 0 L3 dx dψ 2 2 2πx sin b) ψ 2 L L dx 2... where k b) Requiring continuous derivatives at the boundaries yields dψ x 0: kA cos(k 0) kB sin(k 0) kA Ce k 0 kA C dx x L : kA cos kL kB sin kL De L x L : A sin 40. 46: T Ge 2L with G 16 E U0 E 1 U and 0 2m(U 0 E ) 1 L 2 T ln If E 5.5 eV, U 0 10.0 eV, m 9.11 10 31 kg, and T 0.0010 G Then κ 2(9.11 10 31 .
EU
m
k
0
22
2
is consistent and so
kxAψ sin
is a solution of
)1 .40. (Eq
for this case.
40. 14:
According to Eq .40. 17, the wavelength of the electron inside of the square. 40. 3.
m1099.4
)sm10kg)(3.00108(9.11
m)10s)(4.55J1063.6)(805.1(
8
λ)805.1(
So
.
8
805.1
)625.043.2(
λ
.625.0and43.2
.
8
10
831
734
2
2
12
12
2
2
mc
h
L
mL
h
E
hc
EEE
EEEE
mL
h
E
40. 19:
x
mE
Bx
m
A
ψ
2
cos
E2
sin:Eq. (40. 16)
.Eq. (40. 15))(
2
2
cos
22
sin
2
2
222
2
ψ
mE
x
mEmE
Bx
mEmE
A
dx
ψd
40. 20:
),(
xx
DeCe
dx
dψ
ψDeCe
dx
ψd
xx