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42.1: a)
K)J1038.1(3
)eVJ1060.1)(eV109.7(2
3
2
2
3
23
194
k
K
TkTK
K1.6
T
b)
K.600,34
)KJ1038.1(3
)eVJ1060.1()eV48.4(2
23
19
T
c) The thermal energy associated with room temperature (300 K) is much greater than
the bond energy of
2
He
(calculated in part (a)), so the typical collision at room
temperature will be more than enough to break up
.He
2
However, the thermal energy at
300 K is much less than the bond energy of
2
H
, so we would expect it to remain intact at
room temperature.
42.2: a)
.eV0.5
4
1
2
0
r
e
πε
U
b)
.eV2.4)eV5.3eV3.4(eV0.5
42.3: Let 1 refer to C and 2 to O.
nm1128.0,kg10656.2,kg10993.1
0
26
2
26
1
rmm
)carbon(nm0644.0
0
21
2
1
r
mm
m
r
)carbon(nm0484.0
0
21
1
2
r
mm
m
r
b)
;mkg1045.1
2462
22
2
11
rmrmI
yes, this agrees with Example 42.2.
42.4: The energy of the emitted photon is
,eV1001.1
5
and so its frequency and
wavelength are
GHz44.2
)sJ1063.6(
)eVJ1060.1)(eV1001.1(
34
195
h
E
f
.m123.0
Hz)1044.2(
)sm1000.3(
λ
9
8
f
c
This frequency corresponds to that given for a microwave oven.
42.5: a) From Example 42.2,
24623
1
mkg10449.1andJ10674.7meV479.0
IE
srad1003.12givesand
12
1
2
2
1
IEωEKIωK
b)
(carbon)sm3.66)srad1003.1)(m100644.0(
129
111
ωrv
(oxygen)sm8.49)srad1003.1)(m100484.0(
129
222
ωrv
c)
s106.102
12
ωπT
42.6: a)
J10083.2)eV2690.0(
2
1
2
1
20
0
ωE
sm1091.1
kg10139.1
)J10083.2(2
2
gives
2
1
3
26
2
r
0
max
2
maxr0
m
E
vvmE
b) According the Eq. 42.7 the spacing between adjacent vibrational energy levels
is twice the ground state energy:
.
,)
2
1
(
1
hfωEEE
ωnE
nn
n
Thus, using the
E
specified in Example 42.3, it follows that its vibrational
period is
s.1054.1
)eVJ1060.1)(eV2690.0(
)sJ1063.6(1
14
19
34
E
h
f
T
c)
The vibrational period is shorter than the rotational period.
42.7: a)
2
HLi
HLi
2
r
r
mm
mm
rmI
eV.1053.7J1020.1
mkg1069.3
)sJ10054.1(4
4
))13()3()14(4(
2
mkg1069.3
)kg1067.1kg1017.1(
)m1059.1)(kg1067.1)(kg1017.1(
321
247
234
22
34
247
2726
2102726
E
E
II
EEE
b)
.m1066.1λ
J1020.1
s)m1000.3)(sJ1063.6(
λ
4
21
834
E
hc
42.8: Each atom has a mass m and is at a distance
2L
from the center, so the moment
of inertia is
.mkg1021.22)2)((2
24422
mLLm
42.9: a)
I
l
llll
I
E
I
ll
E
I
ll
E
ll
2
22
22
1
2
)(
2
2
)1(
,
2
)1(
b)
.
πI
l
π
E
h
E
f
2
2
ΔΔ
42.10: a)
,mk
hc
E
r
λ
and solving for
,k
m.N205
λ
2
r
2
m
πc
k
42.11: Energy levels are
I
ll
ωnEEE
ln
2
)1(
2
1
2
eV)10395.2()1(eV)269.0(
2
1
4
lln
where the values are from
Examples 42.2 and 42.3.
a)
:2,11,0
lnln
.m10592.4
)eVJ10602.1)(eV2700.0(
)sm10998.2)(sJ10626.6(
λ
eV2700.0)eV10395.2)(4()eV2690.0)(1(
6
19
834
4
E
hc
EEE
if
b)
:1,12,0
lnln
m10627.4λ
)eVJ10602.1)(eV2680.0(
)sm10998.2)(sJ10626.6(
λ
eV2680.0)eV10395.2)(4()eV2690.0)(1(
6
19
834
4
E
hc
EEE
if
c)
:2,13,0
lnln
m.10634.4
)eVJ10602.1)(eV2676.0(
)sm10998.2)(sJ10626.6(
λ
eV2676.0)eV10395.2)(6()eV2690.0)(1(
6
19
834
4
ΔE
hc
EEE
if
42.12:
2where,eV196.0J1014.32
r
20
r
mmmkmkω
has been
used.
42.13: a)
r
2
1
2
m
k
ππ
ω
f
mN963
)kg1015.3kg1067.1(
)]Hz1024.1(2)[kg1015.3()kg1067.1(
2
2
2627
2142627
21
2
21
2
r
k
π
k
mm
πfmm
πfmk
b)
hfωωnωnE
2
1
2
3
eV513.0
J1022.8Hz1024.1sJ1063.6
201434
E
c)
infraredm1042.2
Hz1024.1
sm1000.3
λ
6
14
8
f
c
42.14: a) As a photon,
nm.200.0
eVJ1060.1eV1020.6
sm1000.3sJ1063.6
λ
193
834
E
hc
b) As a matter wave,
nm200.0
eVJ1060.1eV6.37kg1011.92
sJ1063.6
2
λ
1931
34
mE
h
p
h
and
c) as a matter wave,
nm200.0
eVJ1060.1eV0205.0kg1067.12
sJ1063.6
2
λ
1927
34
mE
h
.
42.15: The volume enclosing a single sodium and chlorine atom
3
10
m102.822
329
m104.49
. So the density
.mkg102.16
m1049.4
kg105.89kg1082.3
33
329
2626
ClNa
ρ
V
mm
ρ
42.16: For an average spacing a, the density is
3
amρ
, where m
is the average of the
ionic masses, and so
329
33
2526
3
m1060.3
)mkg1075.2(
2kg1033.1kg1049.6
ρ
m
a
,
and
nm330.0m1030.3
10
a
. b)
The larger (higher atomic number) atoms have the
larger spacing.
42.17:
J1014.2
m1031.9
)sm1000.3()sJ1063.6(
λ
13
13
834
hc
E
eV101.34
6
So the number of electrons that can be excited to the conduction band is
eV12.1
eV1034.1
6
n
6
1020.1
electrons.
42.18: a)
nm227m1027.2
7
E
hc
, in the ultraviolet.
b) Visible light lacks enough energy to excite the electrons into the conduction
band, so visible light passes through the diamond unabsorbed.
c) Impurities can lower the gap energy making it easier for the material to absorb
shorter wavelength visible light. This allows longer wavelength visible light to pass
through, giving the diamond color.
42.19: a) To be detected the photon must have enough energy to bridge the gap width
eV12.1
E
m1011.1
eVJ1060.1eV12.1
sm1000.3sJ1063.6
λ
6
19
834
E
hc
, in the infrared.
b) Visible photons have more than enough energy to excite electrons from the
valence to conduction band. Thus visible light is absorbed, making silicon opaque.
42.20:
sm1017.13
5
rms
mkTv
, as found in Example 42.9. The equipartition
theorem does not hold for the electrons at the Fermi energy. Although these electrons are
very energetic, they cannot lose energy, unlike electrons in a free electron gas.
42.21: a)
Eψ
z
ψ
y
ψ
x
ψ
m
2
2
2
2
2
22
2
where
L
πzn
L
πyn
L
πxn
A
ψ
z
y
x
sinsinsin
,
2
2
2
ψ
L
πn
x
ψ
x
and similarly for
2
2
2
2
and
z
ψ
y
ψ
.
2
22222
2
2
2
2
2
2
mL
πnnn
E
E
ψψ
L
πn
L
πn
L
πn
m
zyx
z
y
x
b) Ground state
2
22
2
3
1
mL
π
Ennn
zyx
,
The only degeneracy is from the two spin states. The first excited state
1
,2,1or1,1,2
2
22
3
2,1,1or
mL
π
E
and the degeneracy is
632
.
The second excited state
2
22
2
9
2,2,1or2,1,2or1,2,2
mL
π
E
and the
degeneracy is
632
.
42.22:
dVψ
2
1
,
2
sinsinsin
3
2
0
2
0
2
0
22
L
A
dz
L
πzn
dy
L
πyn
dx
L
πxn
A
L
z
L
y
L
x
so
23
2 LA
(assuming A to be real positive).
42.23: Density of states:
eV.states105.1
eVJ1060.1Jstates105.9
s)J10054.1(2
)eVJ101.60(eV)5.0)(m101.0(kg))109.11(2(
2
2
22
1940
3342
211921362331
21
32
23
π
Eg
E
π
Vm
Eg
42.24: Equation (42.13) may be solved for
πLmEn
21
rs
2
, and substituting this
into Eq. (42.12), using
VL
3
, gives Eq. (42.14).
42.25: Eq.(42.13):
2
222
rs
2
mL
πn
E
.103.4
)eVJ1060.1()eV7.0()kg109.11(2
)sJ101.054(
010.0
2
7
rs
1931
34
rs
n
π
m
mE
π
L
n
42.26: a) From Eq. (42.22),
.eV94.1
5
3
Fav
EE
b)
.sm108.25
kg109.11
eVJ101.601.942
2
5
31
19
mE
c)
.K1074.3
KJ1038.1
eVJ1060.1eV23.3
4
23
19
F
k
E
42.27: a)
R
π
R
E
kT
π
C
V
eVJ1060.1eV48.52
K300KJ1038.1
2
19
232
F
2
K.molJ194.00233.0 RC
V
b)
3
1068.7
KmolJ25.3
KmolJ0.194
.
c) Mostly ions (see Section 18.4).
42.28: a) See Example 42.10: The probabilities are
67
1037.2,1078.1
, and
5
1051.1
. b) The Fermi distribution, Eq. (42.17), has the property that
EfEEf 1
F
(see Problem (42.46)), and so the probability that a state at the top
of the valence band is occupied is the same as the probability that a state of the bottom of
the conduction band is filled (this result depends on having the Fermi energy in the
middle of the gap).
42.29:
1
1
F
kTEE
e
Ef
.eV20.0J1020.3
1
104.4
1
lnK300KJ1038.1
1
1
ln
20
F
4
23
F
F
EEE
EE
Ef
kTEE
So the Fermi level is 0.20 eV below the conduction band.
42.30: a) Solving Eq. (42.23) for the voltage as a function of current,
.V0645.01
mA3.60
mA40.0
ln1ln
S
e
kT
I
I
e
kT
V
b) From part (a), the quantity
11.12
kTeV
e
, so far a reverse-bias voltage of the
same magnitude,
mA30.31
11.12
1
1
SS
IeII
kTeV
.
42.31:
1
SS
1
1
kTeV
kTeV
e
IeII
a)
580.0
K300KJ101.38
V1050.1C1060.1
23
219
kT
eV
A.0118.0
1
A1025.9
580.0
3
S
e
I
Now for
387.0,V0100.0
kTeVV
mA56.5A1056.51A0118.0
3387.0
eI
b) Now with
580.0,mV0.15
kT
eV
V
A1018.51A0118.0
3580.0
eI
.
If
387.0mV0.10
kT
eV
V
A1077.31A0118.0
3387.0
eI
.
42.32: See Problem (42.7):
248
2
2
mkg1014.7
2
λ2
c
π
h
E
I
.
42.33: a)
mC103.8m102.4C101.60
291019
qdp
b)
C10.31
m102.4
mC103.0
19
10
29
d
p
q
c)
78.0eq
d)
0590
C104.9
m101.6
mC101.5
12
10
30
.eq
d
p
q
This is much less than for sodium chloride (part (c)). Therefore the bond for hydrogen
iodide is more covalent in nature than ionic.
[...]... cm 2.297 cm 0.050 cm E 1.12 1014 Hz The force 42. 42: The vibration frequency is, from Eq (42. 8), f h constant is k (2πf ) 2 mr 777 N m 1 k 1 2k E0 42. 43: En n 2 mr 2 mH E0 1 2(576 N m) (1.054 10 34 J s) 4.38 10 20 J 0.274 eV 2 1.67 10 27 kg This is much less than the H 2 bond energy 42. 44: a) The frequency is proportional to the reciprocal... (1.95) 234 μm 42. 38: From the result of Problem (42. 9), the moment inertia of the molecule is 2l hlλ I 2 6.43 10 46 kg m 2 E 4π c and from Eq (42. 6) the separation is r0 I 0.193 nm mr L2 2l (l 1) 2I 2I E g 0 (l 0), and there is an additional multiplicative factor of 2l + 1 because for each l state there are really 2l 1 ml states with the same energy 42. 39: a) Eex .. .42. 34: The electrical potential energy is U 5.13 eV, and r 42. 35: 1 e2 2.8 10 10 m 4πε0 U a) For maximum separation of Na and Cl for stability: U e2 5.1 eV 3.6 eV 1.60 10 19 J eV 2.40 10 19 J 4πε0 r r (1.60 10 19 C) 2 9.6 10 10 m 19 4πε0 (2.40 10 J ) b) For K and Br : U r 42. 36: e2 4.3 eV 3.5 eV ... 3 5 2m V V 2 dV 42. 51: a) Eav 32 3 π 4 3 2 5m 53 N V 53 P dEtot 32 3 π 4 3 2 N 5m V dV N 8.45 10 28 m 3 V 32 3 π 4 3 (1.054 10 34 J s) 2 p (8.45 10 28 m 3 ) 5 3 5(9.11 10 31 kg) b) 3.80 1010 Pa 3.76 105 atm (!) c) There is a large attractive force on the electrons by the copper ions 42. 52: a) From Problem (42. 51): 53 32 3 π 4 3 2... (6 12) hf 2I 3.00 108 m s c λ 4.40 10 6 m 2 3(3.96 1011 Hz) 6.93 1013 Hz 3 2I f 42. 46: The sum of the probabilities is f ( EF E ) f ( EF E ) 1 e E kT 1 E kT e 1 1 1 e E kT 1 E / kT e 1 1 e E kT 42. 47: Since potassium is a metal we approximate E F E F0 EF 32 3 π 4 3 2 n 2 3 2m But the electron concentration n ... 2.03 eV 31 2(9.11 10 kg) 42. 48: a) First we calculate the number-density of neutrons from the given mass-density: n (7.0 1017 kg m 3 ) / 1.67 10 27 kg neutron ) 4.2 10 44 m 3 Now use Eq 44.21 2 4 1 2 4 2 3 3 π 3 2 n 3 3 3 π 3 (6.63 10 34 J s 2π ) 2 (4.2 10 44 m 3 ) 3 EF 0 1.8 10 11 J 27 2m 2(1.67 10 kg ) b) Set kT EF0 (see Exercise 42. 26) to obtain EF 0 (1.8 ... isotope with the deuterium atom is 12 12 m m (mH mF ) 1 (mF mD ) f f0 F H m m ( m m ) f 0 1 (m m ) F D D F F H Using f 0 from Exercise (42. 13) and the given masses, f 8.99 1013 Hz 42. 45: a) I mr r 2 mH mI r 2 (1.657 10 27 kg) (0.160 10 9 m) 2 mH mI 4.24 10 47 kg m 2 b) Vibration-rotation energy levels are: 1 k 2 h El l... 1035 m 3 400 so relativistic effects 1.67 1033 m 3 42. 54: a) Following the hint, 1 e2 1 e2 k dr d dr 4π r 2 2π 0 r03 0 r r 0 and ω 2k m 1 e2 1.23 10 19 J 0.77 eV, π 0 mr03 where (m 2) has been used for the reduced mass b) The reduced mass is doubled, and the energy is reduced by a factor of 42. 55: a) U 1 4π 0 i j qi q j rij q2 4π 0 2 to 0.54... result of Problem (44-4), these are seen to correspond to transition from levels 8, 7, 6, 5 and 4 to the respective next lower levels 2 Then, 0.410 10 21 J , from which I 2.71 10 47 kg m 2 I 42. 37: a) Pr (44.36) yields I 2.71 10 47 kg m 2 , and so r r I mr I (mH mCl ) mH mCl (2.71 10 47 kg m 2 ) (1.67 10 27 kg 5.81 10 26 kg) (1.67 10 27 kg 5.81 10 26 kg)... 34 J s 2π ) 2 (4.2 10 44 m 3 ) 3 EF 0 1.8 10 11 J 27 2m 2(1.67 10 kg ) b) Set kT EF0 (see Exercise 42. 26) to obtain EF 0 (1.8 10 11 J ) 1.3 1012 K (1.38 10 28 J K ) k 42. 49: a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells So there are 1 8 8 2 atoms per unit cell 2 n 4.66 10 28 atoms m 3 3 9 .
eV.states105.1
eVJ1060.1Jstates105.9
s)J10054.1(2
)eVJ101.60(eV)5.0)(m101.0(kg))109.11(2(
2
2
22
1940
3 342
211921362331
21
32
23
π
Eg
E
π
Vm
Eg
42. 24: Equation (42. 13) may be solved for
πLmEn
21
rs
2
, and substituting this
into Eq. (42. 12), using
VL
3
, gives Eq. (42. 14).
42. 25:. are:
cm.050.0cm2.297cm347.2λ:01
cm.025.0cm1.148cm173.1
λ:12
ll
ll
42. 42: The vibration frequency is, from Eq. (42. 8),
14
1012.1
h
E
f
Hz. The force
constant is
.mN777)2(
r
2
mπfk
42. 43:
H
0
r
2
2
1
2
1
m
k
E
m
k
nE
n