Tài liệu Physics exercises_solution: Chapter 38 doc

38.1:
Hz105.77
m
10
5.20
sm103.00
λ
14
7
8





c
f
eV.2.40J103.84)sm10(3.00s)mkg10(1.28
smkg101.28
m105.20
sJ106.63
λ
19827
27
7
34










pcE
h
p
38.2: a)
eV.107.49J0.0120s)10(20.0W)(0.600
163


Pt
b)
eV.1.90J103.05
λ
19


hc
hf
c)
.1094.3
16

hf
Pt
38.3: a)
Hz.105.91
sJ106.63

eV)J10(1.60eV)10(2.45
20
34
196






h
E
fhfE
b)
m.105.08
Hz105.91
sm103.00
λ
13
20
8





f
c
c)
λ

is of the same magnitude as a nuclear radius.
38.4:
hc
W
hc
P
hf
P
dNdE
dtdE
dt
dN
m)10(2.48)(12.0λ
)(
)(
7


sec.photons101.50
19

38.5:

 hfmv
2
max
2
1
J
10

3.04
eVJ10(1.60eV)(5.1
m102.35
sJ10(6.63
20
19
7
34
))

















c

.sm102.58
kg109.11

J)102(3.04
5
31
20
max






v
38.6:









m102.72
Hz101.45
λ
7
15
0
c
h

hc
hfhfE


eV.1.44J102.30
19


38.7: a)
Hz105.0λ
14
 cf
b) Each photon has energy
J.10.313
19
 hfE
Source emits
sphotons102.3photons/s)10(3.31)sJ(75sosJ75
2019


c) No, they are different. The frequency depends on the energy of each photon and the
number of photons per second depends on the power output of the source.
38.8: For red light
nm700λ

eV1.77
J101.6
1eV
J102.84

m)10(700
)sm10(3.00s)J10(6.626
λ
19
19
9
834



















hc
hf

38.9: a) For a particle with mass,

.4means2.2
1212
2
KKppmpK 
b) For a photon,
.2means2.
1212
EEpppcE



38.10:



hfK
max
Use the information given for
:findtonm400λ


J
10
3.204
eV)J10(1.602eV)(1.10
m10400
)sm10(2.998s)J10(6.626
19
19
9

834
max










Khf

Now calculate
:nm300λfor
max
K
eV2.13J103.418
J103.204
m10300
)sm10(2.998s)J10(6.626
19
19
9
834
max











hfK
38.11: a) The work function
00
eV
λ
eV 
hc
hf

J.107.53
V)(0.181C)10(1.60
m102.54
s)m10(3.00s)J10(6.63
19
19
7
834











The threshold frequency implies


hchc
hf
th
th

th
λ
λ
m.102.64
J107.53
)sm10(3.00s)J10(6.63
λ
7
19
834
th








b)
eV,4.70J107.53
19



as found in part (a), and this is the value from Table
38.1.
38.12:
a) From Eq. (38.4),
V.2.7V2.3
m)10(2.50
s)m10(3.00s)eV10(4.136
λ
1
7
815















hc
e
V
b) The stopping potential, multiplied by the electron charge, is the maximum kinetic
energy, 2.7 eV.
c)
s.m109.7
kg)10(9.11
V)(2.7C)102(1.6022
5
31
19






m
eV
m
K
v
38.13: a)
)sm10(3.00)smkg10(8.24
828


pcE

eV1.54
eVJ101.60
J102.47
J102.47
19
19
19







b)
m.108.05
s)mkg10(8.24
s)J10(6.63
λ
λ
7
28
34








p
hh
p
This is infrared radiation.
38.14: a) The threshold frequency is found by setting V = 0 in Eq. (40.4),
.
0
hf


b)
eV.3.34105.35
m1072.3
λ
19
7
0





hchc
hf

38.15: a)
.eV1.44J102.31
m108.60
)sm10(3.00s)J10(6.63

λ
19
7
834







hc
E

So the internal energy of the atom increases by




eV1.44eV6.52toeV44.1 E
eV.5.08

b)
eV.2.96J104.74
m
10
4.20
s)m10(3.00s)J10(6.63
λ
19

7
834







hc
E

So the final internal energy of the atom decreases to
eV.5.64eV2.96eV2.68





E
38.16: a)
eV.20
1


E
b) The system starts in the n = 4 state. If we look at all paths
to n = 1 we find the 4-3, 4-2, 4-1, 3-2, 3-1, and 2-1 transitions are possible (the last three
are possible in combination with the others), with energies 3 eV, 8 eV, 18 eV, 5 eV, 15
eV, and 10 eV, respectively. c) There is no energy level 8 eV above the ground state

energy, so the photon will not be absorbed. d) The work function must be more than 3
eV, but not larger than
5 eV.
38.17: a)







22
1
2
1
λ
1
n
R
(Balmer series implies final state is n = 2)
nm433m104.33m
)1021(1.10
100
21
100
λ
100
21
25
1

4
1
λ
1
:5
7
7












R
RRnH

b)
Hz106.93
m104.33
sm103.00
λ
14
7
8






c
f
c)
eV.2.87


hfE
38.18: Lyman: largest is
nm,122
)m10(1.097
)34()34(
λ,2
17




R
n
in the ultraviolet.
Smallest is
nm,91.2
1
λ, 
R

n
also ultraviolet. Paschen: largest is
nm,1875
)7144(
λ,4 
R
n
in the infrared. Smallest is
nm,820
9
λ, 
R
n
also
infrared.
38.19:
m105.890
)sm10(3.000s)J10(6.626
λ
7
834
1
2
3







hc
E
g
eV.102.00
.eV2.107J103.371
m105.896
)sm10(3.000s)J10(6.626
λ
eV.2.109J103.375
3
19
7
834
2
19
2
1
2
3
2
1














E
hc
E
g
38.20:
a) Equating initial kinetic energy and final potential energy and solving for the
separation radius r,
m.105.54
C)J10(4.78
C)10(1.60(184)
4
1
)2()92(
4
1
14
6
19
0
0








ε
K
ee
ε
r


b) The above result may be substituted into Coulomb’s law, or, the relation between
the magnitude of the force and the
magnitude of the potential energy in a Coulombic field
is
N.13.8
m)10(5.54
)evJ10(1.6eV)10(4.78
14
196






r
K
F
38.21: a)
)m1050.6(4
C)1060.1(164
4

)82()2(
4
14
0
219
00
21





εrε
ee
r
ε
qq
U

MeV.3.63eV103.63J105.81
613


U
b)
MeV.3.63J105.81
13
212211



UKUKUK
c)
.sm101.32
kg106.64
J)102(5.812
2
1
7
27
13
2






m
K
vmvK
38.22:
Hz.103.09
λ
andnm97.0λso,
)4(
1
1
λ
1
15

2










c
fR
38.23: a) Following the derivation for the hydrogen atom we see that for
3
Be
all we
need do is replace
Then.4by
22
ee
.
eV13.60
16)(Be)H(16
8
)4(1
)(Be
2
3
22

22
2
0
3









n
EE
hn
em
ε
E
nnn
So for the ground state,
eV.218)(Be
3
1


E
b) The ionization energy is the energy difference between the
1and




nn
levels.
So it is just 218 eV for
3
Be
, which is 16 times that of hydrogen.
c)
.
11
)m10(1.74
11
8
)4(
λ
1
2
2
2
1
18
2
2
2
1
32
0
22




















nnnnch
em

So for
m101.31
4
1
1)m10(1.74
λ
1
,1to2
818










nn
m.107.63λ
9

This is 16 times shorter than that from the hydrogen atom.
d)
(H).
4
1
)4(
)(Be
2
22
0
3
nn
r
em
hn
ε
r 



38.24: a), b) For either atom, the magnitude of the angular momentum is


2
h
s.mkg101.05
234


38.25:
,)eV6.13(
2
nE
n

so this state has
.351.16.13 n
In the Bohr model.

nL

so for this state
.smkg103.163
234


L
38.26:

a) We can find the photon’s energy from Eq. 38.8
J.104.58
5
1
2
1
)m10(1.097)sm10(3.00s)J10(6.63
1
2
1
19
22
17834
22


















n
hcRE

The corresponding wavelength is
nm.434λ 
hc
E
b) In the Bohr model, the angular momentum of an electron with principal quantum
number n is given by Eq. 38.10
.
2

h
nL

Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the
following loss in angular momentum (which we would assume is transferred to the
photon):
s.J103.17
2
s)J103(6.63
2
)52(
34
34






π
h
L

However, this prediction of the Bohr model is wrong (as shown in Chapter 41).
38.27: a)
m/s1018.2
)sJ1063.6(2
C)1060.1(
1:
2
1
6
34
0
219
1
2
0








vn
nh

e
v
n

.sm1027.7
3
3
sm1009.1
2
2
5
1
3
6
1
2


v
vh
v
vh
b) Orbital period
4
332
0
2
0
222
0

4
21
22
me
hn
ε
nhe
mehn
v
r
n
n






.s1013.4)3(:3
s1022.1)2(:2
s1053.1
C)1060.1(kg)1011.9(
)sJ1063.6(4
1
153
13
153
12
16
41931

3342
0
1











TTn
TTn
Tn

c) number of orbits
.102.8
s1022.1
s100.1
6
15
8







38.28:
a) Using the values from Appendix F, keeping eight significant figures, gives
.m100973731.1
17 
R
(Note: On some standard calculators, intermediate values in the
calculation may have exponents that exceed 100 in magnitude. If this is the case, the
numbers must be manipulated in a different order.)
b) Using the eight-figure value for
18
101798741.2
λ
gives

 hcR
hc
ER
eV.605670.13J

c)
Using the value for the proton mass as given in Appendix F gives
.m5970so,m100967758.1
117 
 RR
38.29:
kTEE
p
s
ps

e
n
n
)(
3
5
35



But
J10306.3eV66.20
18
5


s
E
J103.14
J10992.2eV70.18
19
18
3




E
E
p

a)
.1015.1
33)K300KJ1038.1()J1014.3(
3
5
2319



e
n
n
p
s
b)
.1039.3
17)K600KJ1038.1()J1014.3(
2319



e
c)
.1082.5
9)K1200KJ1038.1()J1014.3(
2319



e

d) The
s5
state is not highly populated compared to the
p3
state, so very few atoms
are able to make the required energy jump to produce the 632.8 nm light.
38.30:
.
)(
2
2
212232
2/1
23
KTEE
p
p
pp
e
n
n


From the diagram
.J10375.3
m10890.5
s)m10000.3()J10626.6(
λ
19
7

834
1
2/3








hc
E
g

.J1000.4J10371.3J10375.3so
.J10371.3
m10896.5
s)m10000.3()J10626.6(
λ
221919
2/12/3
19
7
834
2
21












E
hc
E
g

.944.0
K).500K/1038.1()J1000.4(
2
2
2322
2/1
2/3



J
p
p
e
n
n
So more atoms are in the

21
2 p
state.
38.31:
.J1088.1
m1006.1
s)m1000.3()sJ1063.6(
λ
20
5
834







hc
E

Total energy in 1 second from the laser
.J1050.7
3
 PtE
So the number of
photons emitted per second is
.1000.4
J101.88
J1050.7

17
20
3







E
E
38.32: 20.66
nm,632λandJ,103.14eV1.96eV18.70eV
19


E
hc
f
c
in good
agreement.
38.33:
min
max
λ
hc
hfeV
AC


m1011.3
)V4000)(C1060.1(
)sm10s)(3.00J1063.6(
λ
10
19
834
min







AC
eV
hc
This is the same answer as would be obtained if electrons of this energy were used.
Electron beams are much more easily produced and accelerated than proton beams.
38.34: a)
kV.8.29V1029.8
λ
3

e
hc
b) The shortest wavelength would correspond
to the maximum electron energy,

,eV
and so
.nm0414.0λ 
eV
hc
38.35: An electron’s energy after being accelerated by a voltage V is just
.eVE

The
most energetic photon able to be produced by the electron is just:
V
λ
e
hc
E
hc

.nm0829.0m1029.8
V)105.1)(C1060.1(
)sm1000.3)(sJ1063.6(
λ
11
419
834








38.36: a) From Eq. (38.23),
,
)(
λ
1cos
mch



and so a)
nm,0.0500nm0542.0λ



.137and,731.0
nm002426.0
nm0042.0
1cos



b)
.3.82.134.0
nm002426.0
nm0021.0
1cos.nm0500.0nm0521.0
λ




c)
,0λ


the photon is undeflected,
1cos


and
.0


38.37:
)cos1(λλ



mc
h
.)λmaximizetois180))(1(1(λλ
max




chosen
mc
h


mc
h2
λ 
.m1014.7m)102(2.426m1065.6λ
111211
max




38.38: a)




m)10426.2()cos)(1λλb)nm.0691.0λ
12

mc(h
eV
hc
.nm0698.0λsom,1011.7)0.45cos1(
13




c)
.keV8.17
λ




hc
E
38.39:
)cos1(λλ



mc
h
mc
h
λ
2
λ;180,λλ 




6
1070.9
λ
2
λ
λ


mc

hΔ
38.40:
The change in wavelength of the scattered photon is given by Eq. 38.23
 
).cos1(λ)cos1(
λλ
Δλ
λ
Δλ


mc
h
mc
h
Thus,
.m1065.2)11(
)100.0)(m/s1000.3)(kg1067.1(
)sJ1063.6(
λ
14
827
34








38.41:
The derivation of Eq. (38.23) is explicitly shown in Equations (38.24) through
(38.27) with the final substitution of
).cos1(λλyieldingλandλ







mc
h
hphp
38.42:
.K1025.7
m10400
Km1090.2m1090.2
3
9
33








m

λ
K
T
38.43:
mm.06.1m1006.1
K728.2
Km1090.2Km1090.2
λ
3
33
m







T
This is in
the microwave part of the electromagnetic spectrum.
38.44:
From Eq. (38.30),
a)




m
c

f
λ
and,mm966.0
K00.3
Km10898.2
λ
3
m
.Hz1010.3
11

Note that a more precise value of the Wien displacement law constant has been used. b)
A factor of 100 increase in the temperature lowers
m
λ
by a factor of 100 to 9.66
m

and
raises the frequency by the same factor, to
.Hz1010.3
13

c) Similarly,
nm966λ
m

.Hz1010.3and
14
f

38.45: a)
lrATAeσH
24
;


41
42823
41
)KmW10671.5)(26.0)(m30.0(m)10(0.20
W100




















Aeσ
H
T
K1006.2
4
T
b)
nm141λ;Km1090.2λ
m
3
m


T
Much of the emitted radiation is in the ultraviolet.
38.46: (a) Wien’s law:
T
k
m

λ
nm97m107.9
K30,000
mK1090.2
λ
8
3






m
This peak is in the ultraviolet region, which is not visible. The star is blue because the
largest part of the visible light radiated is in the
violetblue
part of the visible spectrum
(b)
4
σATP 
(Stefan-Boltzmann law)

m102.8
)K000,30)(4(
Km
W
105.67W)1086.3)(000,100(
9
2
42
826










R
R

12
m106.96
m108.2
8
9
sunstar




RR
(c) The visual luminosity is proportional to the power radiated at visible wavelengths.
Much of the power is radiated nonvisible wavelengths, which does not contribute to the
visible luminosity.
38.47: Eq. (38.32):
for1
2
1but
)1(
λ
2
)(
2
5
2
x
x

xe
e
hc
λI
x
λkThc




.Eq
λ
2
)
λ(λ
2
)(1
45
2

ckT
kThc
hc
λIx

(38.31), which is Rayleigh’s distribution.
38.48:
a) As in Example 38.10, using four-place values for the physical constants,
,80.95
λ


kT
hc
from which
.1044.6
λ)λ(
38
4



σT
I
b) With

T
2000 K and the same values for
,λandλ

37.14
λ

kT
hc
and so
.1054.7
λ)(
6
4




σT
λI
c) With
.1036.1
λ)λ(
and790.4
λ
,K6000
3
4




σT
I
kT
hc
T
d) For these temperatures, the intensity varies strongly with temperature, although for
even higher temperatures the intensity in this wavelength interval would decrease. From
the Wien displacement law, the temperature that has the peak of the corresponding
distribution in this wavelength interval is 5800 K (see Example 38.10), close to that used
in part (c).
38.49:
a) To find the maximum in the Planck distribution:
25
22

5
2
5
2
)1(λ
)(2
)1(
λ
)2(
50
)1(
λ
2
λλ 














λ
e

λhc
e
hc
e
hc
d
d
d
dI


.
λ
λ
where55Solve
λ55
λ)1(5
λ
λ
kT
hc
xex
e
e
x









Its root is 4.965, so
.
)965.4(
λ965.4
λ kT
hc
α

b)
.Km1090.2
K)J1038(4.965)(1.
)sm1000.3)(sJ1063.6(
)965.4(
λ
3
23
834
m







k
hc

T
38.50:
Combining Equations (38.28) and (38.30),
nm.872m1072.8
)KmW1067.5W/m10(6.94
K)m1090.2(
(
Km1090.2
λ
7
1/442826
3
1/4
3
m










σ)I
38.51: a) Energy to dissociate an
AgBr
molecule is just





23
5
106.02
J1000.1
mole1
)mole(
E
E
.eV04.1
eV)J10(1.60
J1066.1
19
19





b)
.m1020.1
J1066.1
)sm1000.3)(sJ1063.6(
λ
6
19
834








E
hc
c)
.Hz1051.2
m101.20
m/s1000.3
λ
14
6
8





c
f
d)
.eV1014.4
eVJ101.60
J1063.6
)Hz1000.1)(sJ1063.6(
7
19

26
834








hfE
e) Even though a 50-kW radio station emits huge numbers of photons, each individual
photon has insufficient energy to dissociate the AgBr molecule. However, the individual
photons in faint visible light do have enough energy.
38.52: a) Assume a non-relativistic velocity and conserve momentum
λ
h
mv


.
λ
m
h
v

b)
2
2
2

2
λ2λ2
1
2
1
m
h
m
h
mmvK








.
c)
.
λ
2
λ
λ
2
2
2
mc
h

hc
m
h
E
K

Recoil becomes an important concern for small
m
and
small

since this ratio becomes large in those limits.
d)
.nm122m1022.1
eV)J10V)(1.60e(10.2
)sm1000.3)(sJ1063.6(
λeV2.10
7
19
834







E
hc
E

.neglectedbecanrecoilsosmallquiteisThis.1042.5
eV2.10
eV1053.5
eV.105.53J1084.8
)m1022.1)(kg1067.1(2
)sJ1063.6(
9
8
827
2727
234












E
K
K
38.53:
Given a source of spontaneous emission photons we can imagine we have a
uniform source of photons over a long period of time (any one direction as likely as any
other for emission). If a certain number of photons pass out though an area A, a distance

D from the source, then at a distance 2D, those photons are spread out over an area
4)2(
2

times the original area A (because of how surface areas of spheres increase).
Thus the number of photons per unit area DECREASES as the inverse square of the
distance from the source.
38.54: a)
,λ
0
E
hc

and the wavelengths are: cesium: 590 nm, copper: 264 nm,
potassium: 539 nm, zinc: 288 nm. b) The wavelenghts of copper and zinc are in the
ultraviolet, and visible light is not energetic enough to overcome the threshold energy of
these metals.
38.55:
a) Plot: Below is the graph of frequency versus stopping potential.
Threshold frequency is when the stopping potential is zero.
Hz104.60Hz
10
11
.
4
89.1
14
15





th
f
b) Threshold wavelength is
.m1052.6
Hz104.60
)sm1000.3(
λ
7
14
8
th





th
f
c
c) The work function is just
J103.05Hz)1060.4)(sJ1063.6(
191434 

th
hf
eV.1.91

d) The slope of the graph is

meh
e
h
f
V
m

0
sJ1058.6
)C1060.1)(HzV1011.4(
34
1915




38.56: a) See Problem 38.4:



Hz)1000.5(
)10.0)(W200(
)(
)(
14
hhf
P
dNdE
dtdE
dt

dN
sec.photons1003.6
19

b) Demand
 r
r
dtdN
So.cmsecphotons1000.1
4
)(
211
2

m.69.3cm6930
)cmsecphotons10(1.004
secphotons1003.6
2/1
211
19













38.57: a) Recall


λ
0
hc
eV










12
0
12
0102
λ
1
λ
1
λλ
)(
e

hc
V
hchc
VVe
b)
.V477.0
m1095.2
1
m1065.2
1
C)10(1.60
)sm10s)(3.00J1063.6(
7711
834
0

















V
So the change in the stopping potential is an increase of 0.739 V.
38.58: From Eq. (38.13), the speed in the ground state is
).sm1019.2(
6
1
 Zv
Setting
10
1
c
v

gives Z = 13.7, or 14 as an integer. b) The ionization energy is
2
Z
E

(13.6
eV), and the rest mass energy of an electron is 0.511 MeV, and setting
100
2
mc
E

gives Z
= 19.4, or 19 as an integer.
38.59:
)cos1(λλ




mc
h

smkg1099.6λ
m09485.0
2
λλso180
24








hp
mc
h

b)
ee
λλ; EhchcEEE 





eV705J10129.1
λλ
λλ
)(
λ
1
λ
1
16
e














hchcE
38.60:
a) The change in wavelength of the scattered photon is given by Eq. 38.23
.nm0781.0)11(
)sm10kg)(3.0010(9.11

s)J10(6.63
m)100830.0(
)cos1(
λλ)cos1(λλ
831
34
9












mc
h
mc
h
b) Since the collision is one-dimensional, the magnitude of the electron’s momentum
must be equal to the magnitude of the change in the photon’s momentum. Thus,
.smkg102smkg1065.1
)m10(
0830.0
1
0781.0

1
)sJ1063.6(
λ
1
λ
1
2323
1934
e



















hp
c) Since the electron is non relativistic

),06.0(


.J10J1049.1
2
1616
2
e
e


m
p
K
38.61: a)
kg.1069.1
207
207
28
21
21






pe
pe
r

mm
mm
mm
mm
m
b) The new energy levels are given by Eq. (38.18) with
e
m
replaced by
.
r
m

.keV53.2
eV1053.2
eV60.13
kg1011.9
1069.1
eV60.13
8
1
1
2
3
231
kg 28
222
4
2
0









































E
n
n
nm
m
hn
em
E
e
rr
n

c)
)sm10(3.00s)J1063.6(
))eVJ10(1.60eV))1053.2(4Ve1053.2(
)(
1
λ
1
834
1933

21





EE
hc

19
m1053.1



m.1055.6λ
10

38.62:
a) The levels are
eV.0.10andeV,0.8eV,0.5eV,0.1
1234
 EEEE
b) We can go from 4-3(4 eV), 4-2(7 eV), and 4-1(9 eV) directly, but also 3-2(3 eV),
3-1(5 eV), and 2-1(2 eV) after starting from 4.
38.63:
a) The maximum energy available to be deposited in the atom is
eV14.5J1033.2
m1055.8
)sm1000.3(s)J1063.6(
λ

18
8
834







hc
E

but the inoization energy of hydrogen is 13.6 eV, so the maximum kinetic energy
is 14.5 eV
eV.0.900eV6.13


b) If some of the electrons were in the









eV4.3
4

eV6.13
state2
2
En
, then we
would expect a maximum kinetic energy of
eV11.1eV3.4eV5.14


, which is exactly
10.2 eV above that measured in part (a), explaining the anomoly.
38.64:
a) In terms of the satellite’s mass
M
, orbital radius
R
and orbital period
T
,








hT
MR
T

MR
h
L
h
n
22
2
4222

Using the given numerical values,


46
1008.1n
b) The angular momentum of the
satellite in terms of its orbital speed
,V
mass, and radius is
,MVRL

so
,)(
22
MRLV 
and its centripetal acceleration is

32
22
R
M

L
R
V
Newton’s law of gravitation can then be expressed as

MGM
L
R
RM
ML
R
GM
earth
2
32
2
2
earth
or,
,2If

nhL 











2
earth
2
2
2
4
kn
MGM
h
nR

c)
,2 nknR



and for the next orbit,
).4(and,1
earth
22
MGMhnRn


Insertion of numerical values from Appendix F and using
n
from part (a) gives
m,105.1
39

R
which is (d) not observable. (e) The quantum and classical orbit do
correspond, either would be correct, but only the classical calculation is useful.
38.65: a) Quantization of angular momentum implies

mr
nh
v
h
nmvrL

2
2
But










41
2
22
22
222
2

2
44 mD
hn
r
Dmr
hn
D
mv
r
r
mv
DrF
n

b) The energy
DrFsDrmvUKE  ince,
2
1
2
1
22
is completely analogous
to
2222
2
1
2
1
So.
2

1
DrDrr
m
D
mEkxUkxF








.
2
2


nh
m
D
mD
nh
DEn

c) Photon energies
 n
h
m
D

nE
h
m
D
nnEEE
fifi
where
22
)(


.0integers

d) This could describe a charged mass attached to a spring, being spun in a circle.
38.66: a)
,))(1(
1
dee
de
d
edr,




mmm
mm
m
mm
and insertion of the numerical

values gives
.999728.0
edr,
mm 
b) Let
R
R
R





where,
)34(
λ,
)34(
λ
0
is the
Rydberg constant evaluated with
,999456.0
pr,
 mm
so








R
m
m
R
R
R
m
m
R
e
dr,
e
pr,
,
3)4(
λand,
Then
nm.033.0λ
1
3
4
λ
dr,
e
pr,
e
0





















m
m
m
m
R
R
R
R
R
38.67: a) The


H
line is emitted by an electron in the
3

n
energy level,
eV.51.1
)3(
eV60.13
2
3


E
The ground state energy is
V,60.13
1
E
so one must
add at least
1.51eV13.60

eV=12.09 eV if the

H
line is to be emitted.
b) The possible emitted photons are
1,2and,13,23




with the wavelengths
given by
.
11eV60.13
)(
1
λ
1
22











if
fi
nnhc
EE
hc
This yields the wavelengths of
658 nm, 103 nm, and 122 nm for the respective photons above.
38.68:
.

kln
)(
1
2
gex
kT)(
1
2
gex












n
n
EE
Te
n
n
EE
J.101.63eV2.10
eV.6.13

eV.4.3
4
eV6.13
18
gex
g
2ex






EE
E
EE
a)
.10
12
1
2


n
n
K.4275
)ln(10)KJ10(1.38
J)1063.1(
1223
18







T
b)
.10
8
1
2


n
n
K.6412
)ln(10)KJ10(1.38
J)1063.1(
823
18






T
c)
.10

4
1
2


n
n
K.12824
)ln(10)KJ10(1.38
J)1063.1(
423
18






T
d) For absorption to take place in the Balmer series, hydrogen must
start
in the
2

n
state. From part (a), colder stars have fewer atoms in this state leading to weaker
absorption lines.
38.69:
The transition energy equals the sum of the recoiling atom’s kinetic energy and
the photon’s energy.

.λ
λ
ktr
ktr
ktr
EE
hc
EE
hc
E
EEE












If the recoil is neglected
tr
E
hc
λ
2
22

λ
)λ(
λ
11
1
1
111
λ


























































hc
E
hc
E
hc
E
E
hc
E
E
E
hc
EEE
hc
EEE
hc
kk
tr
k
tr
k
tr
trktrtrktr
Conservation of momentum, assuming atom initially at rest, yields:
mc
h

mhc
λ
h
m
h
m
P
EP
h
P
k
kk
2
λ
2
λ
λ22λ
2
2
2
2
2
2













b) For the hydrogen atom:
16
827
34
106.6
)sm10kg)(3.00102(1.67
s)J1063.6(
λ







m and
it doesn’t depend on
.
n
38.70: a)
nm.0.1849λsonm,0049.0
2
λ2)cos(1so180 



mc
h

b)
eV.183J1093.2
λ
1
λ
1
17










hcE
This will be the kinetic energy of the electron. c) The kinetic energy is far less than
the rest mass energy, so a non-relativistic calculation is adequate;
.sm1002.82
6
 mKv
38.71:
a) Largest wavelength shift:

m.1085.4

s)m10kg)(3.0010(9.11
s)J1063.6(2
))1(1(
λ
12
831
34







mc
h
b) We want
).cos1(λλ2λλλλ




mc
h
Smallest energy implies largest
,

so

eV.1056.2

2
eV1011.5
2
λ
2
λ1cos
5
52



mchc
E
mc
h

38.72:
a) Power delivered = pIV.
b)
.deliveredPower
dt
dT
mc
dt
dE
mcdTdE

So
.
1

mc
pIV
dt
dE
mc
dt
dT

c) (a) Power
W;10.8V)100.18)(A0600.0)(010.0(
3
 pIV
(b)
.sK332.0
)KkgJkg)(130250.0(
)V10A)(18.0100.60)(01.0(
33




dt
dT
d) A high melting point and heat capacity—tungsten and copper, for example.
38.73: a) The photon’s energy change











12
12
λ
1
λ
1
)(
hcffhE
319.5
J10111.5
m10100.1
1
m101.132
1
)sm10s)(3.00J1063.6(
17
1010
834

















which is a loss for the photon, but which is a gain for the electron. So, the kinetic energy
of the electron is
eV.319.5J10111.5
17


.sm1006.1
kg1011.9
J)1011.5(2
2
7
31
17






m
KE

v
b) If all the energy of the electron is lost in the emission of a photon, then
J105.111
)sm10s)(3.00J1063.6(
λ
17
834





E
hc
nm.3.89m10892.3
9


38.74: a)
),cos1)((λ),cos1)((λ
2211
θmchθmch






and so the overall
wavelength shift is

).coscos2)((λ
21
θθmch 
(b) For a single scattering through angle
).cos1)((λ,
s
θmchθ



For two successive scatterings through an angle of
,2

for each scattering,

).2cos1)((2λ
t

 mch

ts
2
2
s
2
λλand))2cos(1()2(cos1so1)2cos(
))2(cos1(2)(
λand))2(cos1(2cos1




θmchθθ
Equality holds only when
.180



c)
),(500.0)60cos1()(d)).(268.0)0.30cos1(2)( mchmchmchmch 
which is indeed greater than the shift found in part (c).
38.75:
a) The wavelength of the gamma rays is
m101.24
eV)J10(1.60eV)10(1.00
)sm10(3.00s)J10(6.63
λ
12
196
834







i
i
E
hc

and the wavelength of the final visible photons is
m.1000.5
7

So, the increase in
wavelength per interaction (assuming about
26
10
interactions) is
.m1000.5
10
m101.24m1000.5
33
26
127




b)
).smallfor (
22
11)cos1(
λ
2
2



mc

h
mc
h
mc
h











.)10(3.68radians106.42
s)J10(6.63
m)10(5.00)sm10(3.00Kg)102(9.11
λ2
So
911
34
33831










h
mc

c)



seccollisions103.17
sec103.15
collisions10
sec103.15ears10
12
13
26
136
y
.collisionsec1015.3
13

How far does light travel in this time?


)collisionsec10(3.15)sm1000.3(
138
.0.1mmmm0.0945m1045.9
5



38.76: a) The final energy of the photon is
,and,
λ
KEE
hc
E






where K is the
kinetic energy of the electron after the collision. Then,
.
1
)1(
1
λ
1
λ
)1()λ()λ(
λ
2122
2






















cvh
mc
mchc
hc
Khc
hc
KE
hc

)1((
2



mcK
since the relativistic expression must be used for three-figure accuracy).
b)
)).(λ1arccos( mch




c)
 
 
m1043.2,250.0125.11
1
1
1
12
21
2
00.3
80.1




mc
h


s)J10(6.63
(0.250))sm10(3.00kg)10(9.11m)10(5.10

1
mm105.10
λ
34
83112
3









.74.0
m102.43
m)103.34m10(5.10
1arccos
nm.103.34
12
1212
3


















38.77: a)
f
c
e
hc
I
kThc



λbut
)1(
λ
2
)
λ(
λ5
2

)1(

2
)1()(
2
)(
3
5
5
2




kThfkThf
ec
hf
efc
hc
fI

b)














2
0
0
)()λ(
f
c
dffIdI


32
445
23
45
4
32
4
0
3
32
4
0
2
3
15
2
240
)()2(

)2(
240
1)(2
1
)(2
)1(
2
h
c
Tk
c
h
kT
h
c
kT
dx
e
x
hc
kT
ec
dfhf
xkThf












c) The expression



23
445
15
2
c
h
Tk
as shown in Eq. (38.36). Plugging in the values for
the constants we get
428
KmW1067.5 

σ
.
38.78:
,combining;and,,
4
PtEIAPσTI 
hrs.2.45s108.81
K)(473)KmW10(5.67)m10(4.00
J)(100

3
4428264






TA
E
t

38.79: a) The period was found in Exercise 38.27b:
4
332
0
4
me
hnε
T 
and frequency is
just
.
4
1
332
0
4
hn
me

T
f


b) Eq. (38.6) tells us that
).(
1
12
EE
h
f 
So









2
1
2
2
32
0
4
11
8

nnh
me
f

(from Eq.
(38.18)).
If
222
1
2
2
12
)1(
1111
then,1and


nnnn
nnnn
.
4
largefor
2
2
11
1
)11(
1
1
1

332
0
4
3
222
hn
me
fn
n
nnnn



























 
38.80: Each photon has momentum
,
λ
h
p

and if the rate at which the photons
strike the surface is


,dtdN
the force on the surface is
),()λ( dtdNh
and the pressure is
.)()λ( AdtdNh
The intensity is
,)λ()()()( AhcdtdNAEdtdNI 
and comparison of the two expressions gives the pressure as
).( cI
38.81:
Momentum:
)( PpPp
PpPp













PpPp




energy:
EcpEpc






222
)()( mccPcp 






2222
)()()( mccPEcppc 





22222
)()(2))(()( mccppPcppPc 




(24)(2)()(2)()(
222222
p
EccppppEcppPccppcEEcppc 










0)()(2

2



ppPc

Pc
E
hcPcE
PcE
hc
PcE
PcE
PcE
PcEhc
PcEpc
PcE
p
PcEcpc
PcEc
pp
PcEcpEcpcPcp








































)2)(λ(
λ
2
λ
)(λ2
λλ
)(22
)()2(
22
2
222
If
2
2
2222
1)(,









E
mc
EmcEPcmcE



















 
2
2
c
2
1
1
E
m
E

E
mc
PcE

22
)(
2
1











hcE
cm
E
hc
E
hc
EE
mc
λ
4
λ
1
)2(2
)(
λ

4222
1
b) If
J101.60eV1000.1m,106.10λ
906 
 E



















)J106.1(4
)m10(10.6kg)1011.9(
1
J1060.1
λ

9
64231
9
hc
chc

m.107.0856.0)(1m)1024.1(
1516 

c) These photons are gamma rays. We have taken infrared radiation and converted it into
gamma rays! Perhaps useful in nuclear medicine, nuclear spectroscopy, or high energy
physics: wherever controlled gamma ray sources might be useful.