CHAPTER 39
PHOTONS AND MATTER WAVES
CHAPTER 39
Answer to Checkpoint Questions
1.
2.
3.
4.
5.
(b), (a), (d), (c)
(a) lithium, sodium, potassium, cesium; (b) all tie
(a) same; (b) { (d) x rays
(a) proton; (b) same; (c) proton
same
Answer to Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
(a) microwave; (b) x ray; (c) x ray
(a) true; (b) false; (c) false; (d) true; (e) true; (f) false
potassium
only (b)
Positive charge builds up on the plate, inhibiting further electron emission
only (e)
none
The fractional wavelength change for visible light is too small
(a) greater; (b) less
(a) B ; (b) { (d) A
no essential change
electron
p
(a) decreases by a factor of 1= 2; (b) decreases by a factor of 1/2
electron, neutron, alpha particle
extremely small
(d)
(a) increasing; (b) decreasing; (c) same; (d) same
1059
1060
18.
19.
20.
21.
22.
CHAPTER 39
PHOTONS AND MATTER WAVES
proton
(a)
amplitude of reected wave is less than that of incident wave
(a) zero; (b) yes
all tie
Solutions to Exercises & Problems
1E
The energy of a photon is given by E = hf , where h is the Planck constant and f is the
frequency. The wavelength is related to the frequency by f = c, so E = hc=. Since
h = 6:63 10 34 J s and c = 3:00 108 m/s,
hc = 1:99 10
25 J m =
Thus
1:99 10 25 J m
(1:602 10 19 J/eV)(10 9 m/nm) = 1240 eV nm :
nm :
E = 1240 eV
2E
From the result of 1E
nm = 1240 eV nm = 2:11 eV :
E = 1240 eV
589 nm
3E
h = 6:626 10
34 J s =
6:626 10 34 J s = 4:14 10
1:60 10 19 J/eV
15 eV s :
4E
Let E = 1240 eV nm=min = 0:6 eV to get = 2:1 103 nm = 2:1 m. It is in the infrared
region.
5E
nm = 1240 eV nm = 5:9 10 6 eV = 5:9 eV :
E = 1240 eV
21 107 nm
CHAPTER 39
1061
PHOTONS AND MATTER WAVES
6E
Let
1 m v2 = E = hc
ph
2 e
and solve for v:
r
s
v = m2hc = (m2hc
c
e
e c2 )
s
=
2(1240 eV nm)
8
5
(0:511 106 eV)(590 nm) (3:00 10 m/s) = 8:6 10 m/s :
Since v c the non-relativistic formula K = 12 mv2 may be used.
7E
Let R be the rate of photon emission (number of photons emitted per unit time) of the
Sun and let E be the energy of a single photon. Then the power output of the Sun is given
by P = RE . Now E = hf = hc=, where h is the Planck constant, f is the frequency of
the light emitted, and is the wavelength. Thus P = Rhc= and
(550 nm)(3:9 10 W)
45
=
R = P
hc (6:626 10 34 J s)(3:00 108 m/s) = 1:0 10 photons/s :
26
8E
Let the diameter of the laser beam be d, then the corss-sectional area of the beam is
A = d2 =4. From the formula obtained in 7E the rate in question is given by
R = P
A hc(d2 =4)
nm)(5:0 10 3 W)
= (6:626 10 4(633
34 J s)(3:00 108 m/s)(3:5 10 3 m)2
= 1:7 1021 photons/m2 s :
9E
Since = (1; 650; 763:73) 1 m = 6:0578021 10 7 m = 605:78021 nm, the energy is
1240 nm eV = 2:047 eV :
=
E = hc
605:78021 nm
1062
CHAPTER 39
PHOTONS AND MATTER WAVES
10P
nm (100=s) = 1240 eV nm (1:60 10
P = Et = 1240 eV
550 nm
= 3:6 10 17 W :
19 J/eV)(100=s)
11P
(a) Use the formula obtained in 7E: R = P=hc. The bulb emitting light with the longer
wavelength (the 700 nm bulb) emits more photons per unit time. The energy of each photon
is less so it must emit photons at a greater rate.
(b) Let R` be the rate of photon production at the longer wavelength (` ) and let Rs be
the rate of production at the shorter wavelength (s ). Then
(700 nm 400 nm)(400 J/s) = 6:0 1020 photons/s :
R` Rs = (` hcs )P = (1:60
10 19 J/eV)(1240 eV nm)
The result hc = 1240 eV nm, developed in 1E, was used.
12P
(a) The rate at which solar energy strikes the panel is
P = (1:39 kW/m2 )(2:60 m2 ) = 3:61 kW :
(b) The rate at which solar photons are absorbed by the panel is
:61 kW
R = EP = (1240 eV nm=5503nm)(1
:60 10
ph
19 J/eV)
= 1:00 1022 =s :
(c) The time in question is given by
23
6
:
02
10
N
A
t = R = 1:00 1022 =s = 60:2 s :
13P
The total energy emitted by the bulb is E = 93%Pt, where P = 60 W and t = 730 h =
(730 h)(3600 s/h) = 2:628 106 s. The energy of each photon emitted is Eph = hc=. Thus
the number of photons emitted is
Pt =
(93%)(60 W)(2:628 106 s)
26
N = EE = 93%
hc= (1240 eV nm=630 nm)(1:60 10 19 J/eV) = 4:7 10 :
ph
CHAPTER 39
1063
PHOTONS AND MATTER WAVES
14P
The rate at which photons are emitted from the argon laser source is given by R = P=Eph ,
where P = 1:5 W is the power of the laser beam and Eph = hc= is the energy of each
photon of wavelength . Since = 84% of the energy of the laser beam falls with in the
central disk, the rate of photon absorption of the central disk is
P =
(0:84)(1:5 W)
R0 = R = hc=
(1240 eV nm=515 nm)(1:60 10
= 3:3 1018 photons/s :
19 J/eV)
15P
(a) Assume all the power results in photon production at the wavelength = 589 nm.
Let R be the rate of photon production and E be the energy of a single photon. Then
P = RE = Rhc=, where E = hf and f = c= were used. Here h is the Planck constant,
f is the frequency of the emitted light, and is its wavelength. Thus
9 m)(100 W)
P
(589
10
R = hc = (6:63 10 34 J s)(3:00 108 m/s) = 2:96 1020 photons/s :
(b) Let I be the photon ux a distance r from the source. Since photons are emitted
uniformly in all directions R = 4r2 I and
s
r
2:96 1020 photons/s = 4:85 107 m :
r = 4RI =
4(1:00 104 photons/m2 s)
(c) Let n be the photon density and consider a narrow column with cross-sectional area
A and length `, with its axis along the direction of photon motion. At any instant the
number of photons in the column is N = nA`. All of them will move through one end
in time t = `=c, so the photon ux through the end is I = N=At = nA`c=A` = nc. Now
I = R=4r2 , so R=4r2 = nc and
r
s
R =
2:96 1020 photons/s
r = 4nc
= 280 m :
4(1:00 106 photons/m3 )(3:00 108 m/s)
(d) The photon ux is
R = 2:96 1020 photons/s = 5:89 1018 photons/m2 s :
I = 4r
2
4(2:00 m)2
16E
According to Eq. 39-5 the condition for photoelectric eect to occur is hf .
1064
CHAPTER 39
PHOTONS AND MATTER WAVES
(a) For = 565 nm
15 eV s)(3:00 108 m/s)
hc
(4
:
14
10
hf = =
= 2:20 eV :
565 10 9 m
Since potassium > hf > cesium photoelectric eect can occur in cesium but not in potas-
sium at this wavelength.
(b) Now = 518 nm so
(4:14 10 15 eV s)(3:00 108 m/s) = 2:40 eV :
hf = hc
=
518 10 9 m
This is greater than both cesium and potassium so photoelectric eect can now occur for
both metals.
17E
The energy of the most energetic photon in the visible light range (with wavelength of
about 400 nm) is about E = (1240 eV nm=400 nm) = 3:1 eV. So barium and lithium can
be used, since their work functions are both lower than 3:1 eV.
18E
(a) Since Eph = h= = 1240 eV nm=680 nm = 1:82 eV < = 2:28 eV, there is no photoelectric emission.
(b) The longest (cuto) wavelength of photons which will cause photoelectric emission in
sodium is given by Eph = h=max = , or max = h= = (1240 eV nm)=2:28 eV = 544 nm.
This corresponds to the green color.
19E
Use Eq. 39-5 to nd Kmax :
Kmax = hf = (4:14 10
15 eV s)(3:0 1015 Hz)
2:3 eV = 10 eV :
20E
The energy of an incident photon is Eph = hf = hc=, where h is the Planck constant, f is
the frequency of the electromagnetic radiation, and is its wavelength. The kinetic energy
of the most energetic electron emitted is Kmax = Eph = (hc=) , where is the
work function for sodium. The stopping potential Vstop is related to the maximum kinetic
energy by eVstop = Kmax , so eVstop = (hc=) and
eV nm = 170 nm :
= eV hc+ = 5:1240
0 eV + 2:2 eV
stop
CHAPTER 39
1065
PHOTONS AND MATTER WAVES
Here eVstop = 5:0 eV and hc = 1240 eV nm were used.
21E
The speed v of the electron saties Kmax = 12 me v2 = Eph , or
s
s
2
v = 2(Ephm ) = 2(Ephm c2)c
e
e
r
50 eV)(3:00 108 m/s)2 = 6:76 105 m/s :
= 2(5:80 eV 04::511
106 eV
22P
(a) The kinetic energy Kmax of the fastest electron emitted is given by Kmax = hf =
(hc=) , where is the work function of aluminum, f is the frequency of the incident
radiation, and is its wavelength. The relationship f = c= was used to obtain the second
form. Thus
eV nm 4:20 eV = 2:00 eV :
Kmax = 1240
200 nm
(b) The slowest electron just breaks free of the surface and so has zero kinetic energy.
(c) The stopping potential Vstop is given by Kmax = eVstop , so
V = Kmax = 2:00 eV = 2:00 V :
stop
e
e
(d) The value of the cuto wavelength is such that Kmax = 0. Thus hc= = or
1240 eV nm = 295 nm :
=
= hc
4:2 eV
If the wavelength is longer the photon energy is less and a photon does not have sucient
energy to knock even the most energetic electron out of the aluminum sample.
23P
(a) Use Eq. 39-6:
nm) 1:8 eV = 1:3 V :
Vstop = hf e = hc=e = (1240 eV nm=400
e
(b) Use the formula obtained in 21E:
s
r
s
c2
v = 2(Ephm ) = 2eVmstop = 2eVmstop
e
e
e c2
r
:00 108 m/s)2 = 6:8 105 m/s :
= 2e(1:30V)(3
:511 106 eV
1066
CHAPTER 39
PHOTONS AND MATTER WAVES
24P
From Eq. 39-6
hc = 1240 eV nm 1240 eV nm = 1:07 eV :
Kmax = Eph = hc
max
254 nm
325 nm
25P
To nd the longest possible wavelength max (corresponding to the lowest possible energy)
of a photon which can cause photoelectric eect in platinum, put Kmax = 0 in Eq. 39-5
and use hf = hc=. Thus hc=max = . Solve for max :
1240 eV nm = 233 nm :
=
max = hc
5:32 eV
26P
(a) Use the photoelectric eect equation (Eq. 39-5) in the form hc= = + Kmax . The
work function depends only on the material and the condition of the surface and not on
the wavelength of the incident light. Let 1 be rst wavelength described and 2 be the
second. Let Km1 (= 0:710 eV) be the maximum kinetic energy of electrons ejected by light
with the rst wavelength and Km2 (= 1:43 eV) be the maximum kinetic energy of electron
ejected by light with the second wavelength. Then
hc = + K
m1
1
and
hc = + K :
m2
2
Solve these equations simultaneously for 2 .
The rst equation yields = (hc=1 ) Km1 . When this is used to substitute for in the
second equation the result is (hc=2 ) = (hc=1 ) Km1 + Km2 . The solution for 2 is
(1240 eV nm)(491 nm)
1
2 = hc + (hc
=
1 Km2 Km1 ) 1240 eV nm + (491 nm)(1:43 eV 0:710 eV)
= 382 nm :
(b) The rst equation displayed above yields
eV nm 0:710 eV = 1:82 eV :
= hc Km1 = 1240
491 nm
1
CHAPTER 39
1067
PHOTONS AND MATTER WAVES
27P
For the rst and second case (labeled 1 and 2) we have eV01 = hc=1 and eV02 =
hc=2 , from which h and can be solved.
(a)
1:85 eV 0:820 eV
h = e(V11 V2 )1 = (3:00 1017 nm/s)[(300
nm) 1 (400 nm) 1 ]
c(1 2 )
= 4:12 10 15 eV s :
(b)
nm) (1:85 eV)(300 nm) = 2:27 eV :
= 3(V22 V1 1 ) = (0:820 eV)(400
300 nm 400 nm
1
2
(c) Let = hc=max to obtain
1240 eV nm = 545 nm :
=
max = hc
2:27 eV
28P
Vstop (V)
3.0
2.0
a
1.0
b
c
0
4
6
8
10
12
f (10 14 Hz)
(a) According to Eq. 39-6 the slope of the straight line plotted above is equal to h=e.
Measure the slope directly from the plot to obtain
h = ab =
2:0 V 0:75 V
15
e cb (10:5 1014 7:5 1014 ) Hz = 4:17 10 V s ;
1068
CHAPTER 39
PHOTONS AND MATTER WAVES
so h = (4:17 10 15 V s)(1:6 10 19 C) = 6:7 10 34 J s :
(b) From the plot we nd the lowest photon frequency for photoelectric eect to occur to
be f0 = 5:7 1014 Hz. Thus
34 J s)(5:7 1014 Hz)
= 2:4 eV ;
= hf0 = (6:63 10
1:6 10 19 J/C
which compares favourably with the actual value of 2:3 eV.
29P
The number of photons emitted from the laser per unit time is
3W
2
:
00
10
P
R = E = (1240 eV nm=600 nm)(1:60 10
ph
19 J/eV)
= 6:05 1015 =s ;
of which (1:0 10 16 )(6:05 1015 =s) = 0:605=s actually cause photoelectric emissions.
Thus the current is i = (0:605=s)(1:60 10 19 C) = 9:68 10 20 A:
30P
(a) Find the speed v of the electron from r = me v=eB : v = rBe=me . Thus
2 (rB )2 e2
Kmax = 21 me v2 = 21 me rBe
me = 2me
(1:88 10 4 T m)2 (1:60 10 19 C)2 = 3:10 keV :
= 2(9
:11 10 31 kg)(1:60 10 19 J/eV)
(b) The work done is
eV nm 3:10 keV = 14 keV :
W = Eph Kmax = 711240
10 3 nm
31E
(a)
(b)
3:00 108 m/s = 8:57 1018 Hz :
f = c = 35
:0 10 12 m
E = hf = (4:14 10
(c) From Eq. 37-9
15 eV s)(8:57 1018 Hz) = 3:55 104 eV :
:626 10
p = h = 635
:0 10
34 J s
12 m
= 1:89 10
23 kg m/s :
CHAPTER 39
1069
PHOTONS AND MATTER WAVES
32E
(a) The rest energy of an electron is given by E = me c2 . Thus the momentum of the
photon in question is given by
2
p = Ec = mecc = me c
= (9:11 10 31 kg)(3:00 108 m/s) = 2:73 10
22 kg m/s :
We may also express the momentum in terms of MeV/c: p = me c2 =c = 0:511 MeV=c.
(b)
10 34 J s = 2:43 10 12 m = 2:43 pm :
= hp = 2:673:626
10 22 kg m/s
(c)
00 108 m/s = 1:24 1020 Hz :
f = c = 23::43
10 12 m
33E
(a) When a photon scatters from an electron initially at rest the change in wavelength is
given by
= mh c (1 cos ) ;
e
where me is the mass of an electron and is the scattering angle. Now h=me c = 2:43 10 12 m = 2:43 pm, so = (2:43 pm)(1 cos 30 ) = 0:326 pm. The nal wavelength is
0 = + = 2:4 pm + 0:326 pm = 2:7 pm.
(b) Now = (2:43 pm)(1 cos 120 ) = 3:645 pm and 0 = 2:4 pm + 3:645 pm = 6:05 pm.
34P
(a)
(b)
1240 nm eV = 2:43 10 3 nm = 2:43 pm :
=
= hc
E 0:511 MeV
0 = + = + mh c (1 cos )
e
= 2:43 pm + (2:43 pm)(1 cos 90:0 ) = 4:86 pm :
(c)
43 pm = 0:255 MeV :
E 0 = E 0 = (0:511 MeV) 42::86
pm
1070
CHAPTER 39
PHOTONS AND MATTER WAVES
35P
Since the electron is free it is undergoing no acceleration. Therefore we may, without loss
of generality, consider a stationary electron of rest mass me which collides with a photon
of linear momentum p. Prior to the collision the energy of the electron is its rest energy
Ee = me c2 , while the energy of the photon is simply Eph = pc. Thus the total energy of
the electron-photon system before the collision is
Ei = Ee + Eph = me c2 + pc :
After the collision, the photon has been entirely absorbed by the electron which, according
to the conservation of linear momentum, must have acquired a linear momentum equal to
p, the initial linear momentum of the photon. Thus the nal energy of the system after
the collision is
p
Ef = (pc)2 + m2e c4 :
Equate the initial and the nal energies to obtain
p
me c2 + pc = (pc)2 + m2e c4 :
Now square both sides:
(me c2 + pc)2 = (me c2 )2 + 2me pc3 + (pc)2 = (pc)2 + m2e c4 ;
which yields 2me pc3 = 0, or
p = 0:
This means that the only photon that can participate in the collision is the one with zero
linear momentum. The energy of the photon is thus Eph = pc = 0, which means that the
photon is nonexistent and the collision is impossible.
36P
(a)
(b)
= mh c (1 cos ) = (2:43 pm)(1 cos 180 ) = +4:86 pm :
e
hc = (1240 nm eV)
1
1
E = hc
0 0:01 nm + 4:86 pm 0:01 nm = 41 keV :
(c) From conservation of energy K = E = 41 keV.
(d) The electron will move straight ahead after the collision, since it has acquired some of
the foward linear momentum from the photon.
37P
(a) Since the mass of an electron is me = 9:109 10 31 kg, its Compton wavelength is
6:626 10 34 J s
12
C = mh c = (9:109 10
31 kg)(2:998 108 m/s) = 2:426 10 m = 2:43 pm :
e
CHAPTER 39
1071
PHOTONS AND MATTER WAVES
(b) Since the mass of a proton is mp = 1:673 10
27 kg,
its Compton wavelength is
6:626 10 34 J s
C = (1:673 10
27 kg)(2:998 108 m/s) = 1:321 10
15 m = 1:32 fm :
(c) Use the formula developed in 1E: E = (1240 eVnm)=, where E is the energy and is
the wavelength. Thus for the electron
1240 eV nm = 5:11 105 eV = 0:511 MeV :
E = 2:426
10 3 nm
(d) For the proton
1240 eV nm = 9:39 108 eV = 939 MeV :
E = 1:321
10 6 nm
38P
The fractional change is
E = (hc=) = 1 = 1 1 = 1 = E
hc=
0 0
+ 1
= =1 + 1 = (= )(1 1cos ) 1 + 1 :
C
(a) Now = 3:0 cm = 3:0 1010 pm and = 90 so
1
E =
10
E
(3:0 10 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 8:1 10
11 :
(b) Now = 500 nm = 5:00 105 pm and = 90 so
E =
1
6
5
E
(5:00 10 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 4:9 10 :
(c) Now = 25 pm and = 90 so
1
E =
2
E
(25 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 8:9 10 :
(d) Now = hc=E = 1240 nm eV=1:0 MeV = 1:24 10 3 nm = 1:24 pm and = 90 so
1
E =
E
(1:24 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 0:66 :
1072
CHAPTER 39
PHOTONS AND MATTER WAVES
(e) From the calculation above we see that the shorter the wavelength the greater the
fractinal energy change for the photon as a result of the Compton scattering. Since E=E
is virtually zero for microwave and visible light, Compton eect is signicant only in the
x-ray to gamma ray range of the electromagnetic spectrum.
39P
If E is the original energy of the photon and E 0 is the energy after scattering, then the
fractional energy loss is
0
frac = E E E :
Sample Problem 39-4 shows that this is
frac = + :
Thus
= frac = 0:75 = 3 :
1 frac 1 0:75
A 300% increase in the wavelength leads to a 75% decrease in the energy of the photon.
40P
max = mh c (1 cos )
= m2hc
p
p
max
15
8
= 2(4:14 10 eV s)(3:00 10 m/s)
= 2:65 10
938 MeV
15 m = 2:65 fm :
41P
The dierence between the electron-photon scattering process in this problem and the one
studied in the text (the Compton shift, see Eq. 39-11) is that the electron is in motion
relative with speed v to the laboratory frame. To utilize the result in Eq. 39-11, shift to
a new reference frame in which the electron is at rest before the scattering. Denote the
quantities measured in this new frame with a prime ('), and apply Eq. 39-11 to yield
0 = 0 0 = h (1 cos ) = 2h ;
0
me c
me c
where we noted that = (since the photon is scattered back in the direction of incidence).
Now, from the Doppler shift formula (Eq. 38-25) the frequency f00 of the photon prior to
the scattering in the new reference frame satises
s
f00 = c0 = f0 11 + ;
0
CHAPTER 39
1073
PHOTONS AND MATTER WAVES
where = v=c. Also, as we switch back from the new reference frame to the original one
after the scattering
s
s
f = f 0 11 + = c0 11 + :
Solve the two Doppler-shift equations above for 0 and 00 and substitute the results into
the Compton shift formula for 0 :
s
0 = f1 11 + 1
f0
s
1 = 2h :
1 + me c2
Some simple algebra then leads to
s
E = hf = hf0
1 + m2hc2 11 + e
! 1
;
where we used E = hf .
42P
(a) From Eq. 39-11
= mh c (1 cos ) = (2:43 pm)(1 cos 90 ) = 2:43 pm :
e
(b)
= 2:425 pm = 4:11 10 6 :
590 nm
(c) The change in energy for a photon with = 590 nm is given by
E = hc hc
ph
2
15
:00 108 m/s)(2:43 pm)
= (4:14 10 eV s)(3
(590 nm)2
= 8:67 10 6 eV :
For an x ray photon of energy Eph = 50 keV, remains the same (= 2:43 pm) since it is
independent of Eph . The fractional change in wavelength is now
= =
(50 103 eV)(2:43 pm)
2
hc=Eph (4:14 10 15 eV s)(3:00 108 m/s) = 9:78 10 ;
and the change in photon energy is now
1
1
Eph = hc + =
;
hc = E
ph
+ 1+
1074
CHAPTER 39
PHOTONS AND MATTER WAVES
where = =. Plug in Eph = 50 keV and = 9:78 10 2 to obtain Eph = 4:45 keV.
(Note that in this case 0:1 is not close enough to zero so the approximation Eph hc=2 is not as accurate as in the rst case, in which = 4:12 10 6 . In fact if you
were to use this approximation here you would get Eph 4:89 keV, which does not
amount to a satisfactory approximation.)
43P
(a) From Eq. 39-11 = (h=me c)(1 cos ): In this case = 180 so cos = 1, and
the change in wavelength for the photon is given by = 2h=me c. The energy E 0 of the
scattered photon (whose initial energy is E = hc=) is then
E
E 0 = +hc = 1 + E= = 1 + (2h=mE c)(E=hc) = 1 + 2E=m
e
e c2
:0 keV
= 1 + 2(50:050keV)
=0:511 MeV = 41:8 keV :
(b) From conservation of energy the kinetic energy K of the electron is given by K =
E E 0 = 50:0 keV 41:8 keV = 8:2 keV.
44P
From the result of 38P
E = 1 = 0 = = (h=mc)(1 cos ) = hf 0 (1 cos ) :
E 0
0
0
c=f 0
mc2
Here the minus sign indicates that part of the photon energy is lost during the scattering.
45P
The magnitude of the fractional energy change for the photon is given by
Eph E
ph
=
(hc=) hc= =
1 = 1
1
= ;
=
+ + where = 10%. Thus = =(1 ). Plug in this expression for into Eq. 39-11 and
solve for cos :
mc = 1 (mc2 )
=
1
cos = 1 mc
h
h(1 )
(1 )Eph
keV) = 0:716 :
= 1 (1(10%)(511
10%)(200 keV)
The angle is = 44 .
CHAPTER 39
1075
PHOTONS AND MATTER WAVES
46P
The maximum energy of the electron is attained when the loss of energy of the photon is
the greatest:
E =
1
1
=
1
E
(=C )(1 cos ) + 1 max
=2C + 1 :
Conservation of energy then gives
2
Kmax = Emax = =2E + 1 = (hc=E )(mEc=2h) + 1 = m c2E=2 + E :
C
e
e
47P
Use the formula obtained in 46P:
2
(17:5 keV)2
Kmax = E + Em c2 =2 = 17:5 keV
+ 511 keV=2 = 1:12 keV :
e
48P
Rewrite Eq. 39-9 as
h
h cos = p v
cos m m0
1 (v=c)2
and Eq. 39-10 as
h sin = p v
sin :
m0
1 (v=c)2
Square both equations and add up the two sides:
h
m
2 "
1
2
1 cos 0
2 #
+ 10 sin 2
= 1 v(v=c)2 ;
where we used sin2 + cos2 = 1 to eliminate . Now the RHS can be written as
v2 = c2 1
1
1 (v=c)2
1 (v=c)2 ;
so
#
"
v2 = h 2 1 1 cos 2 + 1 sin 2 + 1 :
1 (v=c)2
mc
0
0
Now rewrite Eq. 39-8 as
h 1 1 +1= p 1
mc 0
1 (v=c)2
1076
CHAPTER 39
PHOTONS AND MATTER WAVES
and compare with the previous equation we obtained for [1 (v=c)2 ] 1 . This yields
h 1 1 +1 2 = h
mc 0
mc
2 "
1
2
1 cos 0
2 #
+ 10 sin + 1:
We have so far eliminated and v. Working out the squares on both sides and noting that
sin2 + cos2 = 1, we get
h (1 cos ) :
0 = = mc
49E
(a)
h =
6:63 10 34 J s
= hp = mv
(40 10 3 kg)(1000 m/s) = 1:7 10
35 m :
(b) The de Broglie wavelength of the bullet is much too short for the wave nature of the
bullet to reveal itself through diraction eects.
50E
(a) Substitute the classical relationship between momentum p and velocity v, v = p=me
into the classical denition of kinetic energy, K =p 12 me v2 , to obtain K = p2 =2me . Here
me is the mass of an electron. Solve for p: p = 2me K . The relationship between the
momentum and the de Broglie wavelength is = h=p, where h is the Planck constant.
Thus
= p h :
2me K
If K is given in electron volts then
6:626 p10 34 J s
= p
2(9:109 10 31 kg) (1:602 10 19 J/eV)K
1=2
1=2
9
p (eV) :
= 1:226 10p m (eV) = 1:226 nm
K
K
51E
Use the result of 50E and K = eV , where V = 25:0 kV. Thus
1=2
1=2
:226 nm (eV)1=2
p (eV) = 1:226 nm
p (eV) = 1p
= 1:226 nm
K
eV
e(25 103 V)
= 7:75 10 3 nm = 7:75 pm ;
CHAPTER 39
1077
PHOTONS AND MATTER WAVES
where we noted that 1 eV = 1e 1 V.
52E
(a)
= 38:7 pm :
= hp = p h = p hc 2 = p 1240 eV nm
2me K
2me c K
2(0:511 MeV)(1:00 keV)
(b)
(c)
1240 eV nm = 1:24 nm :
= hc
=
E
1:00 keV
= 9:04 10 4 nm = 904 fm :
= p hc 2 = p 1240 eV nm
mn c K
2(939 MeV)(1:00 keV)
p
53P
Use the result of 50E: = (1:226 nmeV1=2 )= K , where K is the kinetic energy. Solve for
K:
"
1=2 #2 " 1:226 nm (eV)1=2 #2
1
:
226
nm
(eV)
K=
=
= 4:32 10 6 eV :
590 nm
54P
(a) and (b) The momenta of the electron and the photon are the same:
10 34 J s = 3:3 10
p = h = 60:63
:20 10 9 m
24 kg m/s :
The kinetic energy of the electron is
2
:3 10 24 kg m/s)2 = 6:0 10
Ke = 2pm = (32(9
:11 10 31 kg)
e
18 J = 38 eV ;
while that for the photon is
Kph = pc = (3:3 10
24 kg m/s)(3:00 108 m/s) = 9:9 10 16 J = 6:2 keV :
55P
(a)
38 10 J/K)(300 K) = 3:88 10 2 eV = 38:8 meV :
K = 23 kT = 3(1:2(1
:60 10 19 J/eV)
23
1078
CHAPTER 39
PHOTONS AND MATTER WAVES
(b)
1240 nm eV
= p h = p hc 2 = p
2mn K
2(mn c )K
2(939 MeV)(3:88 10 2 eV)
= 1:46 10 10 m = 146 pm :
56P
(a) Solve v from = h=p = h=(mp v):
10 34 J s
v = mh = (1:675 106:6327 kg)(0
:100 10
p
12 m)
= 3:96 106 m/s :
(b) Let eV = K = 12 mp v2 and solve for V :
2
27
:96 106 m/s)2 = 8:18 103 V :
V = m2pev = (1:67 102(1:60kg)(3
10 19 C)
57P
(a) The average de Broglie wavelength is
= hp = p h = p h
= p hc2
2m(3kT=2)
2(mc )kT
2mK
1240 eV nm
=p
= 7:3 10
3(4)(938 MeV)(8:62 10 5 eV/K)(300 K)
11 m = 73 pm ;
while the average separation is
d = p31n =
1
p
3
p=kT
=
r
3
(1:38 10 23 J/K)(300 K) = 3:4 nm :
1:01 105 Pa
(b) Yes, since d.
58P
(a) The momentum of the photon is given by p = Eph =c, where Eph is its energy. Its
wavelength is
eV nm = 1240 nm = 1:24 nm :
= hp = Ehc = 1240
1:00 eV
ph
CHAPTER 39
1079
PHOTONS AND MATTER WAVES
p
The momentum of the electron is given by p = 2me K , where K is its kinetic energy and
me is its mass. Its wavelength is
= hp = p h :
2me K
According to 50E if K is in electron volts this is
p nm = 1:p226 nm = 1:23 nm :
= 1:226
1:00
K
(b) For the photon
eV nm = 1:24 10 6 nm = 1:24 fm :
= Ehc = 11240
:00 109 eV
ph
Relativity theory must be used to calculate the wavelength for the electron. The momentum
p and kinetic energy K of an electron are related by (pc)2 = K 2 + 2Kme c2 . Thus
p
p
pc = K 2 + 2Kme c2 = (1:00 109 eV)2 + 2(1:00 109 eV)(0:511 106 eV)
= 1:00 109 eV :
The wavelength is
1240 eV nm = 1:24 10 6 nm = 1:24 fm :
=
= hp = hc
pc 1:00 109 eV
59P
(a) For the photon
and for the electron
1240 nm eV = 1:24 keV
Eph = hc
=
1:00 nm
2
)2 = (hc=)2 =
1
1240 nm eV
K = 2pm = (h=
2
2me
2me c
2(0:511 MeV) 1:00 nm
e
(b) Now for the photon
nm eV = 1:24 GeV
Eph = 1240
1:00 fm
2
= 1:50 eV :
and for the electron
p
p
K = p2 c2 + (me c2 )2 me c2 = (hc=)2 + (me c2 )2 me c2
s
2
1240
fm
MeV
=
+ (0:511 MeV)2 0:511 MeV
1:00 fm
= 1:24 103 MeV = 1:24 GeV :
1080
CHAPTER 39
PHOTONS AND MATTER WAVES
Note that at short (large K ) the kinetic energy of the electron, calculated with relativistic
formula, is about the same as that of the photon. This is expected since now K E pc
for the electron, which is the same as E = pc for the photon.
60P
(a) The kinetic energy acquired is K = qV , where q is the charge on an ion and V is the
accelerating potential. Thus K = (1:60 10 19 C)(300 V) = 4:80 10 17 J. The mass of a
single sodium atom is, from Appendix F, m = (22:9898 g/mol)=(6:02 1023 atom/mol) =
3:819 10 23 g = 3:819 10 26 kg. Thus the momentum of an ion is
p
p
p = 2mK = 2(3:819 10
26 kg)(4:80 10 17 J) = 1:91 10 21 kg m/s :
(b) The de Broglie wavelength is
6:626 10 34 J s = 3:46 10
= hp = 1:915
10 21 kg m/s
61P
13 m = 346 fm :
p
We need to use the relativistic formula p = (E=c)2 m2e c2 E=c K=c (since E me c2 ). So
1240 eV nm = 0:025 fm ;
= hp hc
=
K
50 GeV
which is about 200 times smaller than the radius of an average nucleus.
62P
(a) Since
K = 7:5 MeV m c2 = 4(932 MeV), we may use the non-relativistic formula
p
p = 2m K . So
1240 nm eV
= 5:2 fm :
= hp = p hc 2 = p
2m c K
2(4)(932 MeV)(7:5 MeV)
(b) Since = 5:2 fm 30 fm, to a fairly good approximation the wave nature of the particle dose not need to be taken into consideration.
63P
The wavelength associated with the unknown particle is = h=p = h=(mv), where p is its
momentum, m is its mass, and v is its speed. The classical relationship p = mv was used.
Similarly, the wavelength associated with the electron is e = h=(me ve ), where me is its
mass and ve is its speed. The ratio of the wavelengths is =e = (me ve )=(mv), so
e e m = 9:109 10 31 kg = 1:675 10
m = vv
e 3(1:813 10 4 )
27 kg :
CHAPTER 39
1081
PHOTONS AND MATTER WAVES
According to Appendix B this is the mass of a neutron.
64P
p
(a) Let = h=p = h= (E=c)2 m2e c2 and solve for K = E me c2 :
K=
s
s
hc
2
+ m2e c4 me c2
2
1240
eV
nm
2
=
10 10 3 nm + (0:511 MeV) 0:511 MeV
= 0:015 MeV = 15 keV :
(b)
1240 eV nm = 1:2 105 eV = 120 keV :
=
E = hc
10 10 3 nm
(c) The electron microscope is more suitable, as the required energy of the electrons is
much less than that of the photons.
65P
The same resolution requires the same wavelength and since the wavelength and particle
momentum are related by p = h=, this means the same particle momentum. The momentum of a 100-keV photon is p = E=c = (100 103 eV)(1:60 10 19 J/eV)=(3:00 108 m/s) =
5:33 10 23 kg m/s. This is also the magnitude of the momentum of the electron.
In the classical approximation, the kinetic energy of the electron is
2
33 10 23 kg m/s)2 = 1:56 10
K = 2pm = (5:2(9
:11 10 31 kg)
e
15 J :
The accelerating potential is thus
56 10
V = Ke = 11::60
10
15 J
19 C
= 9:76 103 V :
p
Relativistically, the kinetic energy of the electron satises eV = K = (me c2 )2 + (pc)2
me c2 , which we may solve for V :
V = 1e
=1
hp
(me c2 )2 + (pc)2
hp
me
c2
i
(5:11 105 eV)2 + (100 103 eV)2
e
= 9:69 103 V :
i
5:11 105 eV
1082
CHAPTER 39
PHOTONS AND MATTER WAVES
66E
(a)
nn = (a + ib)(a + ib) = (a + ib)(a + i b ) = (a + ib)(a ib)
= a2 + iba iab + (ib)( ib) = a2 + b2 ;
which is always real since both a and b are real.
(b)
jnmj = j(a + ib)(c + id)j
= jac + iad + ibc + ( i)2 bdj
= j(ac bd) + i(ad + bc)j
p
= (ac bd)2 + (ad + bc)2
p
= a2 c2 + b2 d2 + a2 d2 + b2 c2 :
But
p
p
jnj jmj = ja + ibj jc + idj = a2 + b2 c2 + d2
p
= a2 c2 + b2 d2 + a2 d2 + b2 c2 ;
so jnmj = jnj jmj.
67P
Plug Eq. 39-17 into Eq. 39-16. Note that
d = d Aeikx + Be ikx = ikAeikx ikBe ikx
dx dx
and
Thus
d2 = d ikAeikx ikBe ikx = k2 Aeikx k2 Beikx :
dx2 dx
d2 + k2 = k2 Aeikx k2 Beikx + k2 Aeikx + Be ikx = 0 :
dx2
68P
Use the Euler formula ei = cos + i sin to re-write (x) as
(x) = 0 eikx = 0 (cos kx + i sin kx)
= ( 0 cos kx) + i( 0 sin kx) = a + ib ;
where a =
0 cos kx
and b =
0 sin kx
are both real quantities.
CHAPTER 39
1083
PHOTONS AND MATTER WAVES
(b)
(x; t) = (x)e i!t = 0 eikx e i!t = 0 ei(kx !t)
= [ 0 cos(kx !t)] + i [ 0 sin(kx !t)] :
69P
The angular wave number k is related to the wavelength by k = 2= and the wavelength
is related to the particle momentum p by = h=p, so k = 2p=h. Now the kinetic energy
K and thepmomentum are related by K = p2 =2m, where m is the mass of the particle.
Thus p = 2mK and
p
2
2mK :
k=
h
70P
For Epot = E0 , Schrodinger's equation becomes
d2 + 82 m (E E ) = 0 :
0
dx2
h2
Substitute
result is
=
0 eikx .
The second derivative is d2 =dx2 = k2 0 eikx = k2 . The
2
k2 + 8h2m (E E0 ) = 0 :
Solve for k and obtain
r
2
p
k = 8h2m [E E0 ] = 2h 2m(E E0 ) :
71P
From Eq. 39-14
j(x; y; z; t)j = =j
=j
=j
Here we used the result of 66E.
(x; y; z) e i!t = j (x; y; z)j e i!t (x; y; z)j jcos( !t) + i sin( !t)j
p
(x; y; z)j (cos !t)2 + (sin !t)2
(x; y; z)j :
1084
CHAPTER 39
PHOTONS AND MATTER WAVES
72P
The wave function is now given by
(x; t) = 0 e i(kx+!t) :
This function describes a plane matter wave traveling in the negative x direction. An example of the actual particles that t this description is a free electron with linear momentum
p =
(hk=2) i and kinetic energy K = p2 =2me = h2 k2 =82 me .
73P
(a) The wave function is now given by
(x; t) =
Thus
0
h
i
ei(kx !t) + e i(kx+!t) =
0e
i!t eikx + e ikx :
i!t eikx + e ikx 2
i!t 2 eikx + e ikx 2
j(x; t)j2 = 0 e
= 0e
= 02 eikx + e ikx 2
= 02 j(cos kx + i sin kx) + (cos kx i sin kx)j2
= 4 02 (cos kx)2
= 2 02 (1 + cos 2kx) :
(b)
4
3
|Ψ(x, t)/ψ0|2
2
1
0
0
2
4
6
kx
8
p
Consider two plane matter waves, each with the same amplitude 0 = 2 and travels in
opposite directions along the x axis. The combined wave is a standing wave:
(x; t) = 0 ei(kx !t) + 0 e i(kx+!t) = 0 eikx + e ikx e i!t
= (2 0 cos kx) e i!t :
CHAPTER 39
1085
PHOTONS AND MATTER WAVES
Thus the squared amplitude of the matter wave is
j(x; t)j2 = (2 0 cos kx)2 e
i!t 2 = 2 2 (1 + cos 2kx) ;
0
which is the same as the function plotted above.
(c) Set j(x; t)j2 = 2 02 (1 + cos 2kx) = 0 to obtain cos(2kx) = 1. This gives
2kx = 2 2 = (2n + 1) ;
Solve for x:
(n = 0; 1; 2; 3; . . . )
x = 14 (2n + 1) :
(d) The most probable positions for nding the particle are where j(x; t)j / (1 + cos 2kx)
reaches its maximum. Thus cos 2kx = 1, or
2kx = 2 2 = 2n ;
Solve for x:
(n = 0; 1; 2; 3; . . . )
x = 21 n :
74E
(a) Since px = py = 0, px = py = 0. Thus from Eq. 39-20 both x and y are innite.
It is therefore impossible to assign a y or z coordinate to the position of an electron.
(b) Since it is independent of y and z the wave function (x) should describe a plane
wave that extends innitely in both the y and z directions. Also from Fig. 39-12 we see
that j(x)j2 extends inifnitely along the x axis. Thus the matter wave described by (x)
extends throughout the entire thhree-dimensional space.
75E
10 34 J s = 2:1 10
p hx = 6:632(50
pm)
24 kg m/s :
76E
Denote the Plank's constant in the imaginary universe as h0 . Then the uncertainty in the
position of the baseball would be
0
h0 =
0:60 J s
x h=2p = 2m
v 2(0:50 kg)(1:0 m/s) = 0:19 m :
1086
CHAPTER 39
PHOTONS AND MATTER WAVES
It would be hard to catch such a baseball, since its size is less than the uncertainty in its
position.
77P
If x = =2 then from Eq. 39-20
2h = p ;
p = hx = =h2 = h=p
were we used h = 2h and p = h=.
For p = 0 the formula above gives p = 0. This is understandable, since now = h=p ! 1
so x = =2 ! 1. So while the measurement of momentum can be exact, it is now
impossible to locate the position x of the particle.
As the measured momentum p increases, so does p. This is also understandable because
as p increases x = =2 = h=p / 1=p decreases, keeping xp = h.
78P
(a)
1240 nm eV = 124 keV :
=
E = hc
10:0 10 3 nm
(b)
1
1
hc
=
E
=
hc
=
E = hc
+ + 1 + =
124 keV
E
=
= 1 + (= )(1
1
cos )
1 + (10:0 pm=2:43 pm)(1 cos 180 )
C
= 40:5 keV :
1
(c) It is impossible to \view" an atomic electron with such a high-energy photon, because
with the energy imparted to the electron the photon would have knocked the electron out
of its orbit.
79P
The probability T that a particle of mass m and energy E tunnels through a barrier of
height U and width L is given by
T = e 2kL ;
where
r
2
k = 8 m(hU2 E ) :
CHAPTER 39
1087
PHOTONS AND MATTER WAVES
For the proton
s
2
k = 8 (1:6726 10
3:0 Mev)(1:6022 10
(6:6261 10 34 J s)2
13 J/MeV)
3:0 MeV)(1:6022 10
(6:6261 10 34 J s)2
13 J/MeV)
27 kg)(10 Mev
= 5:8082 1014 m 1 ;
kL = (5:8082 1014 m 1 )(10 10 15 m) = 5:8082, and
T = e 25:8082 = 9:02 10 6 :
The value of k was computed to a greater number of signicant digits than usual because
an exponential is quite sensitive to the value of the exponent.
The mass of a deuteron is 2:0141 u = 3:3454 10 27 kg, so
s
2
k = 8 (3:3454 10
27 kg)(10 MeV
= 8:2143 1014 m 1 ;
kL = (8:2143 1014 m 1 )(10 10
15 m) = 8:2143,
and T = e 28:2143 = 7:33 10 8 :
80P
Let
T = exp( 2kL) = exp
and solve for E :
"
r
2
2L 8 m(hU2 E )
#
T 2
E = U 21m h4ln
L
2
(1240
eV
nm)(ln
0
:
001)
1
= 6:0 eV 2(0:511 MeV)
4(0:70 nm)
= 5:1 eV :
81P
(a) The rate at which incident protons arrive at the barrier is n = 1:0 kA=1:60 10
6:25 1023 =s. Let nTt = 1 to nd the waiting time t:
t = (nT )
1=
"
r
1 exp 2L 82 mp (U E )
n
h2
#
(0:70 nm) p8(938 MeV)(6:0 eV 5:0 eV)
= 6:25 11023 =s exp 21240
eV nm
111
104
= 3:37 10 s 10 y ;
19 C =
1088
CHAPTER 39
PHOTONS AND MATTER WAVES
which is much, much longer than the age of the universe.
(b) Replace the mass of the proton with that of the electron to obtain the corresponding
waiting time for an electron:
#
" r
2 me (U E )
8
1
t = (nT ) 1 = exp 2L
n
h2
(0:70 nm) p8(0:511 MeV)(6:0 eV 5:0 eV)
= 6:25 11023 =s exp 21240
eV nm
= 2:1 10 19 s :
The enormous dierence between the two waiting times is the result of the dierence
between the masses of the two kinds of particles.
82P
(a) If m is the mass of the particle and E is its energy then the probability it will tunnel
through a barrier of height U and width L is given by
T = e 2kL ;
where
r
2
k = 8 m(hU2 E ) :
If the change U in U is small (as it is) the change in the tunneling probability is given by
Now
Thus
dT U = 2LT dk U :
T = dU
dU
r
r
82 m =
82 m(U E ) =
1
k :
dk = p 1
dU 2 U E h2
2(U E )
h2
2(U E )
U
T = LTk U E :
For the data of Sample Problem 39-7, 2kL = 10:0, so kL = 5:0 and
T = kL U = (5:0) (0:0100)(6:8 eV) = 0:20 :
T
U E
6:8 eV 5:1 eV
There is a 20% decrease in the tunneling probability.
(b) The change in the tunneling probability is given by
and
2kL L = 2kT L
T = dT
dL L = 2ke
T = 2k L = 2(6:67 109 m 1 )(0:0100)(750 10
T
There is a 10% decrease in the tunneling probability.
12 m) =
0:10 :
CHAPTER 39
1089
PHOTONS AND MATTER WAVES
(c) The change in the tunneling probability is given by
dT E = 2Le 2kL dk E = 2LT dk E :
T = dE
dE
dE
Now dk=dE = dk=dU = k=2(U E ), so
T = kL E = (5:0) (0:01000)(5:1 eV) = 0:15 :
T
U E
6:8 eV 5:1 eV
There is a 15% increase in the tunneling probability.
83P
The kinetic energy of the car of mass m moving at speed v is given by E = 12 mv2 ; while
the potential barrier it has to tunnel through is Epot = mgh, where h = 24 m. According
to Eq. 39-21 and 39-22 the tunneling probability is given by T e 2kL , where
k=
r
82 m(Epot
h2
s
1 mv 2 )
2
E ) = 82 m(mgh
h2
s 2(1500 kg)
2 )(24 m) 1 (20 m/s)2
= 6:626
2
(9
:
80
m/s
34
10 J s
2
= 1:2 1038 m 1 :
Thus 2kL = 2(1:2 1038 m 1 )(30 m) = 7:2 1039 . You can see that T e 2kL is almost
zero.
84P
(a) Set Epot = 0 in Eq. 39-15:
d2 + 82 mE = 0 :
dx2
h2
For the wave function (x) = Aeikx + Be ikx
d2 = (ik)2 Aeikx + ( ik)2 Be ikx = k2 (Aeikx + B ikx ) = k2 :
dx2
Plug this into the Schrodinger's equation to obtain
k2
2
(x) + 8hmE
(x) =
2
k2 +
82 mE
h2
(x) = 0 :
1090
CHAPTER 39
Since (x) 6= 0,
PHOTONS AND MATTER WAVES
2
k2 + 8hmE
2 = 0:
Thus
82 mE = 82 m p2 = 42 p2 ;
h2
h2 2m
h2
or k = 2p=h. Here k is arbitrarily set to be positive, which does not aect the generality
of the solution since it contains both e+ikx and e ikx .
(b)
k2 =
j (x)j2 = Aeikx + Be ikx 2
= jA cos kx + iA sin kx + B cos kx iB sin kxj2
= j(A + B ) cos kx + i(A B ) sin kxj2
= [(A + B ) cos kx]2 + [(A B ) sin kx]2
= (A2 + 2AB + B 2 ) cos2 kx + (A2 2AB + B 2 ) sin2 kx
= (A2 + B 2 )(cos2 kx + sin2 kx) + 2AB (cos2 kx sin2 kx)
= A2 + B 2 + 2AB cos 2kx ;
where to reach the last step we used the trigonometric identities cos2 + sin2 = 1 and
cos2 sin2 = cos 2.
(c) Since both A and B are constants j (x)j2 is a function of cos 2kx, which varies between
1 and +1. Thus j (x)j2 varies between A2 + B 2 + 2AB ( 1) = (A B )2 and A2 + B 2 +
2AB (+1) = (A + B )2 .
85P
(a) For Epot = const. > E Eq. 39-15 can be written as
d2 + 82 m(Epot E ) = d2 + k2 = 0 ;
dx2
h2
dx2
where
r
2
E) :
k = 8 m(Ehpot
2
Plug the trial solution (x) = Ce kx into the Schrodinger's equation:
( k)2 Ce kx + k2 Ce kx = 0 :
Thus (x) = Ce kx is indeed a solution to the equation in this region. You can verify that
in general there can be another solution in the form De+kx , where D is another constant.
This solution, however, should be discarted in our case becasue it leads to the unphysical
limit (x) ! 1 at x ! 1.
(b) Since (x) is real,
j (x)j2 = [ (x)]2 = Ce
kx 2 = C 2 e 2kx :