OpenStax College Physics
Instructor Solutions Manual
Chapter 29
CHAPTER 29: INTRODUCTION TO
QUANTUM PHYSICS
29.1 QUANTIZATION OF ENERGY
1.
A LiBr molecule oscillates with a frequency of 1.7  1013 Hz . (a) What is the difference
in energy in eV between allowed oscillator states? (b) What is the approximate value
of n for a state having an energy of 1.0 eV?
Solution (a) E  hf  6.63  1034 J  s 1.7  1013 s 1   1.127  10 20 J
1 eV


2
2
so that 1.127  10 20 J 
  7.04  10 eV  7.0  10 eV
19
1
.
60

10
J


(b) Using the equation E = nhf , we can solve for n :
n
2.
Solution


1.0 eV 1.60 1019 J/eV  1  13.7  14
E 1
 
hf 2 6.63 10 34 J  s 1.7 1013 s 1 2



The difference in energy between allowed oscillator states in HBr molecules is 0.330
eV. What is the oscillation frequency of this molecule?
E  hf  f 


E 0.330 eV  1.60 10 19 J/eV

 7.96 1013 Hz
34
h
6.63 10 J  s
3.
A physicist is watching a 15-kg orangutan at a zoo swing lazily in a tire at the end of a
rope. He (the physicist) notices that each oscillation takes 3.00 s and hypothesizes that
the energy is quantized. (a) What is the difference in energy in joules between allowed
oscillator states? (b) What is the value of n for a state where the energy is 5.00 J? (c)
Can the quantization be observed?
Solution
(a) f 
1
1

 0.333 sec 1 so that
T 3.00 s
E  hf  6.63  1034 J  s 0.333 sec 1   2.21 1034 J
E 1
E 1
5.00 J
1
(b) n 
 
 
  2.26  1034
 34
hf 2 E 2 2.210  10 J 2
(c) No, E is much too small and n is much too large.
29.2 THE PHOTOELECTRIC EFFECT
4.
What is the longest-wavelength EM radiation that can eject a photoelectron from
silver, given that the binding energy is 4.73 eV? Is this in the visible range?
OpenStax College Physics
Solution
Instructor Solutions Manual
KE  hf  BE ; KE  0  hf 


hc


Chapter 29
 BE 

hc
6.63  10 34 J  s 3.00  108 m/s

 2.63  10 7 m  263 nm
19
4.73 eV  1.60 10 J/eV
BE


No, this is UV.
5.
Find the longest-wavelength photon that can eject an electron from potassium, given
that the binding energy is 2.24 eV. Is this visible EM radiation?
Solution
KE  hf  BE ; KE  0  hf 

hc


 BE 

hc
6.63  10 34 J  s 3.00  108 m/s

 5.55  10 7 m  555 nm
2.24 eV  1.60 10 19 J/eV
BE
Yes, it is visible as green light.



6.
What is the binding energy in eV of electrons in magnesium, if the longest-wavelength
photon that can eject electrons is 337 nm?
Solution
KE  hf  BE; KE  0  BE  hf 
BE 
hc

hc

6.63 10 J  s 3.00 10

3.37 10 m 
34
 5.902 10 19 J 
8

m/s

9
1 eV 
1.60 10
19
J
  3.69 eV
7.
Calculate the binding energy in eV of electrons in aluminum, if the longest-wavelength
photon that can eject them is 304 nm.
Solution
The longest wavelength corresponds to the shortest frequency, or the smallest
energy. Therefore, the smallest energy is when the kinetic energy is zero. From the
equation KE  hf  BE  0 , we can calculate the binding energy (writing the
frequency in terms of the wavelength):
hc
BE  hf 


hc 6.63 10 34 J  s 3.00 108 m/s 
BE 


3.04 10 7 m 
 1.000 eV 
 6.543 10 19 J  
  4.09 eV
19
 1.60 10 J 
8.
What is the maximum kinetic energy in eV of electrons ejected from sodium metal by
OpenStax College Physics
Instructor Solutions Manual
Chapter 29
450-nm EM radiation, given that the binding energy is 2.28 eV?
Solution
hf 
hc


6.63  10 J  s3.00  10 m/s 
4.50  10 m
34
8
7
 1.000 eV 
 4.42  10 19 J  
 2.7625 eV
19 
 1.60  10 J 
KE  hf  BE  2.7625 eV  2.28 eV  0.483 eV
9.
Solution
UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons
are bound by 4.82 eV. What is the maximum kinetic energy of the ejected
photoelectrons?
hf 
hc

6.63  10 J  s 3.00  10

1.20  10 m 
34
8
m/s

7
 1.000 eV 
 1.6575  10 18 J  
  10.359 eV
19
 1.602  10 J 
KE  hf  BE  10.359 eV  4.82 eV  5.54 eV
10.
Solution
Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of
0.860 eV from sodium metal. What is the binding energy of electrons to sodium
metal?
hf 
hc


6.63  10 J  s 3.00  10
4.00  10 m 
34
8
m/s

7
 1.000 eV 
 4.9725  10 19 J  
  3.1078 eV
19
 1.60  10 J 
KE  hf  BE  BE  hf  KE  3.1078 eV  0.86 eV  2.25 eV
11.
Solution
UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV
electrons. What is the binding energy of electrons to uranium metal?
hf 
hc


6.63  10 J  s 3.00  10
3.00  10 m 
34
8
m/s

7
 1.000 eV 
 6.63  10 19 J  
  4.14 eV
19
 1.602  10 J 
KE  hf  BE  BE  hf  KE  4.14 eV  0.500 eV  3.64 eV
12.
What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium
metal, given that the binding energy is 2.71 eV? What type of EM radiation is this?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 29
KE  hf  BE 
 1.60  10 19 J 
  7.536  10 19 J
hf  BE  KE  2.71 eV  2.00 eV  4.71 eV  
1 eV


34
8
hc
hc 6.63  10 J  s 3.00  10 m/s
hf 
 

 2.639  10 7  264 nm

hf
7.536  10 19 J
This is ultraviolet radiation.





13.
Find the wavelength of photons that eject 0.100-eV electrons from potassium, given
that the binding energy is 2.24 eV. Are these photons visible?
Solution
KE  hf  BE and c  f so



hc
6.63 10 34 J  s 3.00 108 m/s  1.000 eV 


19 
BE  KE
2.24 eV  0.100 eV
 1.60 10 J 
 5.313 10 7 m  531 nm.

Yes, these photons are visible.
14.
What is the maximum velocity of electrons ejected from a material by 80-nm photons,
if they are bound to the material by 4.73 eV?
Solution
KE  hf  BE 
hc
 BE
λ
 1.60 10 19 J 
6.63 10 34 J  s 3.00 108 m/s






4.73eV
1 eV
8.00 10 8 m





 1.7295 10
18


J


1
2
1
2KE  2  1.729 10 J 
KE  mv 2  v 

 1.95  10 6 m/s

31
2
m
 9.1110 kg 
15.
18
Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a
detection device?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 29
hc
 BE
λ
 1.60  10 19
6.63  10 34 J  s 3.00  10 8 m/s





2.71eV
1 eV
4.20  10 7 m

KE  hf  BE 



 3.997  10
 20


J


J


1
1
2KE  2 3.997  10  20 J  2
5
KE  mv 2  v 

  2.962  10 m/sec
31
2
m
 9.11  10 kg 
2
d
2.50  10 m
t 
 8.44  10 8 s
5
v 2.962  10 m/sec
16.
Solution
A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto
calcium metal. (a) How many electrons per second are ejected? (b) What power is
carried away by the electrons, given that the binding energy is 2.31 eV?
(a) E  hf 



hc 6.63  10 34 J  s 3.00  10 8 m/s

 4.9725  10 19 J
7
λ
4.00  10 m


2.00 103 J/s
 4.02 1015 /s
19
4.9725 10 J
(b) KE  hf  BE  Eγ  BE
N
 1.60  10 19 J 
  1.3245  10 19 J
 4.9725  10 19 J  2.28 eV 
1
eV


15
19
P  NKE  4.022  10 /s 1.3245  10 J  5.33  10 4 W  0.533 mW



17.
(a) Calculate the number of photoelectrons per second ejected from a 1.00-mm2 area
of sodium metal by 500-nm EM radiation having an intensity of 1.30 kW/m 2 (the
intensity of sunlight above the Earth’s atmosphere). (b) Given that the binding energy
is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away
less power than brought in by the photons. Where does the other power go? How can
it be recovered?
Solution
(a) First calculate n, the number of photons per second hitting a square-meter area.
Each photon has energy
6.626 10 34 J  s 2.998 108 m/s   3.973 10 19 J 
E 
5.00 10 7 m 
1.30  10 3 J/s  m 2
 3.272  10 21 / s  m 2
3.973  10 19 J
The number of ejected electrons per second equals the number of photons per
second hitting the sodium metal, or N  nA :
n
OpenStax College Physics

Instructor Solutions Manual

Chapter 29
2

 1m 
15
N  3.272  10 /s  m 1.00 nm 
  3.27  10 /s
 1000 nm 
hc
(b) KE  hf  BE 
 BE
λ
 1.60  10 19 J 
  3.204  10  20 J
 3.973  10 19 J  2.28 eV 
1
eV


21
2
2



P  N  KE  3.272  1015 /s 3.204  10  20 J  1.05  10  4 W  0.105 mW
(c) The other energy goes to the sodium metal to free the electrons (binding energy).
This lost power can be recovered by the spontaneous absorption of electrons by
sodium metal. For each electron absorbed an energy of 2.28eV will be released.
18.
Unreasonable Results Red light having a wavelength of 700 nm is projected onto
magnesium metal to which electrons are bound by 3.68 eV. (a) Use KEe  hf – BE to
calculate the kinetic energy of the ejected electrons. (b) What is unreasonable about
this result? (c) Which assumptions are unreasonable or inconsistent?
Solution
(a) hf 
6.63  10
34


J  s 3.00  108 m/s

7.00  10 7 m


1 eV
  1.776 eV
 2.841  10 19 J 
19
 1.602  10 J 
hc




and since KE  hf  BE  1.776 eV  3.68 eV   1.90 eV
(b) Negative kinetic energy is impossible.
(c) The assumption that the photon can knock the electron free is unreasonable.
19.
Unreasonable Results (a) What is the binding energy of electrons to a material from
which 4.00-eV electrons are ejected by 400-nm EM radiation? (b) What is
unreasonable about this result? (c) Which assumptions are unreasonable or
inconsistent?
Solution
(a) We want to use the equation KE  hf  BE to determine the binding energy, so
we first need to determine an expression of hf . Using E  hf , we know:
hc 6.626  10 34 J  s 2.998  108 m/s 
hf 


4.00  10 7 m


1 eV
  3.100 eV
 4.966  10 19 J 
19
 1.602  10 J  
and since KE  hf  BE : BE  hf  KE  3.100 eV  4.00 eV   0.90 eV
(b) The binding energy is too large for the given photon energy.
(c) The electron’s kinetic energy is too large for the given photon energy; it cannot be
greater than the photon energy.
OpenStax College Physics
Instructor Solutions Manual
Chapter 29
29.3 PHOTON ENERGIES AND THE ELECTROMAGNETIC SPECTRUM
20.
What is the energy in joules and eV of a photon in a radio wave from an AM station
that has a 1530-kHz broadcast frequency?
Solution
E  hf  6.63 10 34 J/s 1.53 10 6 s 1  1.0144 10 27 J  1.0110 27 J



1 eV


 1.0144 10 27 J
 6.340 10 9 eV  6.34 10 9 eV
19 
 1.60 10 J 
21.
(a) Find the energy in joules and eV of photons in radio waves from an FM station that
has a 90.0-MHz broadcast frequency. (b) What does this imply about the number of
photons per second that the radio station must broadcast?
Solution
34
8 1
26
(a) E  hf  6.63 10 J/s 9.00 10 s  5.97 10 J
1 eV


 5.97 10 26 J
 3.73 10 7 eV
19 
 1.60 10 J 
(b) This implies that a tremendous number of photons must be broadcast per second.
In order to have a broadcast power of, say 50.0 kW, it would take
5.00  10 4 J/s
 8.38  10 29 photon/sec .
5.97  10 26 J/photon
22.
Calculate the frequency in hertz of a 1.00-MeV γ -ray photon.
Solution

E  hf  f 


E
1.00  10 6 eV

 2.42  10 20 Hz
15
h 4.14  10 eV  s
23.
(a) What is the wavelength of a 1.00-eV photon? (b) Find its frequency in hertz. (c)
Identify the type of EM radiation.
Solution
(a) E 
24.
Do the unit conversions necessary to show that hc  1240 eV  nm , as stated in the
text.
Solution
 10 9 nm  1.00 eV 

hc  6.63  10 34 J  s 3.00  108 m/s 
 1240 eV  nm
19 
1
m
1.60

10
J




25.
Confirm the statement in the text that the range of photon energies for visible light is
1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.
hc
hc 1240 eV  nm
λ

 1240 nm  1.24 µm
λ
E
1.00 eV
c 3.00 108 m/s
(b) f  
 2.42 1014 Hz
λ 1.24 10 6 m
(c) The radiation is in the infrared part of the spectrum.



OpenStax College Physics
Solution
Instructor Solutions Manual
E min 
hc 1240 eV  nm

 1.63 eV
λ max
760 nm
E max 
hc 1240eV  nm

 3.26 eV
λ min
380 nm
Chapter 29
26.
(a) Calculate the energy in eV of an IR photon of frequency 2.00  1013 Hz . (b) How
many of these photons would need to be absorbed simultaneously by a tightly bound
molecule to break it apart? (c) What is the energy in eV of a γ ray of frequency
3.00  1020 Hz ? (d) How many tightly bound molecules could a single such γ ray
break apart?
Solution
1eV
(a) E  hf  6.63 10 34 J  s 2.00 1013 s 1 
 0.0829 eV
1.60 10 -19 J
10.0 eV
(b)
 121
8.29  10 1 eV
(c) E  hf  4.144  10 16 eV3.00  10 20 s 1   1.242  10 6 eV  1.24 MeV
(d)
1.242 10 6 eV
 1.24 10 5
10.0 eV
27.
Prove that, to three-digit accuracy, h  4.14  10 15 eV  s , as stated in the text.
Solution
 1.00 eV
h  6.63  10 34 J  s 
19
 1.60  10
28.
(a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV
accelerating potential, such as a color TV? (b) What is their frequency?
Solution
1.00 eV
(a) qv  1.60 10 19 C2.50 10 4 V 
 2.50 10 4 eV
19
1.60 10 J
4
E
2.5.0 10 eV
(b) E  hf  f  
 6.04 1018 Hz
15
h 4.14 10 eV  s
29.
What is the accelerating voltage of an x-ray tube that produces x rays with a shortest
wavelength of 0.0103 nm?
Solution
hc
hc 6.63  10 34 J  s 3.00  10 8 m/s
qV  E 
V 

 1.21  10 5 V  121 kV
19
11
λ
qλ
1.60  10 C 1.03  10 m
30.
(a) What is the ratio of power outputs by two microwave ovens having frequencies of
950 and 2560 MHz, if they emit the same number of photons per second? (b) What is
the ratio of photons per second if they have the same power output?





15
15
  4.144  10 eV  s  4.14  10 eV  s
J




OpenStax College Physics
Instructor Solutions Manual
Chapter 29
Solution
(a) Let N be the number of photons per second
P
N hf
N f
2560 MHz
 P  NE  Nhf 1  1 1  1 1 
 2.69 since N1  N 2
P2 N 2 hf 2 N 2 f 2
950 MHz
P
N hf
N
f
950 MHz
(b) 1  1 1  1  2 
 0.371 since P1  P2
P2 N 2 hf 2 N 2
f1 2560 MHz
31.
How many photons per second are emitted by the antenna of a microwave oven, if its
power output is 1.00 kW at a frequency of 2560 MHz?
Solution
Eγ  hf  6.63  10 34 2.56  109 s 1  1.697  10 24 J



1.00  103 J/s
N
 5.89  10 26 photons/se cond
24
1.697  10 J/photon
32.
Some satellites use nuclear power. (a) If such a satellite emits a 1.00-W flux of γ rays
having an average energy of 0.500 MeV, how many are emitted per second? (b) These
γ rays affect other satellites. How far away must another satellite be to only receive
one γ ray per second per square meter?
Solution
5
19
14
(a) Eγ  5.00 10 eV 1.60 10 J/eV  8.00 10 J



1.00 103 J/s
 1.25 1013 photons/s
14
8.00 10 J/photon
(b) Φ N  photons/m 2  N 
4π r 2
N
 N
r  
 4π  N
33.
Solution



1
2
1.25  1013 photon/s 

2 
 4π1 photon/s  m  
1
2
 9.97  10 5 m  997 km
(a) If the power output of a 650-kHz radio station is 50.0 kW, how many photons per
second are produced? (b) If the radio waves are broadcast uniformly in all directions,
find the number of photons per second per square meter at a distance of 100 km.
Assume no reflection from the ground or absorption by the air.
(a) Eγ  hf  6.631034 Js 6.50 105 s 1   4.311028 J
5.00  104 J/s
 1.16  1032 photon/s  1.16  1032 photon/s
4.31  1028 J/photon
(b) To calculate the flux of photons, we assume that the broadcast is uniform in all
directions, so the area is the surface area of a sphere giving:
Then, N 
ΦN 
N
1.16 1032 photons/s

 9.23 10 20 photons/s  m 2
2
2
5
4π r
4π 1.00 10 m


OpenStax College Physics
Instructor Solutions Manual
Chapter 29
34.
How many x-ray photons per second are created by an x-ray tube that produces a flux
of x rays having a power of 1.00 W? Assume the average energy per photon is 75.0
keV.
Solution
Eγ  7.50 10 4 eV 1.60 10 19 J/eV  1.20 10 14 J

N


1.00 J/s
 8.33 1013 photon/s
1.20 10 14 J/photon
35.
(a) How far away must you be from a 650-kHz radio station with power 50.0 kW for
there to be only one photon per second per square meter? Assume no reflections or
absorption, as if you were in deep outer space. (b) Discuss the implications for
detecting intelligent life in other solar systems by detecting their radio broadcasts.
Solution
 N  2 1.1611032 photon/s  2
N
15
  
 r  
(a) Φ N 
  3.04 10 m
2
2
4π r
 4π 1 photon/s  m 
 4π Φ N 
(b) The distance calculated in part (a) is approximately 1/3 ly. Therefore, if radio
stations from intelligent life in other solar systems are to be detected, their
broadcasts would have to have substantial power outputs. Also, since there are
stray radio waves in outer space, their signals would have to be large compared to
the background radio waves. This means it is rather unlikely for us to detect
intelligent life by detecting their radio broadcasts.
1
1


36.
Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range
(typical for incandescent bulbs) with an average wavelength of 580 nm, and that the
photons spread out uniformly and are not absorbed by the atmosphere, how far away
would you be if 500 photons per second enter the 3.00-mm diameter pupil of your
eye? (This number easily stimulates the retina.)
Solution
ΦN 

500
3
π 1.50  10 m

2
 7.073  10 7 photons/s  m
 N
N
ΦN 
 r  
2
4π r
 4π N



5
 1.8110 m  181 km
1
2
2


2.92  1019 photon/s

7
2 
 4π 7.073  10 photon/s  m 

1
2

29.4 PHOTON MOMENTUM
38.
Solution
(a) Find the momentum of a 4.00-cm-wavelength microwave photon. (b) Discuss why
you expect the answer to (a) to be very small.
(a) p 
h 6.63  10 34 J  s

 1.66  10 32 kg  m/s
2
λ
4.00  10 m
OpenStax College Physics
Instructor Solutions Manual
Chapter 29
(b) The wavelength of microwave photons is large, so the momentum they carry is
very small.
39.
Solution
(a) What is the momentum of a 0.0100-nm-wavelength photon that could detect
details of an atom? (b) What is its energy in MeV?
h 6.63 10 34 J  s

 6.63 10 23 kg.m/s
11
λ 1.00 10 m
hc
(b) E 
 pc  6.63  10  23 kg  m/s 3.00  10 8 m/s
λ
(a) p 



1 MeV
13
 1.602  10

  0.124 MeV
J
40.
(a) What is the wavelength of a photon that has a momentum of 5.00  1029 kg  m/s ?
(b) Find its energy in eV.
Solution
h
h
6.63  10 34 J.s
 1.326  10 5 m  13.3 m
(a) Using p  , λ  
 29
λ
p 5.00  10 kg  m/s
(b) Using
E  pc  5.00 10 29 kg  m/s 3.00 108 m/s



1 eV


 1.50 10 20 J  
 9.38 10 -2 eV
19 
 1.60 10 J 
41.
Solution
(a) A γ -ray photon has a momentum of 8.00  1021 kg  m/s . What is its wavelength?
(b) Calculate its energy in MeV.
(a) λ 
h
6.63 10 34 J  s

 8.29 10 14 m
21
p 8.00 10 kgm/s



21
8
(b) E  pc  8.00  10 kgm/s 3.00  10 m/s
1 MeV


 2.40  10 12 J  
 15.0 MeV
13 
 1.602  10 J 
42.
Solution
(a) Calculate the momentum of a photon having a wavelength of 2.50 μm . (b) Find
the velocity of an electron having the same momentum. (c) What is the kinetic energy
of the electron, and how does it compare with that of the photon?
(a) p λ 
h 6.63 10 34 Js

 2.65 10 28 kgm/s
6
λ 2.50 10 m
pe 2.652 10 28 kgm/s

 291 m/s
(b) pe  me ve  ve 
me
9.1110 31 kg
OpenStax College Physics
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Chapter 29
2
1
2
31
2
 26
(c) KE e  mve  0.59.11 10 kg 2.911 10 m/s   3.86  10 J
2
E γ  p γ c  2.652  10 28 kgm/s 3.00  108 m/s   7.96  10 20 J;
Eγ
KE e
43.
Solution
 2.06  10 6
Repeat the previous problem for a 10.0-nm-wavelength photon.
(a) pγ 
h 6.63 10 34 J  s

 6.63 10 26 kg.m/s
8
λ
1.00 10 m
pe 6.63 10 34 kg  m/s

 7.28 10 4 m/s
(b) ve 
31
me
9.1110 kg
2
1
2
31
4
 21
(c) KE e  mve  0.59.11 10 kg 7.277  10 m/s   2.41 10 J
2
Eγ  p γ c  6.63  10 26 kg  m/s 3.00  108 m/s   1.99  10 17 J;
Eγ
KE e
44.
 8.25  10 3
(a) Calculate the wavelength of a photon that has the same momentum as a proton
moving at 1.00% of the speed of light. (b) What is the energy of the photon in MeV?
(c) What is the kinetic energy of the proton in MeV?




Solution (a) p p  mv p  1.67  10 27 kg 1.00  10 2 3.00  10 8 m/s  5.01  10 21 kg  m/s
pγ 
h
h
6.63  10 34 J  s
λ

 1.32  10 13 m
λ
p γ 5.01  10  21 kg  m/s


(b) Eγ  pγ c  5.0110 21 kg  m/s 3.00 108 m/s

 1 MeV 
 1.503 10 12 J  
 9.39 MeV
13 
 1.60 10 J 
(c) KE  1 mv 2  0.5 1.67 10 27 kg 1.00 10 2 3.00 108 m/s
2
 7.515 10 15 J  4.70 10 2 MeV

45.
Solution



2
(a) Find the momentum of a 100-keV x-ray photon. (b) Find the equivalent velocity of a
neutron with the same momentum. (c) What is the neutron’s kinetic energy in keV?
(a) E  pc  p 



E 1.00 105 eV 1.60 10 19 J/eV

 5.33 10 23 kg  m/s
c
3.00 108 m/s
OpenStax College Physics
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Chapter 29
p 5.333  10 23 kg  m/s

 3.19  10 4 m/s
 27
m
1.67  10 kg
(c) KE  1 mv 2  0.51.67  10  27 kg 3.194  10 4 m/s 2
2
1 keV


3
 8.516  10 19 J  
  5.32  10 keV
16
 1.60  10 J 
(b) v 
46.
Take the ratio of relativistic rest energy, E  γmc 2 , to relativistic momentum,
p  γmu , and show that in the limit that mass approaches zero, you find E / p  c .
Solution
E mc 2 c 2


p mu
u
As the mass of particle approaches zero, its velocity u will approach c so that the
E c2
 c , which is consistent
ratio of energy to momentum approaches lim m0 
p c
E
with the equation p  for photons.
c
48.
Unreasonable Results A car feels a small force due to the light it sends out from its
headlights, equal to the momentum of the light divided by the time in which it is
emitted. (a) Calculate the power of each headlight, if they exert a total force of
2.00  10 2 N backward on the car. (b) What is unreasonable about this result? (c)
Which assumptions are unreasonable or inconsistent?
Solution
(a) F 
E  mc 2 and p  mu so
p E / c
E

  P  Fc
t
t
t
2
P  1.00 10 N 3.00 108 N   3.00 10 6 W
(b) This is far too much energy for a car headlight.
(c) The force assumed is much too large, although it would have only a small effect on
the car.
29.6 THE WAVE NATURE OF MATTER
49.
Solution
50.
At what velocity will an electron have a wavelength of 1.00 m?
λ
h
h
h
6.63 10 34 J  s

v

 7.28 10 4 m
31
p mv
mλ 9.1110 1.00 m


What is the wavelength of an electron moving at 3.00% of the speed of light?
OpenStax College Physics
Solution
51.
Solution
52.
Solution
53.
Solution
54.
Solution
λ
Instructor Solutions Manual
h
6.63 10 34 J  s

 8.09 10 11 m
mv 3.00 10 2 3.00 108 m/s



At what velocity does a proton have a 6.00-fm wavelength (about the size of a
nucleus)? Assume the proton is nonrelativistic. (1 femtometer = 10 15 m .)
λ
h
h
h
6.63 10 34 J  s

v

 6.62 107 m/s
27
15
p mv
mλ 1.67 10 kg 6.00 10 m


Solution

What is the velocity of a 0.400-kg billiard ball if its wavelength is 7.50 cm (large
enough for it to interfere with other billiard balls)?
λ
h
h
h
6.63 10 34 J  s

v

 2.2110 32 m/s
p mv
mλ 0.400 kg  7.50 10 2 m


Find the wavelength of a proton moving at 1.00% of the speed of light.
λ
h
h
6.63  10 34 J  s


 1.32  10 -13 m
 27
8
p mv 1.67  10 kg 0.01  3.00  10 m/s



Experiments are performed with ultracold neutrons having velocities as small as 1.00
m/s. (a) What is the wavelength of such a neutron? (b) What is its kinetic energy in
eV?
h
h
6.63  1034 J  s


 3.956  10 7 m  396 nm
p mv 1.675  10 27 kg 1.00 m/s 
1
1
2
(b) KE  mv 2   1.675  10 27 kg 1.00 m/s 
2
2
1 eV


9
 8.375  10  28 J  
  5.23  10 eV
19
 1.602  10 J 
(a) λ 



55.
Chapter 29

(a) Find the velocity of a neutron that has a 6.00-fm wavelength (about the size of a
nucleus). Assume the neutron is nonrelativistic. (b) What is the neutron’s kinetic
energy in MeV?
h
h
h
6.63 10 34 J  s

v

 6.62 107 m/s .
27
15
p mv
mλ 1.67 10 kg 6.00 10 m
2
1
(b) KE  mv 2  0.5 1.67 10 27 kg 6.62 10 7 m/s
2
 1 MeV 
 3.659 10 12 J  
 22.9 MeV
13 
 1.60 10 J 
(a) λ 






OpenStax College Physics
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Chapter 29
56.
What is the wavelength of an electron accelerated through a 30.0-kV potential, as in a
TV tube?
Solution
1
 2qV  2
KE  qV  mv 2  v  

2
 m 
h
h
h
6.63 10 34 J  s
λ 


p mv 2qVm 12
2 1.60 10 19 C 3.00 10 4 V 9.1110 31 kg
1
 



1
2
 7.09 10 12 m
57.
What is the kinetic energy of an electron in a TEM having a 0.0100-nm wavelength?
Solution
h2
(6.63  10 34 J  s) 2
KE 

 1 keV / 1.60  10 16 J  15.1 keV
2
31
11
2
2m
(2) 9.11  10 kg (10 m)
Note that in the previous problem the electron with KE of 30 KeV was found to have
a wavelength of 0.007 nm. To obtain this  of 0.01 nm, the electron would have to
accelerate through a potential difference of (0.007 / 0.010) 2  30 kV  15 kV , which is
consistent with the above result.
58.
(a) Calculate the velocity of an electron that has a wavelength of 1.00 μm . (b)
Through what voltage must the electron be accelerated to have this velocity?
Solution

(a) λ 

h
h
h
6.63 10 34 J  s

v

 7.28 10 2 m/s
31
6
p mv
mλ 9.1110 kg 1.00 10 m



31


1 2
mv
9.11  10 kg 7.278  10 2 m/s 2

(b) qV  mv  V 
2
2q
2 1.60  10 19 C
2
59.
Solution



2
 1.51  10 6 V
The velocity of a proton emerging from a Van de Graaff accelerator is 25.0% of the
speed of light. (a) What is the proton’s wavelength? (b) What is its kinetic energy,
assuming it is nonrelativistic? (c) What was the equivalent voltage through which it
was accelerated?
(a) λ 
h
h
6.63  10 34 J  s


mv m0.250c  1.67  10 27 kg 0.250 3.00  108 m/s




15
 5.29  10 m  5.29 fm
2
1
(b) KE  mv 2  0.51.67  10  27 kg 0.2503.00  10 8 m/s   4.70  10 12 J
2
KE 4.697  10 12 J
(c) KE  qV  V 

 2.94  10 7 V  29.4 MV
19
q
1.60  10 C
60.
The kinetic energy of an electron accelerated in an x-ray tube is 100 keV. Assuming it
OpenStax College Physics
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Chapter 29
is nonrelativistic, what is its wavelength?
Solution



KE  100keV  1.00  10 5 1.60  10 19 J  1.60  10 14 J
KE 
λ
1 2
 2KE 
mv  v  

2
 m 
1
2
h
h
h



p mv 2KEm  12
 mv  2KEm 
1
2
6.63 10 34 J  s
21.60 10
19

C 9.1110
31
kg

1
2
 3.88 10 12 m
61.
Unreasonable Results (a) Assuming it is nonrelativistic, calculate the velocity of an
electron with a 0.100-fm wavelength (small enough to detect details of a nucleus). (b)
What is unreasonable about this result? (c) Which assumptions are unreasonable or
inconsistent?
Solution
(a)  
h
h
h
6.63 10 34 kg.m/s

v

 7.28 1012 m/s
p mv
m 9.1110 31 kg 1.00 10 16 m
(b) It is many times faster than the speed of light (an impossibility).
(c) The non-relativistic assumption is unreasonable at the given wavelength.



29.7 PROBABILITY: THE HEISENBERG UNCERTAINTY PRINCIPLE
62.
(a) If the position of an electron in a membrane is measured to an accuracy of
1.00 μm , what is the electron’s minimum uncertainty in velocity? (b) If the electron
has this velocity, what is its kinetic energy in eV? (c) What are the implications of this
energy, comparing it to typical molecular binding energies?
Solution
h
h
(a) ΔxΔp  4π  Δp  4x  mΔv 
h
6.63  10 34 J  s
Δv 

 57.9 m/s
4x 4π 9.11  10 31 kg 1.00  10 6 m 
1 2
2
31
(b) KE  mv  0.5  9.1110 kg 57.91 m/s 
2
1 eV


9
 1.528 10 27 J  
  9.55 10 eV
19
 1.60 10 J 
(c) From Table 29.1 we see that typical molecular binding energies range from about
1eV to 10 eV; therefore the result in part (b) is approximately 9 orders of
magnitude smaller than typical molecular binding energies.
63.
(a) If the position of a chlorine ion in a membrane is measured to an accuracy of
1.00 μm , what is its minimum uncertainty in velocity, given its mass is 5.86  1026 kg ?
(b) If the ion has this velocity, what is its kinetic energy in eV, and how does this
compare with typical molecular binding energies?
OpenStax College Physics
Instructor Solutions Manual
Chapter 29
Solution (a)
h
h
ΔxΔp 
 Δp 
 m Δv , so
4π
4π x
Δv 
h
6.63  10 34 J  s

 9.00  10  4 m/s
4π mΔ x 4π 5.86  10  26 kg 1.00  10 6 m



2
1 2
 26
4
(b) KE  mv  0.5  5.86 10 kg 9.00 10 m/s 
2
1 eV


 2.375 10 32 J  
 1.48 10 13 eV
19 
 1.60 10 J 
64.
Suppose the velocity of an electron in an atom is known to an accuracy of
2.0  103 m/s (reasonably accurate compared with orbital velocities). What is the
electron’s minimum uncertainty in position, and how does this compare with the
approximate 0.1-nm size of the atom?
Solution
ΔxΔp  Δxm Δv  

h
h
 Δx 
4π
4π m Δv
6.63 10 34 J  s
 2.90  10 8 m  29.0 nm
4π 9.1110 31 kg 2.00  10 3 m/s



Δx 2.90 10 8 m

 290 times larger
D 1.00  10 10 m
65.
The velocity of a proton in an accelerator is known to an accuracy of 0.250% of the
speed of light. (This could be small compared with its velocity.) What is the smallest
possible uncertainty in its position?
Solution
ΔxΔp  Δxm Δv  
Δx 
h

4π
h
6.63  10 34 J  s

 4.21 10 14 m
4π mΔv 4π 1.67  10 27 kg 0.00250  3.00  108 m/s




66.
A relatively long-lived excited state of an atom has a lifetime of 3.00 ms. What is the
minimum uncertainty in its energy?
Solution
h
, so
4
h
6.63 10 34 J  s
1 eV


13
ΔE 

 1.759 10 32 J  
  1.10 10 eV
3
19
4π Δt 4π 3.00 10 s
1.60

10
J.


Et 

67.

20
(a) The lifetime of a highly unstable nucleus is 10 s . What is the smallest
uncertainty in its decay energy? (b) Compare this with the rest energy of an electron.
OpenStax College Physics
Solution
Instructor Solutions Manual
34
(a) Et  h  ΔE  h  6.63 10 J20 s
4
4π Δt 4π 1.00 10 s

Chapter 29

 1 MeV 
2
 5.28 10 15 J  
  3.30 10 MeV
13
1.60

10
J.


34
(b) ΔE  h  6.63 10 J20 s
4π Δt 4π 1.00 10 s


 1 MeV 
 5.28 10 15 J  
 3.30 10 2 MeV
13 
 1.60 10 J 
68.
The decay energy of a short-lived particle has an uncertainty of 1.0 MeV due to its
short lifetime. What is the smallest lifetime it can have?
Solution
 1.60  10 13 J 
  1.60  10 13 J;
ΔE  1.0 MeV 
1
MeV


h
h
6.63  10 34 J  s
ΔEΔt 
 Δt 

 3.30  10 22 s
13
4π
4π ΔE 4π 1.60  10 J


69.
The decay energy of a short-lived nuclear excited state has an uncertainty of 2.0 eV
due to its short lifetime. What is the smallest lifetime it can have?
Solution
 1.60  10 19 J 
  3.2  10 19 J;
ΔE  2.0 eV 
1
eV


h
h
6.63 1034 J  s
ΔEΔt 
 Δt 

 1.6 1016 s
13
4π
4π ΔE 4π3.2 10 J 
70.
What is the approximate uncertainty in the mass of a muon, as determined from its
decay lifetime?
Solution
ΔEΔt  Δmc 2 Δt 


h

4π
h
6.63 10 34 J  s
Δm 

 2.66 10 46 kg
2
2
8

6
4π c Δt 4π 3.00 10 m/s 2.20 10 s

71.


Derive the approximate form of Heisenberg’s uncertainty principle for energy and
time, Et  h , using the following arguments: Since the position of a particle is
uncertain by x   , where  is the wavelength of the photon used to examine it,
there is an uncertainty in the time the photon takes to traverse x . Furthermore, the
photon has an energy related to its wavelength, and it can transfer some or all of this
energy to the object being examined. Thus the uncertainty in the energy of the object
is also related to  . Find t and E ; then multiply them to give the approximate
OpenStax College Physics
Instructor Solutions Manual
Chapter 29
uncertainty principle.
Solution
x 

c
c
hc
 hc   
E 
 Et      h

   c 
x  ct  t 
29.8 THE PARTICLE-WAVE DUALITY REVIEWED
72.
Integrated Concepts The 54.0-eV electron in Example 29.7 has a 0.167-nm
wavelength. If such electrons are passed through a double slit and have their first
maximum at an angle of 25.0 , what is the slit separation d ?
Solution Using the equation d sin   m , m  0,1,2.... , we can solve for the slit separation
10.167 nm  0.395 nm
m
when m  1 for the first order maximum: d 

sin 
sin 25.0
73.
Solution
74.
Solution
Integrated Concepts An electron microscope produces electrons with a 2.00-pm
wavelength. If these are passed through a 1.00-nm single slit, at what angle will the
first diffraction minimum be found?

12
m
 m 
1  1 2.00  10
D sin   m    sin 1 

sin


9
 D 
 1.00  10 m
  0.115


Integrated Concepts A certain heat lamp emits 200 W of mostly IR radiation
averaging 1500 nm in wavelength. (a) What is the average photon energy in joules?
(b) How many of these photons are required to increase the temperature of a person’s
shoulder by 2.0C , assuming the affected mass is 4.0 kg with a specific heat of
0.83 kcal/kg  C . Also assume no other significant heat transfer. (c) How long does
this take?
6.63 10

34

J  s 3.00 108 m/s
 1.3 10 19 J
6

1.5 10 m/s
(b) Q  NE  mcT 
(a) Eγ 
N
hc

mcT 4.00 kg 0.83 kcal/kg  C4186 J/kcal 2.00 C

 2.1  10 23
19
E
1.326  10 J
(c) Q  NE  pt  t 
NE
P

2.096  10 1.326  10
23
200 J/s
19
J
  139 s  1.4  10
2
s
OpenStax College Physics
75.
Instructor Solutions Manual
Chapter 29
Integrated Concepts On its high power setting, a microwave oven produces 900 W of
2560 MHz microwaves. (a) How many photons per second is this? (b) How many
photons are required to increase the temperature of a 0.500-kg mass of pasta by
45.0C , assuming a specific heat of 0.900 kcal/kg  C ? Neglect all other heat
transfer. (c) How long must the microwave operator wait for their pasta to be ready?



Solution (a) E  hf  6.63 10 34 J 2.56 109 s -1  1.70 10 24 J
P
900 J/s

 5.30 10 26 photon/s
 24
E 1.697 10 J/photon
(b) Q  NE  mcT 
mcT 0.500 kg 0.900 kcal/kg. C4186 J/kcal 45.0 C
N

 5.00  10 28
 24
E
1.697  10 J
(c) NE  pt  t 
76.
NE
P

5.00  10 1.697  10
28
900 J/s
24
J
  94.2 s
Integrated Concepts (a) Calculate the amount of microwave energy in joules needed
to raise the temperature of 1.00 kg of soup from 20.0C to 100C . (b) What is the
total momentum of all the microwave photons it takes to do this? (c) Calculate the
velocity of a 1.00-kg mass with the same momentum. (d) What is the kinetic energy of
this mass?
Solution Assume soup has the same specific heat as water.
(a) Q  mcT  1.00 kg 4186 J/kg  C80.0 C  3.35  10 5 J
E 3.349 105 J
(b) E  pc  p  
 1.12 10 3 kg  m/s
8
c 3.00 10 m/s
p
(c) v   1.12  10 3 m/s
m
2
1
(d) KE  mv 2  0.51.00 kg 1.116  10 3 m/s   6.23  10 7 J
2
77.
Integrated Concepts (a) What is  for an electron emerging from the Stanford Linear
Accelerator with a total energy of 50.0 GeV? (b) Find its momentum. (c) What is the
electron’s wavelength?
Solution
(a) E  γmc 2   
E
5.00  10 4 MeV

 9.78  10 4
mc 2
0.511 MeV
9
(b) For such a large  , v  c (within part in 10 )



p  mv  mc  9.78  10 4 9.11  10 31 kg 3.00  10 8 m/s
 2.67  10
17
kg  m/s

OpenStax College Physics
(c)  
78.
Instructor Solutions Manual
Chapter 29
h
6.63  10 34 J  s

 2.48  10 17 m  0.0248 fm
17
p 2.674  10 kg  m/s
Integrated Concepts (a) What is  for a proton having an energy of 1.00 TeV,
produced by the Fermilab accelerator? (b) Find its momentum. (c) What is the
proton’s wavelength?
Solution (a) Using E  mc 2





E
1.00 1012 eV 1.60 10 19 J/eV

mc 2 1.6726 10 27 kg 3.00 108 m/s 2


2
 1.063 103  1.06 103
E (1.00 10 eV)(1.60 10 -19 J/eV)
 5.33 10 16 kg  m/s
(b) p  mc  
8
c
3.00 10 m/s
12
h 6.63 10 34 kg  m/s
(c) p  ,   
 1.24 10 18 m
16

p 5.33 10 kg  m/s
h
79.
Integrated Concepts An electron microscope passes 1.00-pm-wavelength electrons
through a circular aperture 2.00 μm in diameter. What is the angle between two justresolvable point sources for this microscope?
Solution

80.
Integrated Concepts (a) Calculate the velocity of electrons that form the same pattern
as 450-nm light when passed through a double slit. (b) Calculate the kinetic energy of
each and compare them. (c) Would either be easier to generate than the other?
Explain.


1.22 1.22 1.00  10 12 m

 6.10  10 7 rad  3.50  10 5 
6
D
2.00  10 m


h
h
Solution (a)
 

p mv
v
h
6.63  10 -34 kg  m/s

 1.617  10 3 m/s  1.62  10 3 m/s
-31
-7
m
9.109  10 kg 4.50  10 m



hc 6.63 10 J  s 3.00 10
(b) For the photon: E 


4.50 10 m
-34
8
-7
m/s
  4.42 10
-19
J
For the electron:
2
1
1
KE  mv 2    9.109 10 31 kg 1.617 10 3 m/s
2
2
 1.19110 24 J  1.19 10 24 J
The photon energy is 3.71  10 5 times greater.
(c) The light is probably easier to make, because 450 nm light is blue light and



OpenStax College Physics
Instructor Solutions Manual
Chapter 29
therefore easy to make. Creating electrons with 7.43 eV of energy would not
be difficult, but would require a vacuum.
81.
Integrated Concepts (a) What is the separation between double slits that produces a
second-order minimum at 45.0 for 650-nm light? (b) What slit separation is needed
to produce the same pattern for 1.00-keV protons.
Solution
m  1  2.5 6.50  10 7 m
1

2 
 2.30  10 6 m
(a) d sin    m    d 
2
sin

sin
45
.
0







1
1
 2KE  2
(b) KE  mv2  v  
 , so that
2
 m 
h
h


mv 2KEm  12
d
2.5h
2KEm 2 sin 
1

(2)1.60  10


kg 
(2.5) 6.63  10 34 J  s
16

J 1.6726  10 27
1
2
(sin 45.0)
 3.20  10 12 m
82.
Integrated Concepts A laser with a power output of 2.00 mW at a wavelength of 400
nm is projected onto calcium metal. (a) How many electrons per second are ejected?
(b) What power is carried away by the electrons, given that the binding energy is 2.31
eV? (c) Calculate the current of ejected electrons. (d) If the photoelectric material is
electrically insulated and acts like a 2.00-pF capacitor, how long will current flow
before the capacitor voltage stops it?
Solution
(a) E γ  hf 



hc 6.63 10 34 J  s 3.00 108 m/s

 4.9725  10 19 J
λ
4.00 10 7 m


2.00  10 3 J/s
 4.022 1015 /s  4.02 1015 /s
19
4.9725  10 J
(b) KE  hf  BE  Eγ  BE
N
 1.60 10 19 J 
  1.324 10 19 J
 4.972 10 J  2.28 eV 
1 eV


15
19
P  NKE  4.022 10 /s 1.324 10 J  5.33 10 4 W  0.533 mW
19



(c) I  Ne  4.022  10 /s 1.60  10 C  6.44  10-4 A
It
(d) Q  CV  It  V  is the voltage across the capacitor. This V will stop the
C
flow when the energy required to cross the capacitor, eV , equals or exceeds the
kinetic energy of the electrons.
15
-19
OpenStax College Physics
eV 
83.
Instructor Solutions Manual




Chapter 29


eIt
CKE 2.00 10 12 F 1.324 10 19 J
 KE  t 

 2.57 10 9 s
C
eI
1.60 10 19 C 6.435 10 4 A
Integrated Concepts One problem with x rays is that they are not sensed. Calculate
the temperature increase of a researcher exposed in a few seconds to a nearly fatal
accidental dose of x rays under the following conditions. The energy of the x-ray
photons is 200 keV, and 4.00  1013 of them are absorbed per kilogram of tissue, the
specific heat of which is 0.830 kcal/kg  C . (Note that medical diagnostic x-ray
machines cannot produce an intensity this great.)
Solution First, we know the amount of heat absorbed by 1.00 kg of tissue is equal to the
number of photons times the energy each one carry, so:
 1.60  10 19 J 
  1.28 J
Q  NE  4.00  1013 2.00  105 eV 
1 eV


Next, using the equation Q  mct , we can determine how much 1.00 kg tissue is
Q
1.282 J
heated: t 

 3.69  10 4 C
mc 1.00 kg 0.830 kcal/kg  C4186 J/kcal 



84.
Integrated Concepts A 1.00-fm photon has a wavelength short enough to detect some
information about nuclei. (a) What is the photon momentum? (b) What is its energy in
joules and MeV? (c) What is the (relativistic) velocity of an electron with the same
momentum? (d) Calculate the electron’s kinetic energy.
Solution
(a)  
h
h 6.63  10 34 kg  m/s
 p 
 6.63  10 19 kg  m/s
p

1.00  10 15 m



19
8
10
(b) E  pc  6.63  10 kg  m/s 3.00  10 m/s  1.99  10 J
 1 MeV 
 1.99  10 10 J  
 1.24  10 3 MeV
13 
 1.60  10 J 
mu
(c) p  mu 
, so that
1
2
2
1 u 2
c
2
c
 2 2 
p 2 1  u 2  m 2u 2  p 2   p u 2   m 2u 2  u 
1/ 2
c
c 

  mc  2 
 
1  
  p  
mc
Since
is very small, use the binomial expansion:
p
  
  
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c
  mc  2 
 
1  
  p  

1/ 2
 1  mc  2 
 
 c1  
 2  p  



2

9.11 10 31 kg 3.00  108 m/s

 c 1
2

2 6.63  10 19 kg  m/s


  c   8.50 10  
2
(d)   1  u
1
1
8 2 2
2
2
Relativistic kinetic energy




2

  c 1  8.50  10 8




 1.18  107

E   1mc2  1.18 107  1 9.111031 kg 3.00 108 m/s
85.
Chapter 29

2
 9.67 107 J
Integrated Concepts The momentum of light is exactly reversed when reflected
straight back from a mirror, assuming negligible recoil of the mirror. Thus the change
in momentum is twice the photon momentum. Suppose light of intensity 1.00 kW/m 2
reflects from a mirror of area 2.00 m 2 . (a) Calculate the energy reflected in 1.00 s. (b)
What is the momentum imparted to the mirror? (c) Using the most general form of
Newton’s second law, what is the force on the mirror? (d) Does the assumption of no
mirror recoil seem reasonable?
Solution (a) E  IA  1.00  10 3 J/m 2 2.00 m 2   2.00  10 3 J
2 E 22.00  10 3 J 
(b) p 

 1.33  10 -5 kg  m/s
c
3.00  10 8 m/s
p 1.33  10 -5 kg  m/s
(c) F 

 1.33  10 -5 N
t
1.00 s
(d) Yes, with such a small force on the mirror, it will move very little during the time it
takes for the photons to bounce off the mirror.
86.
Integrated Concepts Sunlight above the Earth’s atmosphere has an intensity of
1.30 kW/m 2 . If this is reflected straight back from a mirror that has only a small recoil,
the light’s momentum is exactly reversed, giving the mirror twice the incident
momentum. (a) Calculate the force per square meter of mirror. (b) Very low mass
mirrors can be constructed in the near weightlessness of space, and attached to a
spaceship to sail it. Once done, the average mass per square meter of the spaceship is
0.100 kg. Find the acceleration of the spaceship if all other forces are balanced. (c)
How fast is it moving 24 hours later?
Solution
In one second the incident energy per square meter will be 1.30 kJ.
OpenStax College Physics
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3
(a) p  2 E  21.30  10 J m/s   8.67  10 6 kg  m/s per m 2
c
3.00  108 m/s
p 8.67  10 6 kg  m/s 2 per m 2

 8.67  10 6 N per m 2
t
1.00 s
8.67 10 6 N/m 2
(b) F  ma  a 
 8.67 10 5 m/s 2
2
0.100 kg/m
 3600 s 
(c) v  at  8.67 10 5 m/s 2 24 h 
  7.49 m/s
 1h 
F
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Chapter 29