Física para Ciencias e Ingeniería. Volumen II - Douglas C. Giancoli - 4ta edición [Solucionario]

Chapter 36 The Special Theory of Relativity

92. We consider the motion from the reference frame of the spaceship. The passengers will see the trip
distance contracted, as given by Eq. 36-3a. They will measure their speed to be that contracted
distance divided by the year of travel time (as measured on the ship). Use that speed to find the work
done (the kinetic energy of the ship).

v  l  l0 1 v2 c2  v  1 1  0.9887 c
t0 t0 c
  ct0 2    1.0 ly 2 
  l0    6.6 ly 
 1      1    
    
  

 1 c2 
1  v2 1 mc2
W  K   1mc2  

   
 1
 
1 3.2 104 kg 3.00 108 m s 2  1.6 1022 J
1  0.98872

93. The kinetic energy is given by Eq. 36-10.

 
1 mc2  
1 14,500 kg 3.00 108 m s 2
 1 1

 K   1mc2  
1  v2 c2   1  0.982 
 

 5.31021J

We compare this with annual U.S. energy consumption: 5.3  1021 J  53.
1020 J

The spaceship’s kinetic energy is over 50 times as great.

94. The pi meson decays at rest, and so the momentum of the muon and the neutrino must each have the
same magnitude (and opposite directions). The neutrino has no rest mass, and the total energy must
be conserved. We combine these relationships using Eq. 36-13.

 Ev  pv2c2  mv2c4 1/ 2  pvc ; p  pv  p

p2c2  m2c4 1/ 2  pvc  1/ 2
   E  E  Ev
 m c2  p2c2  m2c4  pc 

     m c2  pc  p2c2  m2c4 1/ 2  m c2  pc 2  p2c2  m2c4

Solve for the momentum.

m 2c4  2m c2 pc  p2c2  p2c2  m2c4  pc  m 2c2  m 2c2
2m

Write the kinetic energy of the muon using Eqs. 36-11 and 36-13.

K  E  mc2 ; E  E  Ev  m c2  pc 

   K 
m c2  pc  mc2  m c2  mc2  m 2c2  m2c2
2m

    2m
m c2  mc2  m 2c2  m 2c2
2m 2m

m  m 2 c2
2m
      c2
2m 2  2mm  m 2  m2 c2 m 2  2mm  m2 
2m  2m

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451

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

95. (a) The relative speed can be calculated in either frame, and will be the same value in both frames.

The time as measured on the Earth will be longer than the time measured on the spaceship, as

given by Eq. 36-1a.

v  xEarth ; tEarth  tspaceship  tspaceship 
tEarth 1 v2 c2
1  xEarth 2
 ctEarth
 



2  xEarth 2 2 2  xEarth 2 2
 c  c
       tEarth  
   tspaceship  tEarth    tspaceship 

 xEarth 2 2 6.0 y 2  2.50 y 2  6.5y
 c
     tEarth  
  tspaceship 

(b) The distance as measured by the spaceship will be contracted.

xEarth xspaceship tspaceship 2.50 y
tEarth tspaceship tEarth 6.5 y
v  
 xspaceship  xEarth  6.0 ly  2.3ly

This is the same distance as found using the length contraction relationship.

96. (a) To observers on the ship, the period is non-relativistic. Use Eq. 14-7b.

T  2 m  2 1.88 kg   0.939 s
k 84.2 N m

(b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run

slow.

TEarth  T 0.939s  2.15s
 1  0.9002

1  v2 c2

97. We use the Lorentz transformations to derive the result.

x    x  vt  x   x  vt ; t    t  vx   t    t  vx 
 c2   c2 

ct 2  x2  c  t  vx 2   x  vt2   2  ct  vx 2  x  vt 
 c2   c  


  2 c2  t2  2ct vx   vx 2   x2  2xvt  vt2 
 c  c 
 


1  t2  v 2 1 x2 
1 v2 c2  c  
    c2  v2      



      1
1  v2 c2 c2 1 v2 c2 t2  1  v2 c2  x2

  1 v2 c2

  1 v2 c2
ct2  x2  ct2  x2

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452

Chapter 36 The Special Theory of Relativity

98. We assume that the left edge of the glass is even with point A when the flash of light is emitted.
There is no loss of generality with that assumption. We do the calculations in the frame of reference
in which points A and B are at rest, and the glass is then moving to the right with speed v.

If the glass is not moving, we would have this “no motion” result.

tv0  tglass  tvacuum  distance in glass  distance in vacuum  d  l d
speed in glass speed in vacuum vglass c

 d  ld  nd  ld  nd ld  l  n 1d
cn c c c c
c

If the index of refraction is n  1, then the glass will have no effect on the light, and the time would

simply be the distance divided by the speed of light.

tn1  tglass  tvacuum  distance in glass  distance in vacuum  d  ld  d ld  l
speed in glass speed in vacuum c c c c

Now, let us consider the problem from a relativistic point of view. The speed of light in the glass

will be the relativistic sum of the speed of light in stationary glass, c n , and the speed of the glass, v,

by Eq. 36-7a. We define  to simplify further expressions.

c v c  v 1  vn   1  vn 
n n 1  c   1  c  
vlight  cv  v  c v   c   v 
in glass 1 nc2 nc n nc  n  nc 
1   


The contracted width of the glass, from the Earth frame of reference, is given by Eq. 36-3a.

dmoving  d 1 v2 c2  d
glass 

We assume the light enters the block when the left edge of the block is at point A, and write simple

equations for the displacement of the leading edge of the light, and the leading edge of the block. Set

them equal and solve for the time when the light exits the right edge of the block.

xlight  vlight t   c t ; xright  d  vt ;
in glass n edge 

c d tglass d n
n    c  nv
xlight  xright

 edge
  tglass   vtglass  

Where is the front edge of the block when the light emerges? Use tglass d n with either

 c  nv

expression – for the leading edge of the light, or the leading edge of the block.

v tlight glass c d n  cd
in glass n   c  nv  c  nv
   xlight   


xright  d  vtglass  d vd n  d  c  nv  vdn   cd
edge      c  nv
 c  nv   c  nv

The part of the path that is left, l     cd nv  , will be traveled at speed c by the light. We express
c

that time, and then find the total time.

l    cd

tvacuum   c  nv

c

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453

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

d n l    cd l d n
 c 
ttotal  tglass  tvacuum  tglass   c  nv   c  nv    c  nv

c

 l  n  1 d cv
c cv
c

We check this for the appropriate limiting cases.

Case 1:  ttotall  n 1 d c  v  l  n  1 d c  c  l
c c c  v c c  c c
vc c

This result was expected, because the speed of the light would always be c.

Case 2:  ttotall  n 1 d c  v  l  n  1 d 1  l n 1d
c c c  v c
v0 c c

This result was obtained earlier in the solution.

Case 3:  ttotall  n 1 d c  v  l
c c c  v c
n 1

This result was expected, because then there is no speed change in the glass.

99. The spreadsheet used for this K (1017 J) 1.2 Classical
problem can be found on the 1.0 Relativistic
Media Manager, with filename 0.8
“PSE4_ISM_CH36.XLS,” on 0.6
tab “Problem 36.99.” 0.4
0.2
0.0 0.2 0.4v /c 0.6 0.8

0

100. (a) We use Eq. 36-98. Since there is motion in two dimensions, we have   1 .
vx2 v 2
1 c2  y

dp c2
dt
   Fˆj  ; dpx 0  px   mvx  p0 ; dp y  F  py  Ft   mvy
F dt dt

Use the component equations to obtain expressions for vx2 and v 2 .
y

    mvx  p0  vx  p0  v 2  p02  p02 1  vx2  v 2   v 2  p02 c2  v 2
m x  2m2 m2 c2 y  x y

c2 m2c2  p02

 mvy  Ft  vy  Ft  v 2  F 2t2  F 2t2 1  v 2  v 2  
m y  2m2 m2 x y 

c2 c2

   v
2  F 2t2 c2  vx2
y m2c2  F 2t2

Substitute the expression for v 2y into the expression for v 2 .
x

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454

Chapter 36 The Special Theory of Relativity

              vx2  p02   F 2t2 c2  v 2 
 c2 x   p02

c2  v 2  p02 m2c2  F 2t2 m2c4  F 2t2vx2 
y m2c2  p02 m2c2  F 2t2

m2c2  p02 m2c2  p02

    vx2 m2c2  p02 m2c2  F 2t2  p02 m2c4  F 2t2vx2 

vx2m4c4  vx2m2c2 p02  vx2F 2t2m2c2  v 2 F 2t 2 p02  p02m2c4  p02 F 2t 2v 2 
x x

p0c
m2c2  p02  F 2t2 1/ 2
 vx2m2c2  vx2 p02  vx2F 2t2  p02c2  vx 

Use the expression for vx to solve for vy.

 p02c2 
 F 2t2  c2  m2c2  p02  F 2t2 

m2c2  F 2t2
      v2F 2t2c2 v 2
y x

m2c2  F 2t2

            F2t2
c2 m2c2  p02  F 2t2  p02c2  F 2t2 c2 m2c2  F 2t2 
m2c2  F 2t2 m2c2  p02  F 2t2 m2c2  F 2t2 m2c2  p02  F 2t2

Ftc
m2c2  p02  F 2t 2 1/ 2
 vy 

The negative sign comes from taking the negative square root of the previous equation. We
know that the particle is moving down.

(b) See the graph. We are v /c 1.0 vx
0.8 (- vy)
plotting vx c and vy c. 0.6
0.4
The spreadsheet used for this 0.2
problem can be found on the
Media Manager, with filename
“PSE4_ISM_CH36.XLS,” on
tab “Problem 36.100.”

0.0

0 1 2 t ( s) 3 4 5

(c) The path is not parabolic, because the vx is not constant. Even though there is no force in the x-
direction, as the net speed of the particle increases,  increases. Thus vx must decrease as time
elapses in order for px to stay constant.

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455

CHAPTER 37: Early Quantum Theory and Models of the Atom

Responses to Questions

1. A reddish star is the coolest, followed by a whitish-yellow star. Bluish stars have the highest
temperatures. The temperature of the star is related to the frequency of the emitted light. Since red
light has a lower frequency than blue light, red stars have a lower temperature than blue stars.

2. The energy radiated by an object may not be in the visible part of the electromagnetic spectrum. The
spectrum of a blackbody with a temperature of 1000 K peaks in the IR and the object appears red,
since it includes some radiation at the red end of the visible spectrum. Cooler objects will radiate
less overall energy and peak at even longer wavelengths. Objects that are cool enough will not
radiate any energy at visible wavelengths.

3. The lightbulb will not produce light as white as the Sun, since the peak of its emitted light is in the
infrared. The lightbulb will appear more yellowish than the Sun, which has a spectrum that peaks in
the visible range.

4. A bulb which appears red would emit very little radiant energy at higher visible frequencies and
therefore would not expose black and white photographic paper. This strategy would not work in a
darkroom for developing color photographs since the photographic paper would be sensitive to light
at all visible frequencies, including red.

5. If the threshold wavelength increases for the second metal, then it has a smaller work function than
the first metal. Longer wavelength corresponds to lower energy. It will take less energy for the
electron to escape the surface of the second metal.

6. According to the wave theory, light of any frequency can cause electrons to be ejected as long as the
light is intense enough. A higher intensity corresponds to a greater electric field magnitude and more
energy. Therefore, there should be no frequency below which the photoelectric effect does not
occur. According to the particle theory, however, each photon carries an amount of energy which
depends upon its frequency. Increasing the intensity of the light increases the number of photons but
does not increase the energy of the individual photons. The cutoff frequency is that frequency at
which the energy of the photon equals the work function. If the frequency of the incoming light is
below the cutoff, the electrons will not be ejected because no individual photon has enough energy
to impart to an electron.

7. Individual photons of ultraviolet light are more energetic than photons of visible light and will
deliver more energy to the skin, causing burns. UV photons also can penetrate farther into the skin,
and, once at the deeper level, can deposit a large amount of energy that can cause damage to cells.

8. Cesium will give a higher maximum kinetic energy for the electrons. Cesium has a lower work
function, so more energy is available for the kinetic energy of the electrons.

9. (a) No. The energy of a beam of photons depends not only on the energy of each individual photon
but also on the total number of photons. If there are enough infrared photons, the infrared beam
may have more energy than the ultraviolet beam.

(b) Yes. The energy of a single photon depends on its frequency: E = hf. Since infrared light has a
lower frequency than ultraviolet light, a single IR photon will always have less energy than a
single UV photon.

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456

Chapter 37 Early Quantum Theory and Models of the Atom

10. Fewer electrons are emitted from the surface struck by the 400 nm photons. Each 400 nm photon has
a higher energy than each 450 nm photon, so it will take fewer 400 nm photons to produce the same
intensity (energy per unit area per unit time) as the 450 nm photon beam. The maximum kinetic
energy of the electrons emitted from the surface struck by the 400 nm photons will be greater than
the maximum kinetic energy of the electrons emitted from the surface struck by the 450 nm photons,
again because each 400 nm photon has a higher energy.

11. (a) In a burglar alarm, when the light beam is interrupted (by an intruder, or a door or window
opening), the current stops flowing in the circuit. An alarm could be set to go off when the
current stops.

(b) In a smoke detector, when the light beam is obscured by smoke, the current in the circuit would
decrease or stop. An alarm could be set to go off when the current decreased below a certain
level.

(c) The amount of current in the circuit depends on the intensity of the light, as long as the
frequency of the light is above the threshold frequency. The ammeter in the circuit could be
calibrated to reflect the light intensity.

12. Yes, the wavelength increases. In the scattering process, some of the energy of the incident photon is
transferred to the electron, so the scattered photon has less energy, and therefore a lower frequency
and longer wavelength, than the incident photon. (E = hf = hc/λ.)

13. In the photoelectric effect the photon energy is completely absorbed by the electron. In the Compton
effect, the photon is scattered from the electron and travels off at a lower energy.

14. According to both the wave theory and the particle theory the intensity of a point source of light
decreases as the inverse square of the distance from the source. In the wave theory, the intensity of
the waves obeys the inverse square law. In the particle theory, the surface area of a sphere increases
with the square of the radius, and therefore the density of particles decreases with distance, obeying
the inverse square law. The variation of intensity with distance cannot be used to help distinguish
between the two theories.

15. The proton will have the shorter wavelength, since it has a larger mass than the electron and

therefore a larger momentum   h p.

16. Light demonstrates characteristics of both waves and particles. Diffraction and interference are wave
characteristics, and are demonstrated, for example, in Young’s double-slit experiment. The
photoelectric effect and Compton scattering are examples of experiments in which light
demonstrates particle characteristics. We can’t say that light IS a wave or a particle, but it has
properties of each.

17. Electrons demonstrate characteristics of both waves and particles. Electrons act like waves in
electron diffraction and like particles in the Compton effect and other collisions.

18. Both a photon and an electron have properties of waves and properties of particles. They can both be
associated with a wavelength and they can both undergo scattering. An electron has a negative
charge and a rest mass, obeys the Pauli exclusion principle, and travels at less than the speed of
light. A photon is not charged, has no rest mass, does not obey the Pauli exclusion principle, and
travels at the speed of light.

19. Opposite charges attract, so the attractive Coulomb force between the positive nucleus and the
negative electrons keeps the electrons from flying off into space.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

20. Look at a solar absorption spectrum, measured above the Earth’s atmosphere. If there are dark
(absorption) lines at the wavelengths corresponding to oxygen transitions, then there is oxygen near
the surface of the Sun.

21. At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light
passes through the gas, photons are absorbed, causing electrons to make transitions to higher states
and creating absorption lines. These lines correspond to the Lyman series since that is the series of
transitions involving the ground state or n = 1 level. Since there are virtually no atoms in higher
energy states, photons corresponding to transitions from n > 2 to higher states will not be absorbed.

22. The closeness of the spacing between energy levels near the top of Figure 37-26 indicates that the
energy differences between these levels are small. Small energy differences correspond to small
wavelength differences, leading to the closely spaced spectral lines in Figure 37-21.

23. There is no direct connection between the size of a particle and its de Broglie wavelength. It is
possible for the wavelength to be smaller or larger than the particle.

24. On average the electrons of helium are closer to the nucleus than the electrons of hydrogen. The
nucleus of helium contains two protons (positive charges), and so attracts each electron more
strongly than the single proton in the nucleus of hydrogen. (There is some shielding of the nuclear
charge by the second electron, but each electron still feels the attractive force of more than one
proton’s worth of charge.)

25. The lines in the spectrum of hydrogen correspond to all the possible transitions that the electron can
make. The Balmer lines, for example, correspond to an electron moving from all higher energy
levels to the n = 2 level. Although an individual hydrogen atom only contains one electron, a sample
of hydrogen gas contains many atoms and all the different atoms will be undergoing different
transitions.

26. The Balmer series spectral lines are in the visible light range and could be seen by early
experimenters without special detection equipment.

27. The photon carries momentum, so according to conservation of momentum, the hydrogen atom will
recoil as the photon is ejected. Some of the energy emitted in the transition of the atom to a lower
energy state will be the kinetic energy of the recoiling atom, so the photon will have slightly less
energy than predicted by the simple difference in energy levels.

28. No. At room temperature, virtually all the atoms in a sample of hydrogen gas will be in the ground
state. Thus, the absorption spectrum will contain primarily just the Lyman lines, as photons
corresponding to transitions from the n = 1 level to higher levels are absorbed. Hydrogen at very
high temperatures will have atoms in excited states. The electrons in the higher energy levels will
fall to all lower energy levels, not just the n = 1 level. Therefore, emission lines corresponding to
transitions to levels higher than n = 1 will be present as well as the Lyman lines. In general, you
would expect to see only Lyman lines in the absorption spectrum of room temperature hydrogen, but
you would find Lyman, Balmer, Paschen, and other lines in the emission spectrum of high-
temperature hydrogen.

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458

Chapter 37 Early Quantum Theory and Models of the Atom

Solutions to Problems

In several problems, the value of hc is needed. We often use the result of Problem 96, hc  1240eVnm.

1. We use Wien’s law, Eq. 37-1.

   (a)
2.90 103 mK  2.90 103 mK  1.06 105 m  10.6 m
P  T
 273 K 

This wavelength is in the far infrared.

   (b)
P  2.90 103 mK  2.90 103 mK  8.29 107 m  829 nm
T
3500 K

This wavelength is in the infrared.

   (c)
2.90 103 mK 2.90 103 mK  6.90 104 m  0.69 mm
P  T
 4.2 K

This wavelength is in the microwave region.

   (d)
P  2.90 103 mK  2.90 103 mK  1.06 103 m  1.06 mm
T
 2.725 K 

This wavelength is in the microwave region.

2. We use Wien’s law to find the temperature for a peak wavelength of 460 nm.

   2.90 103 mK 2.90 103 mK
 T  P  460 109 m  6300 K

3. Because the energy is quantized according to Eq. 37-2, the difference in energy between adjacent
levels is simply E = nhf.

  E  hf  6.631034 Js 8.11013Hz  5.4 1020 J  0.34eV

4. We use Eq. 37-1 with a temperature of 98 F  37C  310 K.

   P 
2.90 103 mK  2.90 103 mK  9.4 106 m  9.4 m
T
310 K

5. (a) Wien’s displacement law says that PT  constant. We must find the wavelength at which

I ,T  is a maximum for a given temperature. This can be found by setting I   0.

I    2 hc2 5   2 hc2    5 
      ehc / kT 
 ehc / kT  1    1 

  hc 
 kT  2
  
ehc / kT  1 5 6   e5 hc / kT   
ehc / kT 1 2 
 2 hc2 
  

 

2 hc2 5  ehc / kT  hc  5  0  5 ehc / kT  5  hc  
ehc / kT  1  kT   kT  
  2 

6

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459

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

ex 5  x  5 ; x  hc
PkT

This transcendental equation will have some solution x = constant, and so hc  constant, and
PkT

so PT  constant . The constant could be evaluated from solving the transcendental equation,

(b) To find the value of the constant, we solve ex 5  x  5, or 5  x  5ex. This can be done

graphically, by graphing both y  5  x and y  5ex on the same set of axes and finding the

intersection point. Or, the quantity 5  x  5ex could be calculated, and find for what value of x

that expression is 0. The answer is x = 4.966. We use this value to solve for h. The spreadsheet

used for this problem can be found on the Media Manager, with filename

“PSE4_ISM_CH37.XLS,” on tab “Problem 37.5.”

hc  4.966 
PkT

  h
 4.966 PTk  4.966 2.90 103 mK 1.38 1023 J K  6.62 1034 Js
c 3.00 108 m s

(c) We integrate Planck’s radiation formula over all wavelengths.

I ,T  d    2 hc2 5  d ; let hc x ;   hc ; d   hc dx
0  ehc / kT  1  kT xkT x2kT
0  

 2 hc2  hc 5 
0  xkT  
 2 hc2 5  2 k 4T 4 x3
 ehc / kT  1  h3c2 x
I ,T d  0  d      hc dx      dx
  x2kT  0  
   0  
   ex 1  e 1



2 k 4   x3  dx  T 4 T 4
h3c2   x  
 0  e 1   

Thus the total radiated power per unit area is proportional to T 4. Everything else in the

expression is constant with respect to temperature.

6. We use Eq. 37-3.

  E  hf  6.626 1034 Js 104.1106 Hz  6.898 1026 J

7. We use Eq. 37-3 along with the fact that f  c  for light. The longest wavelength will have the

lowest energy.

    E1
 hf1  hc  6.631034 Js 3.00 108 m / s  4.85  1019 J  1eV J   3.03 eV
1 410 109 m  1.60 1019 

    E2
 hf2  hc  6.63 1034 Js 3.00 108 m / s  2.65  1019 J  1eV J   1.66 eV
2 750 109 m  1.60 1019 

Thus the range of energies is 2.71019 J  E  4.91019 J or 1.7eV  E  3.0eV .

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460

Chapter 37 Early Quantum Theory and Models of the Atom

8. We use Eq. 37-3 with the fact that f  c  for light.

  
 c  hc  6.631034 Js 3.00 108 m/s  3.27 1012 m  3.3103 nm
f E 1.60 1019 J/eV 380 103eV

Significant diffraction occurs when the opening is on the order of the wavelength. Thus there would
be insignificant diffraction through the doorway.

9. We use Eq. 37-3 with the fact that f  c  for light.

   Emin  hfmin
 f min  Emin  0.1eV 1.60 1019 J eV  2.411013 Hz  2 1013 Hz
h
6.63 1034 Js

 max
 c  3.00 108 m s  1.24 105 m  1105 m
f min 2.411013 Hz

10. We use Eq. 37-5.

 p  h 
 
6.63 1034 Js  1.07 1027 kgm s
6.20 107 m

11. At the minimum frequency, the kinetic energy of the ejected electrons is 0. Use Eq. 37-4a.

K  hfmin  W0  0  f min  W0  4.8 1019 J  7.2 1014 Hz
h 6.631034 Js

12. The longest wavelength corresponds to the minimum frequency. That occurs when the kinetic

energy of the ejected electrons is 0. Use Eq. 37-4a.

K  hfmin  W0  0  f min  c  W0 
max h

    max
ch  3.00 108 m s 6.63 1034 Js  3.36 107 m  336 nm
 W0
3.70eV 1.60 1019 J eV

13. The energy of the photon will equal the kinetic energy of the baseball. We use Eq. 37-3.

  K  hf
 1 mv2  h c    2hc  2 6.631034 Js 3.00 108 m s  3.05 1027 m
2  mv2
0.145kg30.0 m s2

14. We divide the minimum energy by the photon energy at 550 nm to find the number of photons.

     E  nhf  Emin
 n Emin  Emin  1018 J 550  109 m  2.77  3 photons
hf hc 6.63  1034 Js 3.00  108 m s

15. The photon of visible light with the maximum energy has the least wavelength. We use 410 nm as
the lowest wavelength of visible light.

  hfmax
 hc  6.631034 Js 3.00 108 m/s  3.03eV
min 1.60 1019 J/eV 410 109 m

Electrons will not be emitted if this energy is less than the work function.

The metals with work functions greater than 3.03 eV are copper and iron.

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461

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

16. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so the work

function is equal to the energy of the photon.

W0  hf  Kmax  hf  hc  1240 eVnm  2.4 eV
 520 nm

(b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum

kinetic energy. We use Eq. 37-4b to calculate the maximum kinetic energy.

K max  hf  W0  hc  W0  1240 eV nm  2.38eV  0.25eV
 470 nm

V0  K max  0.25eV  0.25 V
e e

17. The photon of visible light with the maximum energy has the minimum wavelength. We use Eq. 37-

4b to calculate the maximum kinetic energy.

K max  hf  W0  hc  W0  1240 eVnm  2.48 eV  0.54 eV
 410 nm

18. We use Eq. 37-4b to calculate the maximum kinetic energy. Since the kinetic energy is much less

than the rest energy, we use the classical definition of kinetic energy to calculate the speed.

K max  hf  W0  hc  W0  1240 eVnm  2.48 eV  0.92 eV
 365 nm

 Kmax
 1 mv 2  v 2 K max  20.92eV 1.60 1019 J eV  5.7 105 m/s
2 m
9.11  1031 kg

19. We use Eq. 37-4b to calculate the work function.

W0  hf  Kmax  hc  Kmax  1240 eVnm  1.70 eV  2.65 eV
 285 nm

20. Electrons emitted from photons at the threshold wavelength have no kinetic energy. We use Eq. 37-

4b with the threshold wavelength to determine the work function.

W0  hc  Kmax  hc  1240 eVnm  3.88 eV.
 max 320 nm

(a) We now use Eq. 36-4b with the work function determined above to calculate the kinetic energy

of the photoelectrons emitted by 280 nm light.

K max  hc  W0  1240 eVnm  3.88 eV  0.55 eV
 280 nm

(b) Because the wavelength is greater than the threshold wavelength, the photon energy is less than

the work function, so there will be no ejected electrons.

21. The stopping voltage is the voltage that gives a potential energy change equal to the maximum

kinetic energy of the photoelectrons. We use Eq. 37-4b to calculate the work function where the

maximum kinetic energy is the product of the stopping voltage and electron charge.

W0  hc  K max  hc  eV0  1240 eVnm  1.84 Ve  3.55 eV
  230 nm

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462

Chapter 37 Early Quantum Theory and Models of the Atom

22. The energy required for the chemical reaction is provided by the photon. We use Eq. 37-3 for the
energy of the photon, where f  c / .

E  hf  hc  1240 eVnm  2.0 eV
 630 nm

Each reaction takes place in a molecule, so we use the appropriate conversions to convert

eV/molecule to kcal/mol.

E   2.0 eV   1.60 1019 J   6.02  1023 molecules   kcal   45 kcal/mole
 molecule   eV   mol   4186 
    J

23. (a) Since f  c  , the photon energy given by Eq. 37-3 can be written in terms of the wavelength

as E  hc  . This shows that the photon with the largest wavelength has the smallest energy.

The 750-nm photon then delivers the minimum energy that will excite the retina.

  E  hc 
 
6.6310–34 Js 3.00 108 m s  1 eV   1.66 eV
750 10–9 m  1.60 10–19J 

(b) The eye cannot see light with wavelengths less than 410 nm. Obviously, these wavelength

photons have more energy than the minimum required to initiate vision, so they must not arrive

at the retina. That is, wavelength less than 410 nm are absorbed near the front portion of the

eye. The threshold photon energy is that of a 410-nm photon.

  E  hc 
 
6.6310–34 Js 3.00 108 m s  1 eV J   3.03eV
410 10–9 m  1.60 10–19 

24. We plot the maximum (kinetic) energy 3.0
of the emitted electrons vs. the
2.5 E = 0.4157 f - 2.3042
frequency of the incident radiation.
R2 = 0.9999
Eq. 37-4b says Kmax  hf  W0. The 2.0

best-fit straight line is determined by Energy (eV) 1.5

linear regression in Excel. The slope 1.0
of the best-fit straight line to the data 0.5
should give Planck’s constant, the x-
0.0
intercept is the cutoff frequency, and
6.0 7.0 8.0 9.0 10.0 11.0 12.0
the y-intercept is the opposite of the
work function. The spreadsheet used Frequency (1014 Hz)

for this problem can be found on the

Media Manager, with filename “PSE4_ISM_CH37.XLS,” on tab “Problem 37.24.”

  (a) h  0.4157 eV 1014Hz 1.60 1019 J eV  6.7 1034 Js

W0 2.3042 eV  5.5 1014 Hz
h 0.4157eV 1014Hz
 (b)
hfcutoff  W0  fcutoff  

(c) W0  2.3eV

25. (a) Since f  c  , the photon energy is E  hc  and the largest wavelength has the smallest

energy. In order to eject electrons for all possible incident visible light, the metal’s work
function must be less than or equal to the energy of a 750-nm photon. Thus the maximum value
for the metal’s work function Wo is found by setting the work function equal to the energy of
the 750-nm photon.

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463

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

    Wo
 hc  6.6310–34 Js 3.00 108 m s  1 eV   1.66 eV
 750 10–9 m  1.60 10–19J 

(b) If the photomultiplier is to function only for incident wavelengths less than 410-nm, then we set

the work function equal to the energy of the 410-nm photon.

    Wo
 hc  6.6310–34 Js 3.00 108 m s  1 eV J   3.03 eV
 410 10–9 m  1.60 10–19 

26. Since f  c  , the energy of each emitted photon is E  hc  . We multiply the energy of each

photon by 1.0 106 s to determine the average power output of each atom. At distance of

r  25 cm , the light sensor measures an intensity of I  1.6 nW 1.0 cm2 . Since light energy emitted

from atoms radiates equally in all directions, the intensity varies with distance as a spherical wave.

Thus, from Section 15–3 in the text, the average power emitted is P  4 r2I . Dividing the total
average power by the power from each atom gives the number of trapped atoms.

4 25cm2 1.6 109 W/cm2

6.631034 Js 3.00 108 m/s / 780 109 m
    NP4 r2I 
Patom nhc  1.0 106 /s

 4.9 107 atoms

27. We set the kinetic energy in Eq. 37-4b equal to the stopping voltage, eV0 , and write the frequency
of the incident light in terms of the wavelength, f  c  . We differentiate the resulting equation

and solve for the fractional change in wavelength, and we take the absolute value of the final
expression.

eV0  hc  W0  e dV0   hc d  d   e dV0    e V0
 2  hc  hc

   
  
1.60 1019 C 550 109 m 0.01V  0.004
6.63 1034 Js 3.00 108 m s

28. We use Eq. 37-6b. Note that the answer is correct to two significant figures.

  h 1  cos  
mec

   
 cos1 1  mec   cos1   9.11  1031kg 3.00  108 m s 1.5 1013 m 
h  1 6.63  1034 Js   20

29. The Compton wavelength for a particle of mass m is h mc.

    (a)
h 6.63 1034 Js  2.43 1012 m
mec  9.111031kg 3.00 108 m s

    (b)
h  6.63 1034 Js  1.32 1015 m
mpc 1.67 1027 kg 3.00 108 m s

(c) The energy of the photon is given by Eq. 37-3.

Ephoton  hf  hc  hc  mc2  rest energy

h mc

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464

Chapter 37 Early Quantum Theory and Models of the Atom

30. We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 37-6b. The
Compton wavelength of a free electron is given in the text right after Eq. 37-6b.

   h 
    mec  1  cos   C 1  cos   2.43103 nm 1  cos 



 (a) a    2.43103 nm 1  cos60  1.22 103 nm

 (b) b    2.43103 nm 1  cos90  2.43103 nm

 (c) c    2.43103 nm 1  cos180  4.86 103 nm

31. (a) In the Compton effect, the maximum change in the photon’s wavelength is when scattering

angle   180o . We use Eq. 37-6b to determine the maximum change in wavelength. Dividing

the maximum change by the initial wavelength gives the maximum fractional change.

  h 1 – cos  
mec

 
   
 h 1 – cos   6.6310–34 Js 1  cos180  8.8 106
mec
9.1110–31kg 3.00 108 m s 550 109 m

(b) We replace the initial wavelength with   0.10 nm.

 
   
 h 1 – cos   6.6310–34 Js 1  cos180  0.049
mec
9.1110–31kg 3.00 108 m s 0.10 109 m

32. We find the change in wavelength for each scattering event using Eq. 37-6b, with a scattering angle
of   0.50o. To calculate the total change in wavelength, we subtract the initial wavelength,

obtained from the initial energy, from the final wavelength. We divide the change in wavelength by

the wavelength change from each event to determine the number of scattering events.

    
 h 1 – cos 0.5o  6.6310–34 J  s 1  cos0.5  9.24 10–17 m  9.24 108 nm
mec
9.1110–31 kg 3.00 108 m s

   0
 hc  6.6310–34 J  s 3.00 108 m s  1.24 10–12 m  0.00124 nm .
E0 1.0 106 eV 1.60 10–19 J eV

n   0  555 nm – 0.00124 nm  6 109 events

9.24 10–8 nm

33. (a) We use conservation of momentum to set the initial momentum of the photon equal to the sum
of the final momentum of the photon and electron, where the momentum of the photon is given

by Eq. 37-5 and the momentum of the electron is written in terms of the total energy (Eq. 36-

13). We multiply this equation by the speed of light to simplify.

h  0    h   pe  hc    hc   E 2  E02
      

Using conservation of energy we set the initial energy of the photon and rest energy of the

electron equal to the sum of the final energy of the photon and the total energy of the electron.

 hc   E0   hc   E
     

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465

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

By summing these two equations, we eliminate the final wavelength of the photon. We then

solve the resulting equation for the kinetic energy of the electron, which is the total energy less

the rest energy.

 hc  2  hc   2
      
2  E0  E 2  E02 +E   E0  E  E2  E02

2  hc  2 2  hc   2  hc   E0 2  E02
           
 E0  2E  E0  E2  E2  E02  E 
  hc  
2  2     E0 

2  hc   E0 2  E02 2 2  hc   E0  E0 2  hc 2
         
K  E  E0    


2 2  hc   E0  2 2  hc   E0  2  hc   E0 
           

2  1240 eVnm 2
 0.160 nm 
  228 eV
2  1240 eVnm  
 0.160 nm   5.11  105 eV 

(b) We solve the energy equation for the final wavelength.

 hc   E0   hc   E
     

hc 1 K 1 1 228eV  1
 E0   hc  1240 eV nm 
   hc      0.160 nm    0.165 nm
   
E

34. First we use conservation of energy, where the energy of the photon is written in terms of the
wavelength, to relate the initial and final energies. Solve this equation for the electron’s final energy.

 hc   mc2   hc  E  E   hc    hc   mc2
           

Next, we define the x-direction as the direction of the initial motion of the photon. We write

equations for the conservation of momentum in the horizontal and vertical directions, where  is the

angle the photon makes with the initial direction of the photon and  is the angle the electron makes.

px : h  pe cos  h cos py : 0  pe sin   h sin
  

To eliminate the variable  we solve the momentum equations for the electron’s momentum, square

the resulting equations and add the two equations together using the identity cos2   sin2   1.

 h  h cos 2   pe cos 2  h sin 2  pe sin 2
      

 pe cos 2  pe sin 2   h  h cos 2   h sin 2
      

pe2   h 2  2h2 cos   h 2
      

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466

Chapter 37 Early Quantum Theory and Models of the Atom

We now apply the relativistic invariant equation, Eq. 36-13, to write the electron momentum in terms

of the electron energy. Then using the electron energy obtained from the conservation of energy

equation, we eliminate the electron energy and solve for the change in wavelength.

h 2 2h2 cos  h 2 E2  m2c4  h   h   2
     c2       
        mc  m2c2
 

=  h 2   h 2  m2c2  2hmc  1  1   h2  m2c2
          

 2h2 cos  2hmc  1  1  h2
     

h cos  mc      h       h 1  cos 
mc

35. The photon energy must be equal to the kinetic energy of the products plus the mass energy of the
products. The mass of the positron is equal to the mass of the electron.

2
products
E  K  m cphoton 
products

 Kproducts  Ephoton  m cproducts 2  Ephoton  2melectronc2  2.67 MeV  2 0.511MeV  1.65 MeV

36. The photon with the longest wavelength has the minimum energy in order to create the masses with
no additional kinetic energy. Use Eq. 37-5.

    max
 hc  hc  h  2 1.67 6.63  1034 Js s  6.62 1016 m
Emin 2mc2 2mc  1027 kg 3.00  108 m

This must take place in the presence of some other object in order for momentum to be conserved.

37. The minimum energy necessary is equal to the rest energy of the two muons.

Emin  2mc2  22070.511MeV  212 MeV

The wavelength is given by Eq. 37-5.

  
 hc  6.631034 Js 3.00 108 m s  5.86 1015 m
E 1.60 1019 J eV 212 106 eV

38. Since v  0.001c, the total energy of the particles is essentially equal to their rest energy. Both

particles have the same rest energy of 0.511 MeV. Since the total momentum is 0, each photon must

have half the available energy and equal momenta.

Ephoton  m c2  0.511MeV ; pphoton  Ephoton  0.511MeV c
electron c

39. The energy of the photon is equal to the total energy of the two particles produced. Both particles

have the same kinetic energy and the same mass.

 Ephoton  2 K  mc2  2 0.375MeV  0.511MeV  1.772 MeV

The wavelength is found from Eq. 37-5.

  
 hc  6.63 1034 Js 3.00 108 m s  7.02 1013 m
E 1.60 1019 J eV 1.772 106eV

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467

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

40. We find the wavelength from Eq. 37-7.

 
 h  h  6.631034 Js s  2.9 1032 m
p mv
 0.23 kg   0.10 m

41. The neutron is not relativistic, so we can use p  mv. We also use Eq. 37-7.

    
 h  h  6.63 1034 Js  4.7 1012 m
p mv 1.67 1027 kg 8.5 104 m s

42. We assume the electron is non-relativistic, and check that with the final answer. We use Eq. 37-7.

    
 h  h  v  h  6.63 1034 Js  3.466 106 m s  0.01155c
p mv m 9.111031kg 0.21109 m

Our use of classical expressions is justified. The kinetic energy is equal to the potential energy

change.

    eV 2
1 9.111031 kg 3.466 106 m s
 K  1 mv 2  2 1.60 1019 J eV  34.2eV
2

Thus the required potential difference is 34 V.

43. The theoretical resolution limit is the wavelength of the electron. We find the wavelength from the

momentum, and find the momentum from the kinetic energy and rest energy. We use the result from

Problem 94. The kinetic energy of the electron is 85 keV.

  hc    6.631034 Js 3.00 108 m s

      K 2  2mc2K 1.60 1019 J eV 85 103eV 2  2 0.511106eV 85103eV

 4.11012 m

44. We use the relativistic expression for momentum, Eq. 36-8.
p  mv  mv  h 
1 v2 c2 1 v2 c2 

6.631034 Js 1  0.982
9.111031kg 0.98 3.00 108 m s
       h1 v2 c2  4.9 1013 m
mv 

45. Since the particles are not relativistic, we may use K  p2 2m. We then form the ratio of the

kinetic energies, using Eq. 37-7.

h2

K  p2  h2 ; e  2me 2  mp  1.67 1027 kg  1840
2m 2m 2 p h2 me 9.11  1031 kg

2mp 2

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468

Chapter 37 Early Quantum Theory and Models of the Atom

46. We assume the neutron is not relativistic. If the resulting velocity is small, our assumption will be

valid. We use Eq. 37-7.

    
 h  h  v  h  6.63 1034 Js  1300 m s  1000 m s
p mv m 1.67 1027 kg 0.3109 m

This is not relativistic, so our assumption was valid.

47. (a) We find the momentum from Eq. 37-7.

p h  6.631034 Js  1.11024 kgm s
 6.0 1010 m

(b) We assume the speed is non-relativistic.

h h v h 6.631034 Js  1.2 106 m s
p mv m 9.111031 kg 6.0 1010 m
      

Since v c  4.04 103, our assumption is valid.

(c) We calculate the kinetic energy classically.

   K c2 2  4.17 106 MeV  4.17 eV
 1 mv2  1 mc2 v  1  0.511 MeV  4.04 103
2 2 2

This is the energy gained by an electron if accelerated through a potential difference of 4.2 V.

48. Because all of the energies to be considered are much less than the rest energy of an electron, we can
use non-relativistic relationships. We use Eq. 37-7 to calculate the wavelength.

K  p2  p 2mK ;   h  h
2m p 2mK

(a)   h  6.63  1034 Js  2.7 1010 m  3 1010 m

   2mK 2 9.111031kg 20eV 1.60 1019 J eV

(b)   h  6.63  1034 Js  8.7 1011 m  9 1011m

   2mK 2 9.111031kg 200eV 1.60 1019 J eV

(c)   h  6.63  1034 Js  2.7 1011m

   2mK 2 9.111031kg 2.0 103eV 1.60 1019 J eV

49. Since the particles are not relativistic, we may use K  p2 2m. We then form the ratio of the

wavelengths, using Eq. 37-7.

h

  h  h ; p  2mp K  me 1
p 2mK e h mp

2me K

Thus we see the proton has the shorter wavelength, since me  mp.

50. The final kinetic energy of the electron is equal to the negative change in potential energy of the
electron as it passes through the potential difference. We compare this energy to the rest energy of
the electron to determine if the electron is relativistic.

 K  qV  1e 33103 V  33103eV

Because this is greater than 1% of the electron rest energy, the electron is relativistic. We use Eq.
36-13 to determine the electron momentum and then Eq. 37-5 to determine the wavelength.

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469

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

E2   K  mc 2 2  p2c2  m2c4  p K 2  2Kmc2
c


h hc 1240 eVnm  0.0066 nm
p 

K 2  2Kmc2
    
33 103eV 2  2 33103eV 511103eV

Because  « 5 cm, diffraction effects are negligible.

51. We will assume that the electrons are non-relativistic, and then examine the result in light of that
assumption. The wavelength of the electron can be found from Eq. 34-2a. The speed can then be
found from Eq. 37-7.

d sin  morder    d sin ;   h  h 
morder p mev

    v 
hmorder  6.63  1034 Js 2  590 m s
med sin 9.11  1031kg 3.0 106 m sin 55

This is far from being relativistic, so our original assumption was fine.

52. We relate the kinetic energy to the momentum with a classical relationship, since the electrons are

non-relativistic. We also use Eq. 37-7. We then assume that the kinetic energy was acquired by

electrostatic potential energy.

K  p2  h2  eV 
2m 2m 2

6.63  1034 Js 2
9.11  1031 kg 1.60 1019 C 0.28  109 m
    Vh2 2 2  19 V
2me 2

53. The kinetic energy is 3450 eV. That is small enough compared to the rest energy of the electron for
the electron to be non-relativistic. We use Eq. 37-7.

        
 h  h  hc 6.63  1034 Js 3.00  108 m / s
p 2mc2K 1/ 2 
2mK 1/ 2 1.60  1019 J/eV 2 0.511  106 eV 3450 eV1/ 2

 2.09  1011m  20.9 pm

54. The energy of a level is En   13.6 eV  .

n2

(a) The transition from n = 1 to n' = 3 is an absorption, because the final state, n' = 3, has a

higher energy. The photon energy is the difference between the energies of the two states.

hf  En  En   13.6 eV  1    1   12.1 eV
 32   12 

(b) The transition from n = 6 to n' = 2 is an emission, because the initial state, n' = 2, has a

higher energy. The photon energy is the difference between the energies of the two states.

hf    En  En   13.6 eV  1    1   3.0 eV
 22   62 

(c) The transition from n = 4 to n' = 5 is an absorption, because the final state, n' = 5, has a

higher energy. The photon energy is the difference between the energies of the two states.

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470

Chapter 37 Early Quantum Theory and Models of the Atom

hf  En  En   13.6 eV   1   1   0.31 eV
 52   42 

The photon for the transition from n = 1 to n' = 3 has the largest energy.

55. To ionize the atom means removing the electron, or raising it to zero energy.

Eionization  0 En  0  13.6 eV   13.6 eV   1.51eV

n2 32

56. We use the equation that appears above Eq. 37-15 in the text.

(a) The second Balmer line is the transition from n = 4 to n = 2.

  hc  1240 eVnm  490 nm

E4  E2   0.85eV  3.4eV

(b) The third Lyman line is the transition from n = 4 to n = 1.

  hc  1240 eVnm eV  97.3nm

 E4  E1   0.85 eV  13.6

(c) The first Balmer line is the transition from n = 3 to n = 2.

For the jump from n = 5 to n = 2, we have

  hc  1240 eVnm  650 nm

 E3  E2  1.5 eV   3.4 eV

57. Doubly ionized lithium is similar to hydrogen, except that there are three positive charges (Z = 3) in

the nucleus. The square of the product of the positive and negative charges appears in the energy
term for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2:

En   Z2 13.6 eV    32 13.6 eV    122 eV 

n2 n2 n2

Eionization  0 E1  0   122eV   122 eV
 
 12 

58. We evaluate the Rydberg constant using Eq. 37-8 and 37-15. We use hydrogen so Z = 1.

1  R  1  1   Z 2e4m  1  1  
    
 n2 n2 8 2 h3c n2 n2
0

12 1.602176 1019 C 4 9.109382 1031kg

8.854188 1012 C2 Nm2 2 6.626069 1034 Js 3 2.997925 108 m
         RZ 2e4m

8 2 h 3c 8 s
0

 1.0974 107 C4 kg  1.0974 107 m1

C4 J3s3 m s
N2 m4

59. The longest wavelength corresponds to the minimum energy, which is the ionization energy:

    
 hc  6.631034 Js 3.00 108 m / s  9.14  108 m  91.4 nm
Eion
1.60 1019 J/eV 13.6eV

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471

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

60. Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the
nucleus. The square of the product of the positive and negative charges appears in the energy term
for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2.

En   Z2 13.6 eV   22 13.6 eV   54.4 eV

n2 n2 n2

We find the energy of the photon from the n = 5 to n = 2 transition in singly-ionized helium.

E  E5  E2   54.4 eV   1    1   11.4 eV
 52   22 

Because this is NOT the energy difference between any two specific energy levels for hydrogen, the

photon CANNOT be absorbed by hydrogen.

61. The energy of the photon is the sum of the ionization energy of 13.6 eV and the kinetic energy of

20.0eV. The wavelength is found from Eq. 37-3.

    hf
 hc  Etotal    hc  6.63  1034 Js 3.00  108 m / s  3.70 108 m  37.0 nm
 Etotal
1.60  1019 J/eV 33.6 eV

62. A collision is elastic if the kinetic energy before the collision is equal to the kinetic energy after the

collision. If the hydrogen atom is in the ground state, then the smallest amount of energy it can
absorb is the difference in the n = 1 and n = 2 levels. So as long as the kinetic energy of the

incoming electron is less than that difference, the collision must be elastic.

K  E2  E1    13.6 eV    13.6 eV   10.2 eV
 4 

63. Singly ionized helium is like hydrogen, except that there are two

positive charges (Z = 2) in the nucleus. The square of the product

of the positive and negative charges appears in the energy term

for the energy levels. We can use the results for hydrogen, if we
replace e2 by Ze2:

En   Z2 13.6 eV    22 13.6 eV    54.4 eV 

n2 n2 n2

E1  54.5eV, E2  13.6eV, E3  6.0eV, E4  3.4eV

64. Doubly ionized lithium is like hydrogen, except that there are

three positive charges (Z = 3) in the nucleus. The square of the

product of the positive and negative charges appears in the 7.65
13.6
energy term for the energy levels. We can use the results for
hydrogen, if we replace e2 by Ze2: 30.6

En   Z2 13.6 eV    32 13.6 eV    122.4 eV 

n2 n2 n2

E1  122 eV, E2  30.6eV, E3  13.6eV,

E4  7.65eV

122

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472

Chapter 37 Early Quantum Theory and Models of the Atom

65. The potential energy for the ground state is given by the charge of the electron times the electric

potential caused by the proton.

    U 2
1 e 9.00  109 Nm2 C2 1.60  1019 C 1eV 1.60 1019 J
  e Vproton  e 4 0 r1  0.529  1010 m

  27.2eV

The kinetic energy is the total energy minus the potential energy.

K  E1 U  13.6eV  27.2eV  13.6eV

66. The value of n is found from rn  n2r1, and then find the energy from Eq. 37-14b.

 rn  n2r1  n 
rn  1 0.10 103 m  972
r1 2

0.529 1010 m

E   13.6 eV    13.6 eV   13.6 eV   1.4 105eV

n2 9722 13752

67. The velocity is found from Eq. 37-10 evaluated for n = 1.

mvrn  nh 
2

    v
 h  2 6.631034 Js  2.190 106 m s  7.30 103c
2 r1me 0.529  1010 m 9.11  1031kg

We see that v  c, and so yes, non-relativistic formulas are justified.

The relativistic factor is as follows.

2 1 2 2
 
1  v  2  v  2.190 106 m s  1  2.66 105
  c   c  3.00 108 m s 
  1  1  1  1     0.99997
2 2

We see that 1  v2 c2 is essentially 1, and so again the answer is yes, non-relativistic formulas are

justified.

68. The angular momentum can be used to find the quantum number for the orbit, and then the energy

can be found from the quantum number. Use Eqs. 37-10 and 37-14b.

  L
 n h  n  2 L  2 5.2731034 kgm2 s  5.000  5
2 h 6.626  1034 Js

En   13.6 eV  Z2   13.6 eV  0.544 eV
n2 25

69. Hydrogen atoms start in the n  1 orbit (“ground state”). Using Eq. 37-9 and Eq. 37-14b, we

determine the orbit to which the atom is excited when it absorbs a photon of 12.75 Ev via collision

with an electron. Then, using Eq. 37-15, we calculate all possible wavelengths that can be emitted as

the electron cascades back to the ground state.

E  EU  EL  EU   13.6 eV  EL  E 
n2

n 13.6 eV  13.6 eV eV  4
EL  E 13.6 eV + 12.75

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473

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

Starting with the electron in the n  4 orbit, the following transitions are possible: n  4 to n  3 ;

n  4 to n  2 ; n  4 to n  1 ; n  3 to n  2 ; n  3 to n  1 ; n  2 to n  1 .

 1 1 1  5.333  105 m –1
 32 42 

 1.097 107 m–1 –     1875 nm

 1 1 1 
 22 42 

 1.097 107 m –1 –  2.057 106 m–1    486.2 nm

 1 1 1   1.028 107 m–1   
 12 42 

 1.097 107 m–1 – 97.23 nm

 1 1 1 
 22 32 

 1.097 107 m –1 –  1.524 106 m–1    656.3 nm

 1 1 1 
 12 32 

 1.097 107 m–1 –  9.751106 m–1    102.6 nm

 1 1 1  8.228 106
 12 22 

 1.097 107 m –1 –  m –1    121.5 nm

70. When we compare the gravitational and electric forces we see that we can use the same expression

for the Bohr orbits, Eq. 37-11 and 37-14a, if we replace Ze2 40 with Gmemp.

r1   h 2 0  h2 4 0 
me Ze 2 4 2me Ze2

6.626 1034 Js 2
6.67 1011 Nm2 / kg2 9.111031kg
      r1
 4 h2  2 1.67 1027 kg
2Gm 2 e m p 4 2

 1.20 1029 m

E1   Z 2e4me    Ze2 2 2 2me  E1   2 2G 2 me 3m p 2
8 02h 2  4 0 h2 h2
 



6.67 1011 Nm2 kg2 2 9.111031 kg 3 2
6.626 1034 Js 2
     2 2
  
1.67 1027 kg  4.22 1097 J

71. We know that the radii of the orbits are given by rn  n2r1. Find the difference in radius for adjacent
orbits.

 r  rn  rn1  n2r1  n 12 r1  n2r1  n2  2n  1 r1  2n 1 r1

If n  1, we have r  2nr1  2n rn  2rn .
n2 n

In the classical limit, the separation of radii (and energies) should be very small. We see that letting

n   accomplishes this. If we substitute the expression for r1 from Eq. 37-11, we have this.

r  2nr1  2nh 2 0
 me2

We see that r  h2, and so letting h  0 is equivalent to considering n  .

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474

Chapter 37 Early Quantum Theory and Models of the Atom

72. We calculate the energy from the light bulb that enters the eye by calculating the intensity of the

light at a distance of 250 m by dividing the power in the visible spectrum by the area of a sphere of

radius 250 m. We multiply the intensity of the light by the area of the pupil to determine the energy

entering the eye per second. We divide this energy by the energy of a photon (Eq. 37-3) to calculate

the number of photons entering the eye per second.

P  Pe  I P  D 2
I  4 l2  D2 /4  16  l



    n  4.0 103 m 2
 Pe  P  D 2  0.03075W 550 109 m  
hc /  16hc  l 16  250 m 
 6.626 1034 Js 3.00 108 m s


 1.0 108 photons/sec

73. To produce a photoelectron, the hydrogen atom must be ionized, so the minimum energy of the

photon is 13.6 eV. We find the minimum frequency of the photon from Eq. 37-3.

   E  hf
 f  E  f min  Emin  13.6 eV 1.60  1019 J eV  3.28 1015 Hz
h h
6.63  1034 Js

74. From Section 35-10, the spacing between planes, d, for the first-order peaks is given by Eq. 35-20,
  2d sin. The wavelength of the electrons can be found from their kinetic energy. The electrons
are not relativistic at the energy given.

K  p2  h2   h  2d sin 
2m 2m 2 2mK

d h   6.631034 Js  8.9 1011m

   2sin 2mK 2sin 38 2 9.111031kg 125eV 1.60 1019 J/eV

75. The power rating is the amount of energy produced per second. If this is divided by the energy per

photon, then the result is the number of photons produced per second.

    Ephoton
 hf  hc ; P  P  860 W 12.2  102 m  5.3  1026 photons s
 Ephoton hc
6.63  1034 Js 3.00  108 m s

76. The intensity is the amount of energy per second per unit area reaching the Earth. If that intensity is

divided by the energy per photon, the result will be the photons per second per unit area reaching the

Earth. We use Eq. 37-3.

Ephoton  hf  hc


    Iphotons
 I sunlight  I sunlight  1350 W m2 550  109 m  3.7 1021 photons sm2
Ephoton hc 6.63  1034 Js 3.00  108 m/s

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475

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

77. The impulse on the wall is due to the change in momentum of the photons. Each photon is absorbed,

and so its entire momentum is transferred to the wall.

 Fon wallt  pwall  pphotons   0  npphoton  npphoton  nh 


  n
 t
 F  6.5  109 N 633  109 m  6.2 1018 photons s
h 6.63  1034 Js

78. We find the peak wavelength from Wien’s law, Eq. 37-1.

   P 
2.90  103 mK  2.90  103 mK  1.1  103 m  1.1mm
T
2.7 K

79. The total energy of the two photons must equal the total energy (kinetic energy plus mass energy) of
the two particles. The total momentum of the photons is 0, so the momentum of the particles must
have been equal and opposite. Since both particles have the same mass and the same momentum,
they each have the same kinetic energy.

 Ephotons  Eparticles  2 mec2  K 

K  1 Ephotons  mec2  0.755MeV  0.511MeV  0.244 MeV
2

80. We calculate the required momentum from de Broglie’s relation, Eq. 37-7.

 p  h 
 
6.63  1034 Js  1.11  1022 kgm/s
6.0  1012 m

(a) For the proton, we use the classical definition of momentum to determine the speed of the

electron, and then the kinetic energy. We divide the kinetic energy by the charge of the proton

to determine the required potential difference.

v  p  1.111022 kgm/s  6.65 104 m/s  c
m 1.67 1027 kg

1.67 1027 kg 6.65 104 m/s 2
2 1.60 1019 C
    V
 K mv2  23 V
e  2e 

(b) For the electron, if we divide the momentum by the electron mass we obtain a speed greater

than 10% of the speed of light. Therefore, we must use the relativistic invariant equation to

determine the energy of the electron. We then subtract the rest energy from the total energy to

determine the kinetic energy of the electron. Finally, we divide the kinetic energy by the

electron charge to calculate the potential difference.

1
22
 E   pc2  m0c2 

1
2 3.00 108 m s 2  9.111031kg 2 4 2
        
 1.111022 kgm/s 3.00 108 m s

 8.85 1014 J

  K  E  m0c2  8.85 1014 J  9.111031kg 3.00 108 m s 2  6.50 1015J

V  K  6.50 1015 J  41 kV
e 1.60 1019 C

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476

Chapter 37 Early Quantum Theory and Models of the Atom

81. If we ignore the recoil motion, at the closest approach the kinetic energy of both particles is zero.

The potential energy of the two charges must equal the initial kinetic energy of the  particle:

 K
U  1 Ze ZAge 
4 0
rmin

       rmin 2
1 Ze ZAge 9.00  109 Nm2 C2 247 1.60  1019 C
 4 0  4.8 MeV 1.60  1013 J MeV  2.8 1014 m
K

82. The electrostatic potential energy is given by Eq. 23-5. The kinetic energy is given by the total

energy, Eq. 37-14a, minus the potential energy. The Bohr radius is given by Eq. 37-11.

U  eV  1 Ze2 1 Ze2 mZe2   Z 2e4m
 40 rn   40 n 2 h 2 0
4n 2h 2 2
0

Z 2e4m

K  E U   Z 2e4m    Z 2e4m   Z 2e4m ; U  4n 2h 2 2  Z 2e4m 8n2 h 2 2  2
  K 0 0

8 2 h 2n 2  4n 2h 2 2 8n2 h 2 2 Z 2e4m 4n 2h2 2 Z 2e4m
0 0 0 0

8n 2h 2 2
0

83. We calculate the ratio of the forces.

 Gmemp  6.67  1011 Nm2 kg2 9.11  1031 kg 2 1.67  1027 kg
  9.00  109 Nm2 C2 1.60  1019 C 2
    Fgravitationalr2   Gme m p 
  Felectric  ke2  ke2
 
 r2 

 4.4  1040

Yes, the gravitational force may be safely ignored.

84. The potential difference gives the electrons a kinetic energy of 12.3 eV, so it is possible to provide

this much energy to the hydrogen atom through collisions. From the ground state, the maximum

energy of the atom is 13.6 eV  12.3 eV  1.3 eV. From the energy level diagram, Figure 37-26,

we see that this means the atom could be excited to the n = 3 state, so the possible transitions when

the atom returns to the ground state are n = 3 to n = 2, n = 3 to n = 1, and n = 2 to n = 1. We

calculate the wavelengths from the equation above Eq. 37-15.

32  hc  1240 eVnm  650 nm

 E3  E2  1.5 eV   3.4 eV

31  hc  1240 eVnm eV   102 nm

 E3  E1  1.5 eV  13.6

21  hc  1240 eVnm eV   122 nm

 E2  E1   3.4 eV  13.6

85. The stopping potential is the voltage that gives a potential energy change equal to the maximum

kinetic energy. We use Eq. 37-4b to first find the work function, and then find the stopping potential

for the higher wavelength.

K max  eV0  hc  W0  W0  hc  eV0
 0

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eV1  hc  W0  hc   hc  eV0   hc  1  1   eV0
1 1  0   1 0 
   

    
6.63  1034 Js 3.00  108 m s  440 1 m  380 1 m   2.70 eV  2.25eV
1.60  1019 J eV   109  109 
 

The potential difference needed to cancel an electron kinetic energy of 2.25 eV is 2.25 V.

86. (a) The electron has a charge e, so the potential difference produces a kinetic energy of eV. The
shortest wavelength photon is produced when all the kinetic energy is lost and a photon is
emitted.

hf max  hc  eV  0  hc which gives 0  hc .
0 eV eV

(b) 0  hc 1240 eV nm  0.038 nm
eV  33  103eV

87. The average force on the sail is equal to the impulse on the sail divided by the time (Eq. 9-2). Since

the photons bounce off the mirror the impulse is equal to twice the incident momentum. We use Eq.

37-5 to write the momentum of the photon in terms of the photon energy. The total photon energy is

the intensity of the sunlight multiplied by the area of the sail

 F m2
 p  2E /c  2 E / t   2IA  2 1350 W/m2 1000  9.0 N
t c
t c 3.00 108 m/s

88. We first find the work function from the given data. A photon energy of 9.0 eV corresponds with a
stopping potential of 4.0 V.

eV0  hf  W0  W0  hf  eV0  9.0eV  4.0eV  5.0eV
If the photons’ wavelength is doubled, the energy is halved, from 9.0 eV to 4.5 eV. This is smaller
than the work function, and so no current flows. Thus the maximum kinetic energy is 0. Likewise,
if the photon’s wavelength is tripled, the energy is only 3.0 eV, which is still less than the work
function, and so no current flows.

89. The electrons will be non-relativistic at that low energy. The maximum kinetic energy of the
photoelectrons is given by Eq. 37-4b. The kinetic energy determines the momentum, and the
momentum determines the wavelength of the emitted electrons. The shortest electron wavelength
corresponds to the maximum kinetic energy.

Kelectron  hc  W0  p2  h2  electron h
 2m 2m 2
2m  hc  W0 
electron   

  6.63  1034 Js  1.2 109 m
1240 eVnm
 360 nm 
   2  
9.11  1031kg   2.4 eV  1.60  1019 J eV

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478

Chapter 37 Early Quantum Theory and Models of the Atom

90. The wavelength is found from Eq. 35-13. The velocity of electrons with the same wavelength (and

thus the same diffraction pattern) is found from their momentum, assuming they are not relativistic.

We use Eq. 37-7 to relate the wavelength and momentum.

d sin  n    d sin  h  h 
n p mv

    v
 hn  6.631034 Js 1
md sin 9.111031kg 0.012 103 m sin 3.5  990 m s

91. (a) See the adjacent figure.
(b) Absorption of a 5.1 eV photon represents a transition
from the ground state to the state 5.1 eV above that,
the third excited state. Possible photon emission
energies are found by considering all the possible
downward transitions that might occur as the electron
makes its way back to the ground state.

6.4eV  6.8eV  0.4eV

6.4eV  9.0eV  2.6eV

6.4eV  11.5eV  5.1eV

6.8eV  9.0eV  2.2eV

6.8eV  11.5eV  4.7eV

9.0eV  11.5eV  2.5eV

92. (a) We use Eq. 37-4b to calculate the maximum kinetic energy of the electron and set this equal to

the product of the stopping voltage and the electron charge.

Kmax  hf  W0  eV0  V0  hf  W0  hc /  W0
e e

V0  1240 eVnm  424 nm  2.28eV  0.65 V

e

(b) We calculate the speed from the non-relativistic kinetic energy equation and the maximum

kinetic energy found in part (a).

 Kmax
 1 mvm2 ax  vmax  2 K max  2 0.65eV 1.60 1019 J eV  4.8 105 m/s
2 m
9.11  1031 kg

(c) We use Eq. 37-7 to calculate the de Broglie wavelength.

h h 6.631034 Js  1.52  109 m  1.5nm
p mv 9.111031kg 4.8 105 m s
    

93. (a) We use Bohr’s analysis of the hydrogen atom, where we replace the proton mass with Earth’s

mass, the electron mass with the Moon’s mass, and the electrostatic force Fe  ke2 with the
r2

gravitational force Fg  GmE mM . To account for the change in force, we replace ke2 with
r2

GmEmM . With these replacements, we write expressions similar to Eq. 37-11 and Eq. 37-14a
for the Bohr radius and energy.

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479

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

rn  h2n2 
4 2mke2

6.626 1034 Js 2
6.67 1011 Nm2 / kg2 7.35 1022 kg
      rn
 h2n2  2 5.98 1024 kg n2
4 2GmM2 mE 4 2

  n2 5.16 10129 m

En   2 2e4mk 2 
n2h2

6.67 1011 Nm2 / kg2 2 5.98 1024 kg 2 7.35 1022 kg 3
n2 6.626 1034 Js 2
       En 2 2
  2 2G2mE2 mM3 
n2h2

  2.84  10165 J
n
2

(b) We insert the known masses and Earth–Moon distance into the Bohr radius equation to
determine the Bohr state.

n 4 2GmM2 mE rn
h2

     4 2 6.67 1011 Nm2 / kg2 7.35 1022 kg 2 5.98 1024 kg 3.84 108 m
  6.626 1034 Js 2

 2.73 1068
Since n 1068, a value of n  1 is negligible compared to n. Hence the quantization of

energy and radius is not apparent.

94. We use Eqs. 36-13, 36-11, and 37-7 to derive the expression.

 p2c2  m2c4  E 2 ; E  K  mc2  p2c2  m2c4  K  mc2 2  K 2  2mc2K  m2c4 

p2c2 h2c2 h2c2 hc
2 K 2  2mc2K
 K 2
 2mc 2 K    2    K 2  2mc2K

95. As light leaves the flashlight it gains momentum. This change in momentum is given by Eq. 31-20.

Dividing the change in momentum by the elapsed time gives the force the flashlight must apply to

the light to produce this momentum. This is equal to the reaction force that light applies to the

flashlight.

p  U  P  3.0 W s  1.0 108 N
t ct c 3.00 108 m

96. (a) Since f  c  , the energy of each emitted photon is E  hc  . We insert the values for h and

c and convert the resulting units to eVnm.

    E  hc  6.626 10–34 Js 2.998 108 m s
  
1eV 1.602 10–19J  1240 eVnm
10–9m 1nm
 in nm

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480

Chapter 37 Early Quantum Theory and Models of the Atom

(b) Insert 650 nm into the above equation.

E  1240 eVnm  1.9 eV
650 nm

97. (a) We write the Planck time as tP  G h c , and the units of tP must be T .

tP  G h c  T    L3   ML2   L   L 2  M   T 2  
    T 
 MT 2   T 

There are no mass units in T , and so   , and T   L5  T 3  . There are no

length units in T , and so   5 and T   T 3 5  T 2 . Thus   1   and
2

   5 .
2

tP  G1/ 2h1/ 2c5/ 2  Gh
c5

    (b)
tP  Gh 6.67 1011 N  m2 kg2 6.63 1034 Js  1.35 1043s
c5  3.00 108 m s 5

(c) We write the Planck length as P  G h c , and the units of P must be L.

P  G h c  L   L3   ML2   L   L 2  M   T 2  
    T 
 MT 2   T 

There are no mass units in L, and so   , and L  L5  T 3  . There are no

time units in L, and so   3 and L  L5 3  L2 . Thus   1   and
2

   3 .
2

tP  G1/ 2h1/ 2c3/ 2  Gh
c3

    (d)
P  Gh 6.67 1011 N  m2 kg2 6.63 1034 Js  4.05 1035 m
c3  3.00 108 m s 5

98. For standing matter waves, there are nodes at the two walls. For the ground state (first harmonic),

the wavelength is twice the distance between the walls, or l 1  (see Figure 15-26b). We use Eq.
2

37-7 to find the velocity and then the kinetic energy.

l  1     2l ; p  h  h ; K  p2  1  h 2  h2
2  2l 2m 2m  2l 8ml2



For the second harmonic, the distance between the walls is a full wavelength, and so l  .

l  p h h ; K  p2  1  h 2  h2
 l 2m 2m  l  2ml2

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481

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

99. (a) Apply conservation of momentum before and after the emission of the photon to determine the

recoil speed of the atom, where the momentum of the photon is given by Eq. 37-7.

 0  h  mv
  
 v  h  85 6.63 10–34 Js  6.0 10–3 m s
m 1.66 10–27 kg 780 10–9 m

(b) We solve Eq. 18-5 for the lowest achievable temperature, where the recoil speed is the rms

speed of the rubidium gas.

    v  2
3kT mv 2 85 1.66 10–27 kg 6.0 10–3 m s
m  T  3k  3 1.38 10–23 J K  1.2 10–7 K  0.12 K

100. Each time the rubidium atom absorbs a photon its momentum decreases by the momentum of the

photon. Dividing the initial momentum of the rubidium atom by the momentum of the photon, Eq.

37-7, gives the number of collisions necessary to stop the atom. Multiplying the number of

collisions by the absorption time, 25 ns per absorption, provides the time to completely stop the

atom.

   n 
mv  mv 8u 1.66 1027 kg/u 290 m s 780 109 m  48,140
h h
 6.63 1034Js

T  48,14025ns  1.2 ms

101. (a) See the adjacent graphs. I (,T ) (1012 kg/m/s3) 6 2700 K
(b) To compare the intensities, the 5 3300 K
two graphs are numerically 4
integrated from 400 nm to 760 3 1600 2000
nm, which is approximately the 2
range of wavelengths for visible 1
light. The result of those
integrations is that the higher

temperature bulb is about 4.8

times more intense than the

lower temperature bulb. 0

The spreadsheet used for this 0 400 800 1200

 (nm)

problem can be found on the Media

Manager, with filename “PSE4_ISM_CH37.XLS,” on tab “Problem 37.101.”

102. Planck’s radiation formula I T  was calculated for a temperature of 6000 K, for wavelengths

from 20 nm to 2000 nm. A plot of those calculations is in the spreadsheet for this problem. To
estimate the % of emitted sunlight that is in the visible, this ratio was calculated by numeric
integration. The details are in the spreadsheet.

700 nm

% visible  I ,T d  0.42
400 nm
 2000 nm

 I ,T d
20 nm

So our estimate is that 42% of emitted sunlight is in the visible wavelengths. The spreadsheet used

for this problem can be found on the Media Manager, with filename “PSE4_ISM_CH37.XLS,” on

tab “Problem 37.102.”

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482

Chapter 37 Early Quantum Theory and Models of the Atom

103. (a) For the photoelectric effect experiment, Eq. 37-4b can be expressed as Kmax  hf W0. The
maximum kinetic energy is equal to the potential energy associated with the stopping voltage,

so Kmax  eV0. We also have f  c  . Combine those relationships as follows.

Kmax  hf  W0  eV0  hc  W0  V0  hc 1  W0
 e  e

A plot of V0 vs. 1 should yield a straight line with a slope of hc and a y-intercept of  W0 .
 e e

(b) The graph is shown, with a linear 1.0
regression fit as given by Excel.
V 0 = 1.24(1/) - 2.31
hc 0.8 R2 = 1.00
e
(c) The slope is a   1.24Vm, V 0 (Volts) 0.6

and the y-intercept is b  2.31V. 0.4

The spreadsheet used for this 0.2
problem can be found on the
Media Manager, with filename 0.0
“PSE4_ISM_CH37.XLS,” on tab
“Problem 37.103.” 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6

1/ (m-1)

(d) b   W0  2.31V  W0  2.31eV
e

  (e)
h  ea  1.60 1019 C 1.24 106 Vm  6.611034 Js
c 3.00 108 m s

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483

CHAPTER 38: Quantum Mechanics

Responses to Questions

1. (a) A matter wave ψ does not need a medium as a wave on a string does. The square of the wave
function for a matter wave ψ describes the probability of finding a particle within a certain
spatial range, whereas the equation for a wave on a string describes the displacement of a piece
of string from its equilibrium position.

(b) An EM wave also does not need a medium. The equation for the EM wave describes the way in
which the amplitudes of the electric and magnetic fields change as the wave passes a point in
space. An EM wave represents a vector field and can be polarized. A matter wave is a scalar
and cannot be polarized.

2. According to Bohr’s theory, each electron in an atom travels in a circular orbit and has a precise
position and momentum at any point in time. This view is inconsistent with the postulates of
quantum mechanics and the uncertainty principle, which does not allow both the position and
momentum to be known precisely. According to quantum mechanics, the “orbitals” of electrons do
not have precise radii, but describe the probability of finding an electron in a given spatial range.

3. As mass increases, the uncertainty in the momentum of the object increases, and, from the
Heisenberg uncertainty principle, the uncertainty in the position of the object decreases, making the
future position of the object easier to predict.

4. Planck’s constant is so small that on the scale of a baseball the uncertainties in position and
momentum are negligible compared with the values of the position and momentum. If visible light is
being used to observe the baseball, then the uncertainty in the baseball’s position will be on the order
of the wavelength of visible light. (See Section 38-3.) A baseball is very large compared to the
wavelength of light, so any uncertainty in the position of the baseball will be much smaller than the
extent of the object itself.

5. No. According to the uncertainty principle, if the needle were balanced the position of the center of
mass would be known exactly, and there would have to be some uncertainty in its momentum. The
center of mass of the needle could not have a zero momentum, and therefore would fall over. If the
initial momentum of the center of mass of the needle were exactly zero, then there would be
uncertainty in its position, and the needle could not be perfectly balanced (with the center of mass
over the tip).

6. Yes, some of the air escapes the tire in the act of measuring the pressure and it is impossible to avoid
this escape. The act of measuring the air pressure in a tire therefore actually changes the pressure,
although not by much since very little air escapes compared to the total amount of air in the tire.
This is similar to the uncertainty principle, in which one of the two factors limiting the precision of
measurement is the interaction between the object begin observed, or measured, and the observing
instrument.

7. Yes. In energy form, the uncertainty principle is ΔEΔt > h/2 . For the ground state, Δt is very large,
since electrons remain in that state for a very long time, so ΔE is very small and the energy of the
state can be precisely known. For excited states, which can decay to the ground state, Δt is much
smaller, and ΔE is corresponding larger. Therefore the energy of the state is less well known.

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484

Chapter 38 Quantum Mechanics

8. If Planck’s constant were much larger than it is, then the consequences of the uncertainty principle
would be noticeable with macroscopic objects. For instance, attempts to determine a baseball’s
speed would mean that you could not find its position very accurately. Using a radar gun to find the
speed of a pitcher’s fastball would significantly change the actual course of the ball.

9. According to Newtonian mechanics, all objects have an exact position and momentum at a point in
time. This information can be used to predict the future motion of an object. According to quantum
mechanics, there is unavoidable uncertainty in the position and momentum of all objects. It is
impossible to exactly determine both position and momentum at the same time, which introduces
uncertainly into the prediction of the future motion of the object.

10. If you knew the position precisely, then you would know nothing about the momentum.

11. No. Some of the energy of the soup would be used to heat up the thermometer, so the temperature
registered on the thermometer would be slightly less than the original temperature of the soup.

12. No. However, the greater the precision of the measurement of position, the greater the uncertainty in
the measurement of the momentum of the object will be.

13. A particle in a box is confined to a region of space. Since the uncertainty in position is limited by the
box, there must be some uncertainty in the particle’s momentum, and the momentum cannot be zero.
The zero point energy reflects the uncertainty in momentum.

14. Yes, the probability of finding the particle at these points is zero. It is possible for the particle to pass
by these points. Since the particle is acting like a wave, these points correspond to the nodes in a
standing wave pattern in the box.

15. For large values of n, the probability density varies rapidly between zero and the maximum value. It
can be averaged easily to the classical result as n becomes large.

16. As n increases, the energy of the corresponding state increases, but ΔE/E approaches zero. For large
n, the probability density varies rapidly between zero and the maximum value and is easily averaged
to the classical result, which is a uniform probability density for all points in the well.

17. As the potential decreases, the wave function extends into the forbidden region as an exponential
decay function. When the potential drops below the particle energy, the wave function outside the
well changes from an exponential decay function to an oscillating function with a longer wavelength
than the function within the well. When the potential is zero, the wavelengths of the wave function
will be the same everywhere. The ground state energy of the particle in a well becomes the energy of
the free particle.

18. The hydrogen atom will have a greater probability of tunneling through the barrier because it has a
smaller mass and therefore a larger transmission coefficient. (See Equations 38-17a and 38-17b.)

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485

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

Solutions to Problems

1. We find the wavelength of the neutron from Eq. 37-7. The peaks of the interference pattern are

given by Eq. 34-2a and Figure 34-10. For small angles, we have sin  tan.

  h  h ; d sin  m, m  1,2, ... ; y  l tan
p 2m0 K

sin  tan  m  y  y  m l , m  1, 2, ... 
d l d

       y
 l  d hl  6.0 104 m 6.631034 Js 1.0 m
d 2m0 K 2 1.67 1027 kg 0.030eV 1.60 1019 J eV

 2.8 107 m

2. We find the wavelength of a pellet from Eq. 37-7. The half-angle for the central circle of the

diffraction pattern is given in Section 35-4 as sin  1.22 , where D is the diameter of the opening.
D

Assuming the angle is small, the diameter of the spread of the bullet beam is d  2l tan  2l sin.

  h  h ; d  2l tan  2l sin  2l 1.22  2l 1.22h 
p mv D Dmv

    l 
Dmvd  3.0 103 m 3.0 103 kg 150 m s0.010 m  8.31027 m
2.44h
2.44 6.631034 Js

This is almost 1012 light years.

3. The uncertainty in the velocity is given. Use Eq. 38-1 to find the uncertainty in the position.

   x
     1.055  1034 Js 5.3  1011 m
p mv
1.67 1027 kg 1200 m s 

4. The minimum uncertainty in the energy is found from Eq. 38-2.

   E
   1.055  1034 Js  1.055  1026 J  1eV J   6.59 108 eV  107 eV
t 1  108 s  1.60 1019 

5. The uncertainty in position is given. Use Eq. 38-1 to find the uncertainty in the momentum.

    p
 mv    v    1.055  1034 Js  4454 m s  4500 m s
x mx 9.111031kg 2.6 108 m

6. The uncertainty in the energy is found from the lifetime and the uncertainty principle.

E   h ;E  hv  hc
 t 2t 

h 500 109 m
3.00 108 m s 10 109s
  E 2t    2  2.65 108  3108
hc 2 ct
E


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486

Chapter 38 Quantum Mechanics

E  hc  dE   hc d  E   hc    E   E   
 2 2  E 

The wavelength uncertainty is the absolute value of this expression, and so   3108


7. The uncertainty in the energy is found from the lifetime and the uncertainty principle.

   E
   1.055  1034 Js  1eV J   5.49  1011 eV
t 12 106s  1.60 1019 

E  5.49 1011eV  1.0 1014
E 5500 eV

8. (a) We find the wavelength from Eq. 37-7.

 
 h  h  6.631034 Js s  3.11034 m
p mv
0.012 kg180m

(b) Use Eq. 38-1 to find the uncertainty in momentum

 py
   1.055  1034 Js  1.6 1032 kgm s
y
 0.0065 m 

9. The uncertainty in the position is found from the uncertainty in the velocity and Eq. 38-1.

    xelectron
     1.055 1034 Js  1.4 103 m
p mv
9.111031kg 95m s 8.5 104

   xbaseball
     1.055 1034 Js  9.3 1033 m
p mv
0.14 kg95m s 8.5104

 x melectron

 x  m baseball
baseball 0.14 kg  1.5 1029
electron 9.111031 kg

The uncertainty for the electron is greater by a factor of 1.5  1029.

10. We find the uncertainty in the energy of the muon from Eq. 38-2, and then find the uncertainty in the

mass.

E   ; E  mc2 
t

  m
   1.055  1034 Js   4.7955  1029 J  1eV   3.00  1010 eV c2
c2t c2 2.20  106 s  c2   1.60 1019 J 

11. We find the uncertainty in the energy of the free neutron from Eq. 38-2, and then the mass

uncertainty from Eq. 36-12. We assume the lifetime of the neutron is good to two significant

figures. The current experimental lifetime of the neutron is 886 seconds, so the 900 second value is

certainly good to at least 2 significant figures.

   E
  ; E  mc2  m    1.055  1034 Js  1.3 1054 kg
t c2t
3.00 108 m s 2 900s

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487

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

12. Use the radius as the uncertainty in position for the electron. We find the uncertainty in the

momentum from Eq. 38-1, and then find the energy associated with that momentum from Eq. 36-13.

   p
   1.055  1034 Js  1.055  1019 kgm s.
x 1.0  1015 m

If we assume that the lowest value for the momentum is the least uncertainty, we can estimate the

lowest possible energy.

1/ 2 2 c2 1/ 2
   E  m02c4 
p2c2  m02c4   p 


 2 3.00  108 m s 2  9.11  1031 kg 2 4 1/ 2
        
1.055  1019 kgm s 3.00  108 m s

 3.175  1011 J  1MeV   200 MeV
 1.60 1013J 

13. (a) The minimum uncertainty in the energy is found from Eq. 38-2.

   E
   1.055  1034 Js  1.055  1026 J  1eV J   6.59 108 eV  107 eV
t 1  108 s  1.60 1019 

(b) The transition energy can be found from Eq. 37-14b. Z = 1 for hydrogen.

En   13.6eV Z2  E2  E1    13.6 eV  12    13.6 eV  12   10.2 eV
n2  22   12 
  

E  6.59 108eV  6.46 109  108
E2  E1 10.2 eV

(c) The wavelength is given by Eq. 37-3.

E  hv  hc 


  
 hc  6.63 1034 Js 3.00 108 m s  1.22 107 m  122 nm  100 nm
E
10.2 eV   1.60 1019 J 
 eV 
 

Take the derivative of the above relationship to find .

  hc  d   hc dE     hc E   E 
E E2 E2 E

  E
  E  122 nm 6.46 109  7.88 107 nm  106nm

14. We assume the electron is non-relativistic. The momentum is calculated from the kinetic energy,
and the position uncertainty from the momentum uncertainty, Eq. 38-1. Since the kinetic energy is

known to 1.00%, we have K K  1.00 102.

p 2mK ; dp  2m  2 1   2mK  p  2mK K  1 2mK K
dK  K  2K 2K 2 K

 x
     1.055  1034 Js
p
K 2 9.11  1031 kg 3.50 keV 1.60  1016 J keV
     2 K
1 2mK 1 1.00 102
2

 6.61  1010 m

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488

Chapter 38 Quantum Mechanics

15. Let us assume that the electron has an initial x momentum px , so that it has a wavelength of
  h px . The maxima of the double-slit interference pattern occur at locations satisfying Eq. 34-
2a, d sin  m, m  0,1, 2,. If the angles are small, then we replace sin by  , and so the
maxima are given by   m  d . The angular separation of the maxima is then    d , and the

angular separation between a maximum and the adjacent minimum is    2d . The separation of

a maximum and the adjacent minimum on the screen is then yscreen  l 2d , where l is the
distance from the slits to the detection screen. This means that many electrons hit the screen at a
maximum position, and very few electrons hit the screen a distance l 2d to either side of that

maximum position.

If the particular slit that an electron passes through is known, then y for the electrons at the

location of the slits is d 2. The uncertainty principle says py      h . We assume
yslits d
1 d
2

that py for the electron must be at least that big. Because of this uncertainty in y momentum, the

py h
px
electron has an uncertainty in its location on the screen, as yscreen   yscreen  l d  l .
l h d



Since this is about the same size as the separation between maxima and minima, the interference

pattern will be “destroyed.” The electrons will not be grouped near the maxima locations. They will

instead be “spread out” on the screen, and no interference pattern will be visible.

16. We are given that 1  x,t  and 2  x,t  are solutions to the Schrödinger equation. Substitute the

function A1  x,t   B2  x,t  into the Schrödinger equation.

 2 2  A1  B2   U  x  A1  B2    2 2  A1   2 2  B2   U  x  A1  U  x  B2
2m x 2 2m 2m
x2 x2

  A 2  2 1  B 2 22  AU  x 1  BU  x 2
2m x2 2m x2

 A  2 21 U  x  1   B   2 22 U  x   2 
 2m x2   2m x2 
  

 A i 1   B  i  2   i   A1  B2 
t   t  t

So, since 2 2  A1  B 2   U  x   A1  B 2   i   A1  B 2  , the combination
 2m x 2 t

A1  x,t   B2  x,t  is also a solution to the time-dependent Schrödinger equation.

17. (a) Substitute   x,t   Aeikxt into both sides of the time-dependent Schrödinger equation, Eq.

38-7, and compare the functional form of the results.

2 2 U0  2 2 Aeikxt  U0 Aeikxt   2k2  U  Aei  kx t 
 2m x2  2m x2  
 2m 0 

 Aeikx t    Aeikxt
t t
i  i

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

Both sides of the equation give a result of constant Aeikxt, and so t   x,t   Aeikxt is a

valid solution, if the constants are equal.

Now repeat the process for   x,t   Acoskx  t .

 2 2 U0   2 2 A cos  kx  t  U0 A cos  kx  t 
2m x2 2m
x2

  2k 2 U0  A cos  kx  t
 2m 
 

i   i A cos  kx  t   i Asin kx  t 
t
t

Because coskx  t   sin kx  t for arbitrary values of x and t,   x,t   Acoskx  t  is

NOT a valid solution.

Now repeat the process for   x,t   Asin kx  t .

 2 2 U0   2 2 Asin kx  t  U0 Asin kx  t 
2m x2 2m
x2

  2k 2 U0  Asin kx  t
 2m 
 

i   i Asin kx  t   i Acoskx  t 
t
t

Because coskx  t   sin kx  t for arbitrary values of x and t,   x,t   Asin kx  t  is

NOT a valid solution.

(b) Conservation of energy gives the following result.

E  K U  p2 U0 ; p h  hk  k    2k 2 U0
2m  2 2m

We equate the two results from the valid solution.

 2k2  U  Aei  kx t    Aeikxt  2k 2 U0  
  2m
 2m 0 

The expressions are the same.

18. The wave function is given in the form   x  Asin kx.

(a)   2  2  3.142 1010 m  3.11010 m
k 2.0 1010 m1

(b) p h  6.631034 Js  2.110 1024 kgm s 2.11024 kgm s
 3.142 1010 m

(c) v  p  2.110 1024 kgm s  2.3106 m s
m 9.111031 kg

2.110 1024 kgm s 2
2 9.111031kg
     (d)p2
K  2m  1  15eV
1.60 1019 J eV

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490

Chapter 38 Quantum Mechanics

19. The general expression for the wave function of a free particle is given by Eq. 38-3a. The particles

are not relativistic.

    (a)
k  2  2 p  mv  9.111031kg 3.0 105 m s  2.6 109 m1
 h  1.055  1034 Js

     Asin  2.6 109 m1 x  B cos  2.6 109 m1 x

    (b)
k  2  2 p  mv  1.67 1027 kg 3.0 105 m s  4.7 1012 m1
 h  1.055  1034 Js

     Asin  4.7 1012 m1 x  B cos  4.7 1012 m1 x

20. This is similar to the analysis done in Chapter 16 Section 6 for beats. Referring to Figure 16-17, we
see the distance from one node to the next can be considered a wave packet. We add the two wave
functions, employ the trigonometric identity for the sine of a sum of two angles, and then find the

distance between nodes. The wave numbers are related to the wavelengths by k  2 . Since


1  2, it is also true that k1  k2  1  kavg. We define k  k1  k2.

2
and so k1  k2

 1  2  Asin k1x  Asin k2 x  Asin k1x  sin k2x  2 Asin  1  k1  k2  x  cos  1  k1  k2  x
2 2

 2 Asin kavg x cos  1  k  x 
2

The sum function will take on a value of 0 if 1  k  x  n  1  , n  0,1,2. The distance between
2 2

these nodal locations is found as follows.

x  2 n  1   2n 1 , n  0,1,2.  x  2n 1 1  2n 1  2
2 k
k k k
k

Now use the de Broglie relationship between wavelength and momentum.

p  h  k  k  p ; x  2  2   xp  h
  k p

21. The minimum speed corresponds to the lowest energy state. The energy is given by Eq. 38-13.

h2 mvm2 in h 6.631034 Js  1.8 106 m s
8ml2 2ml 9.111031kg 0.20 109 m
  Emin  1  vmin   2
2

22. We assume the particle is not relativistic. The energy levels are given by Eq. 38-13, and the wave
functions are given by Eq. 38-14.

En  h2n2  p2  pn  hn ; n  2 sin  n x   n  kn  2 
8ml2 2m 2l l  l  l n

n  2l  h , which is the de Broglie wavelength
n pn

23. (a) The longest wavelength photon will be the photon with the lowest frequency, and thus the

lowest energy. The difference between energy levels increases with high states, so the lowest

energy transition is from n = 2 to n = 1. The energy levels are given by Eq. 38-13.

En  n2 h2  n2E1
8ml2

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491

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

    
 c  hc  hc  hc E1  6.631034 Js 3.00 108 m s  4.6 108 m
 E E2  E1 4E1 
39.0eV 1.60 1019 J eV

(b) We use the ground state energy and Eq. 38-13.

E1  h2 
8ml2

l h   6.631034 Js  2.0 1010 m

   8mE1 8 9.111031kg 9.0eV 1.60 1019 J eV

24. The energy levels for a particle in a rigid box are given by Eq. 38-13. Use that equation, evaluated

for n = 4 and n = 1, to calculate the width of the box. We also use Eq. 37-3.

 E hc h2
 hv    E4  E1  8ml2 42  12 

   l 
15h  15 6.631034 Js 340 109 m  1.2 109 m
8mc 8 9.111031kg 3.00 108 m s

25. We assume the particle is not relativistic. The energy levels give the kinetic energy of the particles

in the box.

E1  h2  p12  p12  h2  p1  h  p 2 p1  h
8ml2 2m 4l2 2l l

xp  l h  h
l

This is consistent with the uncertainty principle.

26. The longest wavelength photon will be the photon with the lowest frequency, and thus the

lowest energy. The difference between energy levels increases with high states, so the lowest energy

transition is from n = 2 to n = 1. The energy levels are given by Eq. 38-13.

 E hc h2
 hv    E2  E1  8ml2 22  12 

  l 
3h  3 6.631034 Js 610 109 m  7.4 1010 m
8mc 8 9.111031kg 3.00 108 m s

27. The energy levels for a particle in an infinite potential well are given by Eq. 38-13. The wave

functions are given by Eq. 38-14 with A  2 .
l

6.63 1034 Js 2
2.0 109 m 2 1.60 1019 J eV
    E1h2   9.424 102 eV  0.094 eV
8ml2 8 9.11  1031 kg

 E2  22 E1  4 9.424 102 eV  0.38eV
 E3  32 E1  9 9.424 102 eV  0.85eV
 E4  42 E1  16 9.424 102 eV  1.5eV

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492

Chapter 38 Quantum Mechanics

2  n  2   
l  l  2.0 nm  2.0 nm 
   n   
sin x ; 1  sin x  1.0 nm1/ 2 sin  1.6 nm1 x

        2  1.0 nm1/2 sin  3.1nm1 x ;  3  1.0 nm1/ 2 sin  4.7 nm1 x ;

    4  1.0 nm1/2 sin  6.3nm1 x

28. The wave functions for an infinite square well are given by Eq. 38-14.

n  Asin  n x  ; n 2  A2 sin 2  n x 
 l   l 

(a) The maxima occur at locations where  n 2  A2.

sin 2  n x  1  n x  m  1   , m  0,1,2,n 1 
 l  l 2

xmax   2m  1  l, m  0,1, 2, n 1
 2n 

The values of m are limited because x  l.
(b) The minima occur at locations where  n 2  0.

sin2  n x   0  n x  m , m  0,1,2,n  xmin  m l, m  0,1, 2,n
 l  l n

29. The energy levels for a particle in a rigid box are given by Eq. 38-13. We substitute the appropriate

mass in for each part of the problem.

(a) For an electron we have the following:

6.63 1034 Js 2
2.0 1014 m 2 1.60 1013 J MeV  940 MeV
    Eh2n2 
8ml2 8 9.11  1031 kg

(b) For a neutron we have the following:

6.63 1034 Js 2
2.0 1014 m 2 1.60 1013 J MeV
    Eh2n2   0.51MeV
8ml2 8 1.675 1027 kg

(c) For a proton we have the following:

6.63 1034 Js 2
2.0 1014 m 2 1.60 1013 J MeV
    Eh2n2   0.51MeV
8ml2 8 1.6731027 kg

30. The energy released is calculated by Eq. 38-13, with n = 2 for the initial state and n = 1 for the final

state.

3 6.63 1034 Js 2
1.0 1014 m 2 1.60 1013 J
        E  E2  E1 22  11 h2 
8ml2 8 1.67 1027 kg MeV

 6.17 MeV

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493

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

31. (a) The ground state energy is given by Eq. 38-13 with n = 1.

6.631034 Js 2 1

1.66 1027 kg u 4.0 103 m 2
      E1h2n2
8ml2  1.60 1019 J eV

n1 832 u

 4.0411019 eV  4.0 1019 eV

(b) We equate the thermal energy expression to Eq. 38-13 in order to find the quantum number.

1 kT  h2n2 
2 8ml2

       n  2
kTm l  2 1.38 1023 J K 300 K32 u 1.66 1027 kg u 4.0 103 m
h 6.63 1034 Js

 1.789 108  2 108

(c) Use Eq. 38-13 with a large-n approximation.

E  En1  En  h2 n  12  n 2   h2 2n 1  2n h2  2nE1
8ml2  8ml2 8ml
2

   2 1.789 108 4.0411019 eV  1.4 1010 eV

32. Because the wave function is normalized, the probability is found as in Example 38-8. Change the

variable to   n x , and then d  n dx.

ll

x2 2 2 x2 n x dx 2 l n x2 l 2 0.65n
x1 l l l n n x1 n
sin2 sin2  sin2 

x1 l 0.35n
   Pn dx   d  d

 2 0.65n
n 1   1 sin 2 0.35n
2 4

(a) For the n = 1 state we have the following:

P  2  1   1sin 2 0.65  2  1  0.30   1  sin 1.3  sin 0.70   0.5575  0.56
 2 0.35  2 4
4

(b) For the n = 5 state we have the following:

P  2 1   1sin 2 3.25n  2  1 1.5   1  sin 6.5  sin 3.5   0.2363  0.24
5 2 1.75n 5 2 4
4

(c) For the n = 20 state we have the following:

P  2 1   1 sin 2 13  1  1 6  1  sin 26  sin14   0.30
20 2 7 10 2 4
4

(d) The classical prediction would be that the particle has an equal probability of being at any

location, so the probability of being in the given range is P  0.65nm  0.35nm  0.30. We see
1.00 nm

that the probabilities approach the classical value for large n.

33. Consider Figure 38-9, copied here. To consider the problem with

the boundaries shifted, we would not expect any kind of physics to

change. So we expect the same wave functions in terms of their

actual shape, and we expect the same energies if all that is done is to

change the labeling of the walls to x   1 l and x  1 l. The
2 2

mathematical descriptions of the wave functions would change

because of the change of coordinates. All we should have to do is

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494

Chapter 38 Quantum Mechanics

shift the origin of coordinates to the right by 1 l. Thus we might expect the following wave
2

functions and energies.

n  2 sin  n x  
l  l 

n  2 sin  n x  1 l  2 sin  n x  1 n 
l  l 2 l  l 2 

n 1: 1  2 sin   x  1    2 cos   x  ; E1  h2
l  l 2  l  l  8ml2

n  2: 2  2 sin  2 x      2 sin  2 x  ; E2  4h2
l  l  l  l  8ml2

n  3: 3  2 sin  3 x  3    2 sin  3 x 1      2 sin  3 x  1  
l  l 2  l  l 2  l  l 2 

 2 cos  3 x  ; E3  9h2
l  l  8ml2

n  4: 4  2 sin  4 x  2   2 sin  4  ; E4  16h2
l  l  l  l  8ml2

For any higher orders, we simply add another 2 of phase to the arguments of the above functions.
They can be summarized as follows.

n odd:    1n 1 / 2  2 cos  n x  , En  n2h2
 l  l  8ml2

n even:    1n / 2  2 sin  n x  , En  n2h2
 l  l  8ml2

Of course, this is not a “solution” in the sense that we have not derived these solutions from the

Schrödinger equation. We now show a solution that arises from solving the Schrödinger equation.

We follow the development as given in Section 38-8.

As suggested, let   x  Asin kx   . For a region where U  x  0, k  2mE (Eq. 38-11a).
2

The boundary conditions are   1 l  A sin   1 k l     0 and  1 l  Asin  1 k l     0. To
2 2 2 2

guarantee the boundary conditions, we must have the following:

A sin   1 k l     0   1 kl    m ; Asin  1 k l     0  1 kl    n
2 2 2 2

Both n and m are integers. Add these two results, and subtract the two results, to get two new

expressions.

1 kl   n    1 n  m ; k  1n  m
2 2
l
 1 kl   m
2

2k 2 2 2 n  m2 h 2 n  m 2
2m
So again we have an energy quantization, with E   2ml2  8ml 2 . Note that

m  n is not allowed, because this leads to k  0,   n , and   x  0.

Next we normalize the wave functions. We use an indefinite integral from Appendix B-4.

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495

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

  x  Asin kx    Asin 1 n  m x  1 n  m 
 l 2 

  1 l  1 dx
2  l
 2
dx  A2 sin2  n  m x  1 n  m 1
 2

 1 l
2

let   1n  m x  1 n  m  dx   n l d
2
l  m

x  1 l    n ; x   1 l    m
2 2

A2l n A2l A2 l n  m A2 l
2
n  m sin2  d n  m n  m 2
  2 dx    1   1sin 2 n 
m 2 m
1  4



A2  2  A  2
ll

This is the same as in Section 38-8. Finally, let us examine a few allowed cases.

n  1, m  0: k  ,   1   1,0  2 sin   x  1    2 cos   x  ; E1,0  h2
l 2 l  l 2  l  l  8ml2

n  2, m  0 : k  2 ,      2,0  2 sin  2 x     2 sin  2 x  ; E2,0  4h2
l l  l  l  l  8ml2

n  3, m 0: k  3 ,   3    3,0  2 sin  3 x  3    2 sin  3 x  1   
l 2 l  l 2  l  l 2 

 2 sin  3 x  1     2 cos  3 x  ; E3,0  9h2
l  l 2  l  l  8ml2

n  4, m  0 : k  4 ,   2   4,0  2 sin  4 x  2   2 sin  4  ; E4,0  16h2
l l  l  l  l  8ml2

These are the same results as those obtained in the less formal method. Other combinations of m and
n would give essentially these same results for the lowest four energies and the associated wave
functions. For example, consider n  4, m  1.

n  4, m 1: k  3 ,   5    4,1  2 sin  3 x 5    2 sin  3 x  1  
l 2 l  l 2  l  l 2 

 2 cos  4  ; E4,1  9h2
l  l  8ml2

We see that  4,1   3,0 and that both states have the same energy. Since the only difference in the

wave functions is the algebraic sign, any physical measurement predictions, which depend on the
absolute square of the wave function, would be the same.

34. We choose the zero of potential energy to be at the bottom of the well. Thus in free space, outside

the well, the potential is U0  56eV. Thus the total energy of the electron is E  K  U0  236 eV.
(a) In free space, the kinetic energy of the particle is 180 eV. Use that to find the momentum and

then the wavelength.

K  p2  p 2mK  h 
2m 

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496

Chapter 38 Quantum Mechanics

     
h  2 6.63 1034 Js eV 1/ 2  9.15 1011m
2mK
9.111031kg 180eV 1.60 1019 J

(b) Over the well, the kinetic energy is 236 eV.

     
h  2 6.63 1034 Js eV 1/ 2  7.99 1011m
2mK
9.111031kg 236eV 1.60 1019 J 

(c) The diagram is qualitatively the same as Figure 38-14,

reproduced here. Notice that the wavelength is longer when

the particle is not over the well, and shorter when the

particle is over the well.

35. We pattern our answer after Figure 38-13.

36. (a) We assume that the lowest three states are bound in the well, so that 3
E  U0. See the diagrams for the proposed wave functions. Note
that, in the well, the wave functions are similar to those for the 2
infinite well. Outside the well, for x  l, the wave functions are 1

drawn with an exponential decay, similar to the right side of Figure x0
38-13.

(b) In the region x  0,   0 .

In the well, with 0  x  l, the wave function is similar to that of a
free particle or a particle in an infinite potential well, since U = 0.

Thus   Asin kx  B sin kx , where k  2mE . xl


In the region x  l,   DeGx , where G  2m U0  E  .



37. We will consider the “left” wall of the square well, using Figure 38-12,l and assume that our answer
is applicable at either wall due to the symmetry of the potential well. As in Section 38-9, let
  CeGx for x  0, with G given in Eq. 38-16. Since the wave function must be continuous,

  x  0  C. The energy of the electron is to be its ground state energy, approximated by Eq. 38-

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497

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

13 for the infinite well. If that energy is much less than the depth of the well, our approximation will

be reasonable. We want to find the distance x for which   x  0.010 0.

6.63 1034 Js 2
0.16 109 m 2 1.60 1019 J eV
    Eh2  14.73eV  U0
8ml2 8 9.11  1031 kg

 x  CeGx  eGx  0.010  x  ln 0.010   ln 0.010
 0 C 2m U0  E 
G

 x 
1.055  1034 Js ln 0.010  2.0 1011m

   2 9.111031kg 2000eV 14.73eV 1.60 1019 J eV

The wave function will be 1.0% of its value at the walls at a distance of 2.0 1011m  0.020 nm
from the walls.

38. We use Eqs. 38-17a and 38-17b.

T  e2Gl  G   ln T ; G 2m U0  E   ln T 
2l 2l
2

    E 2 2
2 ln T 2 1.055  1034 Js  ln 0.00050  1
 U0  2m   2l   14eV  2 9.111031 kg     1019 
 0.85 109 m  
2  1.60 J eV 

 14eV  0.76eV  13.24eV  13eV

39. We use Eqs. 38-17a and 38-17b to solve for the particle’s energy.

T  e2Gl  G   ln T ; G 2m U0  E   ln T 
2l 2l
2

    E 2 2
2 ln T 2 1.055  1034 Js ln 0.010  1
 U0  2m   2l  18eV  2 9.111031kg    
   0.55 109 m   
 2  1.60  1019 J eV 

 17.33eV  17eV

40. We use Eqs. 38-17a and 38-17b to solve for the transmission coefficient, which can be interpreted in
terms of probability. For the mass of the helium nucleus, we take the mass of 2 protons and 2
neutrons, ignoring the (small) binding energy.
Proton:

    G 
2m U0  E  2 1.67 1027 kg 20.0 MeV 1.60 1013 J MeV  9.799 1014 m1

 1.055  1034 Js

   2Gl  2 9.799 1014 m1 3.6 1015 m  7.056 ; Tproton  e2Gl  e7.056  8.6 104
   Helium: Mass = 2mproton  2mneutron  2 1.6731027 kg  2 1.675 1027 kg  6.70 1027 kg.

    G 
2m U0  E  2 6.70 1027 kg 20.0 MeV 1.60 1013 J MeV  1.963 1015 m1

 1.055  1034 Js

   2Gl  2 1.963 1015 m1 3.6 1015 m  14.112 ; THe  e2Gl  e14.112  7.4 107

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498

Chapter 38 Quantum Mechanics

41. (a) The probability of the electron passing through the barrier is given by Eqs. 38-17a and 38-17b.

T  e2l 2mU0 E 



       2l
2m U0  E  2 0.25 109 m 2 9.111031kg 1.2 eV 1.60 1019 J eV  2.803

 1.055  1034 Js

T  e2.803  6.063 102  6.1%

(b) The probability of reflecting is the probability of NOT tunneling, and so is 93.9% .

42. The transmitted current is caused by protons that tunnel through the barrier. Since current is directly
proportional to the number of charges moving, the transmitted current is the incident current times
the transmission coefficient. We use Eqs. 38-17a and 38-17b.

T  e2l 2mU0 E 



       2l
2m U0  E  2 2.8 1013 m 2 1.67 1027 kg 1.0 MeV 1.60 1013 J eV  122.86

 1.055  1034 Js

T  e122.86  logT  122.86log e  53.357  T  1053.357  4.4 1054

 I  I0T  1.0 mA 4.4 1054  4.4 1054 mA

43. The transmission coefficient is given by Eqs. 38-17a and 38-17b.

(a) The barrier height is now 1.02 70eV  71.4 eV.

       2l
2m U0  E  2 0.10 109 m 2 9.111031kg 21.4 eV 1.60 1019 J eV

 1.055  1034 Js

 4.735

T  e2l 2mU E   e4.735  8.782 103 ; T  8.782 103  88  12% decrease
T0 0.010


(b) The barrier width is now 1.020.10 nm  0.102 nm.

       2l
2m U0  E  2 0.102 109 m 2 9.111031 kg 20eV 1.60 1019 J eV

 1.055  1034 Js

 4.669

T  e2l 2mU0 E   e4.669  9.382 103 ; T  9.382 103  93.8  6.2% decrease
T0 0.010


44. We assume that the wave function inside the barrier is given by a decaying exponential, so

  x   AeGx.

 T l 2 2
 x  0 2 AeGl
  x   A2  e2Gl

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499

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

45. (a) We assume that the alpha particle is at the outer edge of the nucleus. The potential energy is

electrostatic potential energy, and is found from Eq. 23-10.

2 90 1.60 1019 C 2  1MeV 
     Usurface
 1 Q1Q2  8.988 109 N  m2 C2 8 1015 m  1.60 1013J 
4  rsurface of
nucleus

 32.36 MeV  32 MeV

(b) The kinetic energy of the free alpha particle is also its
total energy. Since the free alpha has 4 MeV, by
conservation of energy the alpha particle had 4 MeV of
potential energy at the exit from the barrier. See the
diagram, a copy of Figure 38-17, modified to show U =
0 inside the barrier, and stated in part (c).

 41  Q1Q2  41  Q1Q2 rsurface of
rexit from rsurface of
 Uexit nucleus rexit from

rexit from barrier
barrier nucleus barrier

rsurface of 

nucleus

 Usurface
rexit from

barrier

U surface 32.36 MeV
U exit 4 MeV
 rexit from

barrier
 rsurface of  8fm  64.72 fm

nucleus

r  rexit from  rsurface of  64.72 fm  8fm  56.72 fm  57 fm
barrier nucleus

(c) We now model the barrier as being rectangular, with a

width of rbarrier  1  56.72 fm   18.9 fm. The barrier exists
3

at both boundaries of the nucleus, if we imagine the

nucleus as 1-dimensional. See the diagram (not to scale).

We calculate the speed of the alpha particle and use that “out” “in” “out”
to find the frequency of collision with the barrier.

E  K  1 mv2 
2

   v 
2E  2 4 MeV 1.60 1013 J MeV 19 fm 19 fm
m
4 1.67 1027 kg 16 fm

 1.38 107 m s  1.4 107 m s

Note that the speed of the alpha is less than 5% of the speed of light, so we can treat the alpha

without using relativistic concepts. The time between collisions is the diameter of the nucleus

(16 fm) divided by the speed of the alpha particles. The frequency of collision is the reciprocal

of the time between collisions.

f  v  1.38 107 m s  8.625 1020 collisions s 8.6 1020 collisions s
d 16 1015 m

If we multiply this collision frequency times the probability of tunneling, T, then we will have

an estimate of “effective” collisions/s, or in other words, the decays/s. The reciprocal of this

effective frequency is an estimate of the time the alpha spends inside the nucleus – the life of

the uranium nucleus.

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500