BW 2 Ch. 28 CQ 2-5, 7-17, 20, 22, 23, 26 – 35.
Q28.1. Reason: As the crystal is compressed, the spacing d between planes of atoms decreases. The formula giving
the Bragg condition is Equation 28.1: 2d cosm  m. As d decreases, cos m must increase. But cos increases as 
decreases. So 1 decreases.
Assess: It seems reasonable that as the atoms get closer together 1 decreases.
Q28.2. Reason: Electrons leave the photoelectron emitter with a certain amount of kinetic energy (depending on the
energy of the photon that was responsible for their emission). When electrons leave the emitter they are in an electric field
(between the emitter and the collector) and hence also have an electric potential energy. Due to the geometry of the
emitter, the electrons are generally headed toward the collector, which may be at a positive or negative electric potential
with respect to the emitter. If the collector is at a positive electric potential with respect to the emitter, the electrons will
experience a positive acceleration, a gain in kinetic energy, and a loss in electric potential energy as they travel to the
collector. If the collector is at a negative electric potential with respect to the emitter, the electrons will experience a
negative acceleration, a loss in kinetic energy, and a gain in electric potential energy as they travel to the collector. Since
the only force acting on the electron is a conservative force, energy is conserved and K = –U.
Let’s agree to give the collector a negative electric potential with respect to the emitter. Let’s also agree to adjust this
negative electric potential until no electrons make it to the collector. To do this we can monitor the collector current and
when the collector current is zero, no electrons make it to the collector. This means that we have momentarily stopped the
electrons and at that instant all their kinetic energy has been traded for electric potential energy. Since the electrons have
been stopped, we can say K  KF  Ki  0  Ki   Ki (the minus tells us the change in kinetic energy is negative—i.e., it
is decreasing).
If we agree that the emitter is at zero electric potential energy, then the change in electric potential energy is
U  U F  Ui  U F  0  U F. This electric potential energy is related to the electric potential and the charge on the electron
by U F  eVstop.
Combining these expressions we have Ki  eVstop. Clearly the stopping potential is a measure of the kinetic energy of the
photo-electrons.
Assess: Since the electric potential of the collector has been adjusted to stop all the electrons, the stopping potential is
correct for only the most energetic electrons. So we should say Ki max  eVstop. This is consistent with Equation 28.7 in the
text.
Q28.3. Reason: (a) For all materials that emit photo-electrons, the electrons need a minimum amount of energy in order
to be released (called the work function). Since a single photon gives all its energy to a single electron, photons must have
a minimum energy and hence frequency (called the threshold frequency).
(b) Increasing the light intensity increases the number of incident photons. Increasing the number of incident photons
increases the number of electrons emitted. Since the moving electrons amount to an electric current, the more electrons,
the more current.
(c) If the anode is positive, it attracts all electrons to the anode. A further increase in V does not cause any more
electrons to reach the anode and thus does not cause a further increase in the current I.
(d) As V becomes more negative, fewer electrons have the energy to reach the anode. Hence a decrease in current.
(e) The greater the light intensity, the greater the number of photons reaching the photoelectron emitting surface. Since a
single photon interacts with a single electron, the energy of the emitted electron depends on the energy of the single
incident photon it interacts with, not the number of photons—it can only interact with one photon.
Classical physics can explain some aspects of the photoelectric effect, but, as mentioned in the text, not all of the aspects
we see. Classically, we do not expect a sharp threshold frequency f 0. We also do not expect the current to appear instantly
when the light is turned on. Classically, we expect that as the light intensity is increased, so is the energy absorbed by the
surface electrons. Hence, the number of photoelectrons emitted, or the photocurrent, is expected to increase
proportionately with light intensity. Classically, we expect the more intense light will give more energy to the
photoelectrons and hence their stopping potential will be increases—this does not happen. Classical physics cannot
explain A and E.
Assess: Only the quantum hypothesis can account for all the characteristics of the photoelectric effect. Einstein’s
explanation of the photoelectric effect won him a well-deserved Nobel Prize.
Q28.4. Reason: According to classical physics, if you shine light on the photoelectron emitting surface it will collect
energy and when it has collected enough energy it will emit an electron. The frequency of the light doesn’t matter. It may
take a different amount of time to absorb the required amount of energy but there is nothing in classical physics to suggest
that there is a threshold frequency. As a result, according to classical physics a plot of photoelectron current as a function
of incident light frequency will appear as shown in the figure.
Assess: Classical theory also predicts that it will take a long time for the surface to collect enough energy to emit an
electron. In the experiment the emission of photoelectrons is essentially instantaneous (once light above the threshold
frequency is incident on the photoelectron emitter).
Q28.5. Reason: (a) A positive anode attracts all of the electrons to the anode. Once all of the electrons reach the
anode, a further increase in V does not cause any further increase in the current.
(b) If an electron has enough kinetic energy when it leaves the cathode it might just reach the anode. As V decreases,
fewer electrons are able to reach the anode until V  Vstop when no electrons reach the anode and the current stops.
(c) Electrons are still emitted, but they are not attracted to the anode and are in fact repelled by it. The electrons that are
emitted do not have sufficient kinetic energy to reach the anode.
Assess: These explanations all fit with Einstein’s quantum hypothesis.
Q28.6. Reason: There is an electric field between the cathode and the anode. When electrons are emitted, the electric
field forces them to the anode.
Assess: This chapter is a good review of earlier concepts dealing with electric fields, electric potential, and electric
potential energy.
Q28.7. Reason: As light shines on the metal spacecraft surface, the photoelectric effect takes place and since electrons
are emitted, the metal and hence space craft obtain a positive charge.
Assess: A surface that is deficient in electrons, has a positive charge.
Q28.8. Reason: The kinetic energy of the emitted electrons is related to the energy of the incident photons and the
work function of the metal by K  Ephoton  Eo. The kinetic energy of the electron is related to its speed by K  mv 2 /2, and
the energy of the photon is related to its wavelength by Ephoton  hc/ . Combining these and solving for v we obtain
v  (2/m)[(hc/ )  Eo ]1/ 2.
Examining this function we see that if the wavelength is constant, then as the work function increases, the speed of the
electrons decreases. So electrons from metal 1 have lower speed.
Assess: It makes sense that if the electron requires more energy to get out of the metal, it will have less energy once it is
out.
Q28.9. Reason: (a) If the current is zero when V  1.0 V that means the frequency of the light is less than the
threshold frequency; f  f0. So no electrons are ejected in the first place. Doubling the intensity of the light means
doubling the number of photons hitting the metal, but the photons still aren’t energetic enough to eject electrons; that is,
the energy of the photons is less than the work function.
So no matter how many photons of this energy one shines on gold there won’t be a photocurrent.
(b) Since no electrons are emitted in the first place, then increasing V won’t make any difference either.
Assess: Instead of computing energy it is also possible to think in terms of frequency and compare f to f0 .
Q28.10. Reason: The energy and wavelength of a photon are related by E  hc/ . The power, rate of arrival of
photons, and the wavelength are related by R  N/t  P /(hc).
By examination of these relationships we see that since 3 (800 nm)  2 (600 nm)  1 (400 nm), then E1  E2  E3 and
N3  N2  N1.
Assess: This course will enhance your ability to examine a function and answer questions of this nature.
Q28.11. Reason: The light quanta are discrete packets of energy. The fundamental unit of light energy is the photon.
The energy is quantized.
Assess: This is equivalent to saying that energy is not continuously dividable. Many aspects of the universe appear to be
discrete (chunky) rather than continuous.
Q28.12. Reason: The intensity of a beam of photons may be determined by
I
Etotal NEphoton N (hf )  N  hc 


  

At
At
At
 t  A 
By investigating this function, we see that when the wavelength of the light does not change, the intensity of the light may
be increased by increasing the number of photons per second ( N /t ).
The correct choice is D.
Assess: Notice that we started with four different concepts (intensity, total energy of the beam, energy of each photon,
and the relationship between frequency and wavelength) and built the function needed to answer the question. This course
will help you develop this skill.
Q28.13. Reason: Since the investigator is measuring a current, we know the wavelength is below threshold and
photoelectrons are being emitted. For the initial situation, every photon striking the surface is emitting an electron with a
specific amount of energy. Since the photons are already below threshold wavelength, reducing the wavelength by a factor
of two will not affect the number of electrons emitted—but it will increase their kinetic energy. Since the number of
electrons emitted is not changed, the current will not be changed. Since the kinetic energy of the electrons is increased, the
stopping potential will be increased.
Assess: Any physical change that impacts the number of electrons emitted per unit time will impact the current. Any
physical change that impacts the kinetic energy of the emitted electrons will impact the stopping potential.
Q28.14. Reason: Equation 28.8,   h/(mv), shows that fast electrons have a shorter wavelength leading to less
diffraction spreading and better resolution.
Assess: This fact is used daily in electron microscopes. Fast electrons have a shorter wavelength than visible light so we
can resolve finer detail.
Q28.15. Reason: Both particles accelerate through the same potential difference and so have the same kinetic energy.
Combine Equation 28.8 with v  2K /m.

h
h


mv m 2K /m
h
2Km
This shows that the particle with the smaller mass corresponds to the larger wavelength. Thus, the electron has the larger
de Broglie wavelength.
Assess: Equation 28.8 shows that the wavelength is inversely proportional to the momentum.
Q28.16. Reason: The de Broglie wavelength and the speed of the associated particle are related by   h/p  h/(mv).
As the neutron travels upward, it will slow down and, according to the preceding expression, the associated de Broglie
wavelength will increase.
Assess: The answer is obtained by investigating the relationship between the two quantities of interest.
Q28.17. Reason: With a very weak electron beam we can send only one electron at a time through the apparatus and
still observe interference; we conclude that each electron goes through both slits. Thus D is correct.
Assess: Interference is a wavelike phenomenon so the electrons act like waves and go through both slits.
The other choices can be eliminated. For example, two electrons in proximity to each other do repel, but that doesn’t cause
interference; this eliminates A.
Q28.18. Reason: The momentum of an electron is related to the associated de Broglie wavelength by p  h/ . In the
worst-case scenario, the uncertainty in the momentum of an electron cannot be greater than the momentum of the electron,
hence p  p  h/. Actually this is a best-case scenario for the uncertainty in the position, since the uncertainty in the
position is inversely proportional to the uncertainty in the momentum. According to the uncertainty principle, the uncertainty
in the position is related to the uncertainty in the momentum by
x 
 px

h /(2)  2 m


 0.32 m
h /
2
2
The uncertainty in the position of the electron will be 0.32 m, which is less than the slit width (1 m) so the electron can
pass through the slit.
Assess: This problem allows us to conclude that a particle can get through a slit that is smaller than the wavelength of the
associated de Broglie wavelength.
Q28.19. Reason: The allowed energies of a particle in a box are given in Equation 28.13: En  (hn)2/(8mL2 ).
(a) From the equation we see that for the energies to be large L, the length of the box should be very small.
(b) Based on the answer to part (a), the system with the smallest dimension would be likely to have the largest energies;
that is, the proton in the nucleus.
Assess: While it is true that the mass of the proton is larger than the mass of the electron and m is also in the denominator
(which would argue the other way), the size difference is much greater than the mass difference, and the L is squared.
Q28.20. Reason: The wavelength of the matter wave is related to the length of the box by n  2L/n (n  1, 2, 3, …),
where n (the quantum number) is the number of half wavelengths in the box (in this case n  4). Knowing the length of
the box and the quantum number n, we can determine the wavelength, then the momentum, and then the speed. However
there is no information available to tell us the direction of motion of the particle.
Assess: Theory provides no method to determine the direction of the particle in the box.
Q28.21. Reason: The allowed energies of a particle in a box are given in Equation 28.13: En  (hn)2/(8mL2 ).
Solve that equation for L.
L
(hn)2
hn

8mEn
8mEn
We are given E1  2 eV and E2  50 eV.
La h(1)/ 8mE1 1 E2 1 50 eV
25




 2 5
Lb h(2)/ 8mE2 2 E1 2 2 eV
2
Assess: We see that the length of box a is 2.5 times the length of box b. Again, ratios are the way to go. So many things
(including units) cancel that way.
Q28.22. Reason: The de Broglie wavelength is related to the speed of the particle by  /h/p  h/mv. As the particle
slows, the wavelength will increase. The figure shows a qualitative sketch of this situation.
Assess: Since the speed and wavelength are inversely proportional, this sketch seems reasonable.
Q28.23. Reason: (a) It represents a changing speed because the wavelength is changing. The longer wavelength
corresponds to a slower speed.
(b) The shorter wavelength on the left means the particle is moving faster on the left; the longer wavelength on the right
means the particle is moving slower there.
Assess: This question reinforces the idea that faster particles have a shorter wavelength.
Q28.24. Reason: Energy of a photon is related to the wavelength by E  hc / . Energy of the electrons in the first
excited state are related to the length of the molecules in the cone by E1  h2 / (8mL2 ). Since the wavelength of red light is
greater, energy of the red light photons is less than energy of the blue light photons. Since a single photon interacts with a
single electron and gives it all its energy, electrons excited by red light photons have less energy that electrons excited by
blue light energy. As a result the length of the molecules in the red cones must be longer that the length of the molecules in
the blue cone.
Assess: An analysis of these equations shows that the greater the wavelength, the less the energy of the photon, the less
the energy of the electron, the longer the molecules containing the oscillating electrons.
Q28.25. Reason: The uncertainty principle states that xp  h / (2 ). The uncertainty in the momentum can’t be any
larger than the momentum. When we put people back together, the momentum is zero or nearly zero—this means the
uncertainty in the position is very large—people will be scrambled.
Assess: Movie makers (especially science fiction movie makers) need good technical advice in order to produce a correct
product.
Q28.26. Reason: The relationship between the wavelength and energy of light is
  hc /E  (6.63  1034 J s)(3.0  108 m/s)(1.0 eV/(1.6  1019 J))/(1.7 eV)  730 nm
The correct choice is B.
Assess: This is a reasonable wavelength for 1.7 eV photons.
Q28.27. Reason: When f increases so does E  hf . And so will Kmax  E  E0.
The correct choice is B.
Assess: To get more electrons (choice A) we would need to increase the intensity of the light.
Q28.28. Reason: The intensity of light is related to the number of photons incident per unit of time by the following
expression:
Etotal NEphoton N (hf )  N  hf 


   
At
At
At
 t  A 
This expression informs us that the intensity may be increased by increasing the number of photons. Since for every
photon an electron is emitted (assuming we are above the threshold frequency), when the number of incident photons is
increased the number of electrons emitted is increased. The correct choice is A.
Assess: More photons means more energy, more power, greater intensity.
I
Q28.29. Reason: From Einstein’s hypotheses about the photoelectric effect we see that each photon deposits all its
energy to one electron. The threshold frequency indicates that photons whose energy is too low ( f  f0, remember
E  hf ) do not have enough energy to eject an electron. The correct choice is A.
Assess: Even a very intense beam of such light will not eject any electrons because each photon has too little energy to
eject one electron.
Q28.30. Reason: The energy of each visible light photon is Ev  hc/v. The energy of each radio wave photon is
Er  hc/r. The number of radio wave photons required to provide the same energy as one visible light photon is
determined by
Nr  Ev /Er  (hc/v )/(hc/r )  r /v  1.0 m/(5.0  107 m)  2.0  106
The correct choice is D.
Assess: The total energy in the beam is the energy of each photon times the number of photons.
Q28.31. Reason: We must recall that E  hf , so the 100 MHz FM photons each have more energy than the
1000 kHz AM photons. Since both stations emit the same amount of energy each second, the AM station must emit more
photons per second since each AM photon has less energy than one of the FM photons. The correct choice is B.
Assess: The two radio stations are not charged by the photon on their electric or power bill, but by how much energy
they use, so their electric bills would be the same.
Q28.32. Reason: The energy of an electron accelerated through an electrical potential difference V is E  eV . If all this
energy is given to an x-ray, the wavelength of the x-ray is
  hc / E  hc / (eV )  (4.14 1015 eV  s)(3.0 108 m/s) / ((1e)(5.0 103 V))  0.25 nm
The correct response is A.
Assess: The x-ray cannot have more energy than the kinetic energy of the electron.
Q28.33. Reason: The power of the laser beam is related to the wavelength of the beam photons by
P
E NEphoton Nhc
N P
(5.0 103 W)(633 109 m)


 

 1.6  1016 / s
t
t
t
t
hc (6.63 1034 J  s)(3.0 108 m/s)
Assess: Since the power is small, we expect a small number of photon per second. While 10 16 may seem like a large
number to you, it is insignificant compared to the number of molecules in a mole.
Q28.34. Reason: The double-slit experiment for light was explained in Chapter 17. The fringes get closer together
when the wavelength is shorter. Such is the case in this question for electrons; doubling the speed of the electrons
increases their momentum and decreases their wavelength. The correct choice is A.
Assess: The de Broglie wavelength is inversely proportional to the momentum.
Q28.35. Reason: The relationship between energy and wavelength for a photo is E  hc/ . From the diagram we see
that EP  EQ  ER, therefore P  Q  R . The correct choice is A.
Assess: A photons wavelength is inversely proportional to its energy.