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Rutgers Analytical Physics 750:228, Spring 2016
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7. Photons, Electrons, Matter Waves, and Atoms
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7. Photons, Electrons, Matter Waves, and Atoms
Due: 11:59pm on Sunday, March 20, 2016
To understand how points are awarded, read the Grading Policy for this assignment.
Does the Photoelectric Effect Depend on the Properties of the Incident Light?
Description: Short conceptual problem on the effect of increasing light intensity and frequency in an experiment on photoelectric effect. Based on Young/Geller
Conceptual Analysis 28.1.
Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.
Part A
In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the metal are held constant. Assuming that the light
incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?
Check all that apply.
Hint 1. How to approach the problem
In the photoelectric effect, some an incident photon's energy is used to remove an electron from the metal, while the remainder becomes the electron's kinetic
energy. The energy required to remove the electron from the metal is called the work function of the metal, and it is a constant for a given material. The stopping
potential is the electric potential needed for an electron at rest to have as much electric potential energy as the kinetic energy of the electron ejected from the
metal. The stopping potential can be measured directly from the experiment.
If the intensity of the incident light is increased, but its frequency is kept constant, the number of photons striking the metal per unit time increases, but the energy
of individual incident photons remains constant. How does this affect the kinetic energy of the emitted electrons and their number per unit time?
Hint 2. Find the kinetic energy of the emitted electrons
In the photoelectric effect, some of the incident photon's energy is used to remove an electron from the metal, while the remainder becomes the electron's
kinetic energy . If the frequency of the incident photon is , and the work function of the metal is , which of the following expressions gives the maximum kinetic
energy of the emitted electron,
? In these expressions is Planck's constant.
Hint 1. Energy of a photon
The energy of a photon of frequency
is given by
,
where
is Planck's constant.
Hint 2. Work function
Recall that the work function of a metal is the energy required to remove an electron from that metal.
ANSWER:
As you found out, the maximum kinetic energy of the emitted electron does not depend on the intensity of the incident light. If the intensity of the incident light
is varied, how does this affect the maximum speed of the emitted electron?
ANSWER:
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The work function of the metal decreases.
The number of electrons emitted from the metal per second increases.
The maximum speed of the emitted electrons increases.
The stopping potential increases.
When the intensity of the incident light is increased, while its frequency is held constant, more photons will hit the metal per unit time and a larger number of electrons
will be emitted per unit time. Experimentally, this corresponds to higher maximum currents observed between the anode and cathode.
Part B
In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface
causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased?
Check all that apply.
Hint 1. How to approach the problem
In the photoelectric effect, some of the incident photon's energy is used to remove an electron from the metal, while the remainder becomes the electron's kinetic
energy. Therefore, if the photon's energy increases, so does the electron's kinetic energy because the energy required to remove an electron from the metal (the
work function) is a constant for any given material. Also, recall that the stopping potential can be used as an indirect measurement of the maximum kinetic energy
of the emitted electrons. Thus, if the electron's energy increases, the stopping potential increases as well.
ANSWER:
The work function of the metal increases.
The number of electrons emitted from the metal per second increases.
The maximum speed of the emitted electrons increases.
The stopping potential increases.
Alternative Exercise 38.52
Description: A hydrogen atom undergoes a transition from the n =5 to the n = 2 state. (a) What is the wavelength of the photon that is emitted? (b) What is the energy of
this photon. (c) If the angular momentum is conserved and if the Bohr model is used to...
A hydrogen atom undergoes a transition from the
5 to the
2 state.
Part A
What is the wavelength of the photon that is emitted?
ANSWER:
= 434
Part B
What is the energy of this photon.
ANSWER:
= 2.86
Part C
If the angular momentum is conserved and if the Bohr model is used to describe the atom, what must the angular momentum be of the photon that is emitted?
ANSWER:
= 3.17×10−34
Exercise 38.9
Description: When ultraviolet light with a wavelength of ## nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to
be ## eV. (a) What is the maximum kinetic energy of the photoelectrons when light of...
When ultraviolet light with a wavelength of 400.0
Typesetting math:
falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10
.
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Part A
What is the maximum kinetic energy of the photoelectrons when light of wavelength 285.0
falls on the same surface?
ANSWER:
= 2.35
=
Exercise 38.16
Description: X rays are produced in a tube operating at ## kV. After emerging from the tube, x rays with the minimum wavelength produced strike a target and undergo
Compton scattering through an angle of 45.0 degree(s). (a) What is the original x-ray...
rays are produced in a tube operating at 21.0
scattering through an angle of
.
. After emerging from the tube,
rays with the minimum wavelength produced strike a target and undergo Compton
Part A
What is the original -ray wavelength?
Express your answer with the appropriate units.
ANSWER:
= 59.2
=
Part B
What is the wavelength of the scattered
rays?
Express your answer with the appropriate units.
ANSWER:
= 59.9
=
Part C
What is the energy of the scattered
rays?
ANSWER:
= 20.7
=
Exercise 38.5
Description: A photon has momentum of magnitude p. (a) What is the energy of this photon? Give your answer in joules. (b) What is the energy of this photon? Give your
answer in electron volts. (c) What is the wavelength of this photon? (d) In what region of ...
A photon has momentum of magnitude
.
Part A
What is the energy of this photon? Give your answer in joules.
ANSWER:
=
= 2.45×10−19
Part B
What is the energy of this photon? Give your answer in electron volts.
ANSWER:
Typesetting math:
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= 1.53
=
Part C
What is the wavelength of this photon?
ANSWER:
= 8.11×10−7
=
Part D
In what region of the electromagnetic spectrum does it lie?
ANSWER:
gamma rays
radio waves
infrared
ultraviolet
visible light
Exercise 38.6
Description: The photoelectric threshold wavelength of a tungsten surface is 272 nm. (a) Calculate the maximum kinetic energy of the electrons ejected from this
tungsten surface by ultraviolet radiation of frequency 1.45 * 10^15 (Hz). Express the answer in...
The photoelectric threshold wavelength of a tungsten surface is 272
.
Part A
Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency
electron volts.
. Express the answer in
ANSWER:
= 1.44
Problem 38.21
Description: (a) An electron inside a hydrogen atom is confined to within a space of 0.110 nm. What is the minimum uncertainty in the electron's velocity? (hbar = 1.055 *
10^-34 (J * s), m_el = 9.11 * 10^-31 (kg))...
Part A
An electron inside a hydrogen atom is confined to within a space of 0.110 nm. What is the minimum uncertainty in the electron's velocity? (
)
,
ANSWER:
5.26 × 105 m/s
7.50 × 105 m/s
5.26 × 107 m/s
5.26 × 109 m/s
7.50 × 107 m/s
Exercise 38.13
Description: Protons are accelerated from rest by a potential difference of DeltaV and strike a metal target. (a) If a proton produces one photon on impact, what is the
minimum wavelength of the resulting x rays? (b) Find the minimum wavelength if ##-keV...
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Protons are accelerated from rest by a potential difference of 4.50
and strike a metal target.
Part A
If a proton produces one photon on impact, what is the minimum wavelength of the resulting x rays?
ANSWER:
= 2.76×10−10
=
Part B
Find the minimum wavelength if 4.50-
electrons are used instead?
ANSWER:
= 2.76×10−10
=
Part C
Why do x-ray tubes use electrons rather than protons to produce x rays?
ANSWER:
3721 Character(s) remaining
Electron beams are much more easily produced and
accelerated than proton beams.
Exercise 38.14
Description: (a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x-rays with a wavelength of
lambda? (b) What is the shortest wavelength produced in an x-ray tube operated at a voltage...
Part A
What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x-rays with a wavelength of 0.160
?
ANSWER:
=
= 7760
Also accepted:
= 7760,
= 7750, h*2.998*10^8/(lambda*1.60*10^-19) = 7760,
= 7760
Part B
What is the shortest wavelength produced in an -ray tube operated at a voltage of 35.0
?
ANSWER:
= h*c/(V*e)*10^9 = 3.55×10−2
Also accepted: h*c/(V*1.60*10^-19)*10^9 = 3.55×10−2, h*2.998*10^8/(V*e)*10^9 = 3.54×10−2, h*2.998*10^8/(V*1.60*10^-19)*10^9 = 3.55×10−2, h*c/(V*e)*10^9 =
3.55×10−2
Exercise 38.18
Description: A photon with wavelength lambda scatters from an electron that is initially at rest. (a) What must be the angle between the direction of propagation of the
incident and scattered photons if the speed of the electron immediately after the...
A photon with wavelength 0.1375
scatters from an electron that is initially at rest.
Part A
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math:
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What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is
?
ANSWER:
= acos(1-3.00*10^8*(lambda*9.11/10^31*v)^2/6.626*10^34/(2*6.626/10^34*3.00*10^8-lambda*9.11/10^31*v^2)) = 99.6
Also accepted: acos(1-2.998*10^8*(lambda*9.108/10^31*v)^2/6.626*10^34/(2*6.626/10^34*2.998*10^8-lambda*9.108/10^31*v^2)) = 99.6, acos(1-3.0*10^8*
(lambda*9.108/10^31*v)^2/6.626*10^34/(2*6.626/10^34*3.0*10^8-lambda*9.108/10^31*v^2)) = 99.6, acos(1-2.998*10^8*(lambda*9.11/10^31*v)^2/6.626*10^34
/(2*6.626/10^34*2.998*10^8-lambda*9.11/10^31*v^2)) = 99.6, acos(1-3.00*10^8*(lambda*9.11/10^31*v)^2/6.626*10^34/(2*6.626/10^34*3.00*10^8-lambda*9.11
/10^31*v^2)) = 99.6
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