Nuclear & Particle Physics - Photoelectric

Light of wavelength 500 nanometers is incident on sodium, with work function 2.28 electron volts. What is the maximum kinetic energy of the ejected photoelectrons?

A. 0.03 eV
B. 0.2 eV
C. 0.6 eV
D. 1.3 eV
E. 2.0 eV

(GR9677 #42)

Solution:

Einstein’s photoelectric equation:  |eV| = hv − W = ^hc/_λ − W

h = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ = 500 nm = 5 × 10⁻⁷ m
W = 2.28 eV

KE = [(4.1 × 10⁻¹⁵)(3 × 10⁸)/(5 × 10⁻⁷ )] − 2.28 
=  [(4.1 × 3 / 5)(10⁻¹⁵ × 10⁸ × 10⁷)] − 2.28
=  2.46 − 2.28 = 0.18 ≈  0.2 eV

Answer: B

Nuclear & Particle Physics - Photoelectric

Questions 31-33 refer to the following apparatus used to study the photoelectric effect.

In this apparatus, the photocathode and the collector are made from the same material. The potential V of the collector, measured relative to ground, is initially zero and is then increased or decreased monotonically. The effect is described by Einstein’s photoelectric equation:  |eV| = hv − W

When the photoelectric equation is satisfied and applicable to this situation, V is the

A. Negative value at which the current stops
B. Negative value at which the current starts
C. Positive value at which the current stops
D. Positive value at which the current starts
E. Voltage induced when the light is on

(GR8677 #31)

Solution:

Stopping potential, V:

• The point at which no electrons make it to the collector.
• The electron charge is negative and the potential V must be a negative quantity in order to make eV positive overall.
• It is related to the maximum kinetic energy of the ejected electrons by eV = KE_max = ½ m (v_max)²

Answer: A

Nuclear & Particle Physics - Photoelectric

Questions 31-33 refer to the apparatus used to study the photoelectric effect (see GR8677 #31).

The photoelectric equation is derived under the assumption that

1. Electrons are restricted to orbits of angular momentum nħ, where n is an integer
2. Electrons are associated with waves of wavelength λ = h/p, where p is momentum
3. Light is emitted only when electrons jump between orbits
4. Light is absorbed in quanta of energy E = hv
5. Light behaves like a wave

(GR8677 #32)

Solution:

According to the classical Maxwell wave theory of light, the average energy carried by an emitted electron should increase with the intensity of the incident light.

However, in photoelectric case, the energies of the emitted electrons are independent of the intensity of the incident radiation.

Einstein resolved this paradox by proposed that the incident light consisted of individual quanta, called photons, that interacted with the electrons in the metal like discrete particles, rather than as continuous waves.

Answer: D

Nuclear & Particle Physics - Photoelectric

Questions 31-33  refer to the apparatus used to study the photoelectric effect (see GR8677 #31).
The quantity W in the photoelectric equation is the

1. Energy difference between the two lowest electron orbits in the atoms of the photocathode
2. Total light energy absorbed by the photocathode during the measurement
3. Minimum energy a photon must have in order to be absorbed by the photocathode
4. Minimum energy required to free an electron from its binding to the cathode materials
5. Average energy of all electrons in the photocathod

(GR8677 #33)

Solution:

Einstein photoelectric equation: |eV| = hv −W

eV = kinetic energy of the electron as it accelerates through the medium between the cathode and collector
W = work function; energy to free the electron from the metal
hv = energy given by the light source.

Answer: D

Nuclear & Particle Physics - Photon Interactions

The figure above show the photon interaction cross sections for lead in the energy range where the Compton, photoelectric, and pair production processes all play a role. What is the correct identification of these cross sections?

A. 1 = Photoelectric, 2 = Compton, 3 = Pair production
B. 1 = Photoelectric, 2 = Pair production, 3 = Compton
C. 1 = Compton, 2 = Pair production, 3 = Photoelectric
D. 1 = Compton, 2 = Photoelectric, 3 = Pair production
E. 1 = Compton, 2 = Photoelectric, 3 = Compton

(GR8677 #85)

Solution:

Photoelectric Effect:

• Low-energy photon (visible/UV/soft X-rays) interacts with matter
• Results: emission of electron

Compton Scattering:

• High-energy photon (X-ray/Gamma ray) interacts with electron
• Results: emission of inelastic electron and lower energy photon

→ Energy Range: Photoelectric Effect Compton Scattering

Pair production:

• High-energy photon (X-ray/gamma ray) interacs with nucleus
• Results: elementary particle and its antiparticle

→ Energy Range:  Compton Scattering (electron target) Pair production (nucleus target)

→  Energy Range:  Photoelectric Compton Pair production

→ 1 = photoelectric, 2 = pair production, 3 = Compton

Answer: B