The Photoelectric Effect AP Physics 2 FRQ – Exam Style Questions etc.
The Photoelectric Effect AP Physics 2 FRQ
Unit 15: Modern Physics
Weightage : 15–18%
Exam Style Practice Questions ,The Photoelectric Effect AP Physics 2 FRQ
Question
(10 points, suggested time 20 minutes)
In each trial of a photoelectric experiment, a scientist uses a device to shine light of a single frequency on two different metals, 1 and 2. The device can emit light with frequency \(f_{A}\), \(f_{B}\), \(f_{C}\). Each frequency of light is used to test both metals.
The scientist determines the minimum de Broglie wavelength \(\lambda _{e}\) of the electrons ejected from the metal in each trial of the experiment. The following table summarizes the results of the experiment. For each trial, the scientist analyzes only the electrons with the minimum de Broglie wavelength.
(a) In a coherent, paragraph-length response, indicate which frequency \(f_{A}\), \(f_{B}\), \(f_{C}\), s greatest and which frequency is least. Justify your answer using physics principles.
(b) Calculate the maximum kinetic energy of the electrons ejected from Metal 1 in Trial 1. Assume that the momentum p of an ejected electron can be described by the classical definition p = mv.
(c) Indicate whether the work function of Metal 1 is greater than, less than, or equal to the work function of Metal 2. Justify your answer by referring to the table of results.
▶️Answer/Explanation
1(a)Example Response
An electron will be ejected if the incident photon has an energy greater than the work function of the metal. Because no electrons were ejected using \(f_{b}\), the corresponding photon energy, and, therefore, frequency must be the least. A photon with greater frequency will result in an ejected electron with more kinetic energy; the kinetic energy of an electron is inversely related to the de Broglie wavelength of the electron. Because the de Broglie wavelength of electrons ejected by light of frequency \(f_{A}\) is greater than those ejected by light of frequency \(f_{C}\), \(f_{A}\) must be less than \(f_{C}\). Therefore, \(f_{C}\) is the greatest.
1(b)Example Solution
\(\lambda _{e} = \frac{h}{p} = \frac{h}{mv}\)
\(v =\frac{h}{m\lambda _{e}}\)
\(k = \frac{1}{2}{mv2} = \frac{1}{2}m\left ( \frac{h}{m\lambda _{e}} \right )^{2}\)
\(k = \frac{^{h2}}{2m\lambda _{e}^{2}}\)
\(k = \frac{\left ( 6.63 *10^{-34}J . s \right )^{2}}{2\left ( 9.11* 10 ^{-31}kg\right )\left ( 6.9 * 10^{-10}m \right )^{2}} = 5 * 10^{-19} J\)
1(c)Example Response
The work function of Metal 1 is less than the work function of Metal 2. When light of the same frequency is incident on both metals, the electron ejected by Metal 1 has a smaller de Broglie wavelength than that of Metal 2, so an electron ejected from Metal 1 has more kinetic energy. The work function is the difference between the photon energy and the maximum kinetic energy. Since the photon energy is the same but the maximum kinetic energy is larger for Metal 1, the difference between the energies, and thus the work function, is smaller for Metal 1.
Question: (12 points, suggested time 25 minutes)
Monochromatic light of frequency f shines on a metal, as shown above. The frequency of the light is varied, and for some frequencies electrons are emitted from the metal. The maximum kinetic energy Ke of the emitted electrons is measured as a function of the frequency of the light.
(a)
i. Based on conservation of energy, the relationship between Ke and f is predicted to be Af = B + Ke when f > f0 and Ke = 0 when f ≤ f0, where A and B are positive constants. A graph of this relationship is shown below. Indicate which aspects of the graph correspond to A and B. Also, explain the physical meaning of A, B, and f0 .
ii. Explain the physical meaning of the horizontal section of the graph between the origin and f0 .
iii. A second metal with different properties than the first metal is now used. On the figure below, the dashed lines are the same lines shown in the previous graph. Sketch lines on the figure below that could represent the data for the second metal. Explain one difference between the two graphs.
(b) The figure below shows an electroscope. A sphere is connected by a vertical bar to the leaves, which are thin, light strips of material. The sphere, leaves, and bar are all made of metal. The electroscope initially has a negative charge, so the leaves are separated.
i. Ultraviolet (UV) light shines on the sphere, causing the leaves of the electroscope to move closer together. Explain why this happens.
ii. Green light then shines on an identical negatively charged electroscope. No movement of the leaves is observed. Explain why the green light does not make the leaves move, while the UV light does.
(c) The brightness of the green light is increased until the intensity (power per unit area) is the same as that of the UV light. What aspect of the green light changes when its brightness is increased? Would shining the brighter green light on the electroscope result in movement of the leaves? Explain why or why not.
Answer/Explanation
Ans:
(a)
i.
A is represented by the slope of the graph B is represented by the y-intercept of the graph A is equal to Planck’s constant and determines the energy of a photon given it’s frequency B is the work function, or amount of energy required to release an election from the nucleus of one of the metal atoms. F0 is the minimum frequency required to have enough energy to overcome the work function.
ii.
The horizontal section represents the range of frequencies that don’t have enough energy to overcome the work function, therefore the election does not have enough energy to break away from the nucleus. Since the election is never emmited, Ke cannot be measured.
iii.
The horizontal section for the graph of a different metal could be different because it could have a lower work function.
(b)
i.
Due to the photoelectric effect, when light hits the sphere, some elections in the sphere absorb enough energy from the photons to be emmited. With less total elections, there is less total negative charge, reducing the repulsive electrostatic force between the leaves, reducing their separation.
ii.
The UV light has a higher frequency than the green light, therefore higher energy. The green light does not give the elections enough energy to overcome the work function of the metal, or energy required to emmit the elections. Therefore no charge is lost because no elections are lost. The UV light has enough energy to overcome the work function.
(c)
When the brightness of the green light is increased its intensity is increased, meaning more photons are released per unit area. The increase in brightness would not result in movement of the leaves. The change in intensity only change the amount of photons preset and does not change the energy per photon. Each photon would still lack the energy to overcome the work function meaning no elections would be released. Meaning the leaves wouldn’t move.
Question: (10 points, suggested time 20 minutes)
Light and matter can be modeled as waves or as particles. Some phenomena can be explained using the wave model, and others can be explained using the particle model.
(a) Calculate the speed, in m/s, of an electron that has a wavelength of 5.0 nm.
(b) The electron is moving with the speed calculated in part (a) when it collides with a positron that is at rest. A positron is a particle identical to an electron except that its charge is positive. The two particles annihilate each other, producing photons. Calculate the total energy of the photons.
(c) A photon approaches an electron at rest, as shown above on the left, and collides elastically with the electron. After the collision, the electron moves toward the top of the page and to the right, as shown above on the right, at a known speed and angle. In a coherent, paragraph-length response, indicate a possible direction for the photon that exists after the collision and its frequency compared to that of the original photon. Describe the application of physics principles that can be used to determine the direction of motion and frequency of the photon that exists after the collision.
Answer/Explanation
Ans:
\(\lambda =\frac{h}{p}\) 5 × 10-9 m = \(\frac{6.63\times 10^{-34}J.5}{(9.11\times 10^{-31}kg)V}\) p = mv
4.555× 10-39 V = 6.63 × 10-34
V = 14554.3 m/s
(b)
E = Δmc2 Δm = 2 (9.11 × 10-31 ) = 1.822 × 10-34
Po = Pf
14554 (me) = mv
E= 1.822 × 10-30 (3×108)2
E = 1.64 × 10-13 J
(c)
The initial momentum has to equal to final momentum in all directions. The frequency of the original will be greater than the final frequency as energy must be conserved and it depends on frequency. The photon will travel back towards the bottom of the page as to conserve momentum of the vertical axis but the angle will be less than the election to conserve horizontal momentum.