Chapter 5 · Photoelectric effect: practice questions
Unit 2 — Introduction to Modern Physics · Physics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 72 questions on this lesson, each with an explanation.
Athe total energy of all the free electrons in the metal.
Bthe minimum energy required to liberate an electron from the metal surface.
Cthe maximum kinetic energy of an emitted photoelectron.
Dthe energy carried by any incident photon.
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Correct answer: (B) the minimum energy required to liberate an electron from the metal surface.
Explanation: The work function is the least energy needed to free an electron from the metal surface.
Question 2Easy
Einstein's photoelectric equation is correctly written as…
Ahν = Ew + ½mv²_max
Bhν = Ew − ½mv²_max
CEw = hν + ½mv²_max
Dhν = ½mv²_max − Ew
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Correct answer: (A) hν = Ew + ½mv²_max
Explanation: The photon energy equals the work function plus the maximum kinetic energy of the emitted electron.
Question 3Easy
For monochromatic light of a fixed frequency above threshold, increasing the intensity of the light causes the number of photoelectrons emitted per second to…
Aincrease.
Bvanish.
Cremain constant.
Ddecrease.
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Correct answer: (A) increase.
Explanation: A higher intensity delivers more photons per second, so more photoelectrons are emitted per second.
Question 4Medium
Light whose photons have an energy of 5 eV falls on a metal of work function 2 eV. The maximum kinetic energy of the emitted photoelectrons is…
A2.5 eV
B3 eV
C7 eV
D10 eV
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Correct answer: (B) 3 eV
Explanation: KE_max = photon energy − work function = 5 − 2 = 3 eV.
Question 5Medium
A photon has a frequency of 6 × 10¹⁴ Hz. Its energy is... (h = 6.63 × 10⁻³⁴ J·s)
A6.63 × 10⁻²⁰ J
B1.1 × 10⁻¹⁹ J
C3.98 × 10⁻¹⁹ J
D2.49 × 10⁻¹⁸ J
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Correct answer: (C) 3.98 × 10⁻¹⁹ J
Explanation: E = hν = 6.63 × 10⁻³⁴ × 6 × 10¹⁴ = 3.98 × 10⁻¹⁹ J.
Question 6Medium
A metal has a work function of 3.3 × 10⁻¹⁹ J. Its threshold frequency is… (h = 6.6 × 10⁻³⁴ J·s)
A2 × 10¹⁵ Hz
B5 × 10¹⁴ Hz
C5 × 10¹⁵ Hz
D2 × 10¹⁴ Hz
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Correct answer: (B) 5 × 10¹⁴ Hz
Explanation: νc = Ew / h = (3.3 × 10⁻¹⁹)/(6.6 × 10⁻³⁴) = 5 × 10¹⁴ Hz.
Question 7Medium
The maximum kinetic energy of the photoelectrons emitted from a surface is 3.2 × 10⁻¹⁹ J. The stopping potential is... (e = 1.6 × 10⁻¹⁹ C)
A2 V
B4 V
C0.5 V
D5.12 × 10⁻³⁸ V
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Correct answer: (A) 2 V
Explanation: Vₛ = KE_max / e = 3.2 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2 V.
Question 8Hard
Light of wavelength 400 nm falls on a metal whose threshold wavelength is 500 nm and photoelectrons are emitted. If the wavelength is changed to 600 nm and at the same time the light intensity is doubled, then…
Amore photoelectrons are emitted, each with less kinetic energy
Bthe same number of photoelectrons is emitted, each with less kinetic energy
Cno photoelectrons are emitted at all
Dtwice as many photoelectrons are emitted with the same kinetic energy
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Correct answer: (C) no photoelectrons are emitted at all
Explanation: 600 nm is longer than the 500 nm threshold, so each photon's energy is below the work function; no electron is released whatever the intensity, since intensity only changes the number of photons.
Question 9Hard
When light of frequency 10¹⁵ Hz falls on a metal of work function 3.4 × 10⁻¹⁹ J, the stopping potential of the emitted photoelectrons is… (h = 6.6 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)
A2 V
B4 V
C1 V
D3.2 V
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Correct answer: (A) 2 V
Explanation: KE_max = hν − Ew = 3.2 × 10⁻¹⁹ J, and Vₛ = KE_max/e = (3.2 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2 V.
Question 10Hard
Photoelectrons of maximum kinetic energy 1.6 × 10⁻¹⁹ J are emitted from a metal of work function 3.2 × 10⁻¹⁹ J. The frequency of the incident light is... (h = 6.63 × 10⁻³⁴ J·s)
A2.41 × 10¹⁴ Hz
B7.24 × 10¹⁴ Hz
C1.06 × 10¹⁵ Hz
D4.83 × 10¹⁴ Hz
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Correct answer: (B) 7.24 × 10¹⁴ Hz
Explanation: hν = KE_max + Ew = 4.8 × 10⁻¹⁹ J, so ν = 4.8 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 7.24 × 10¹⁴ Hz.