In an experiment on the photoelectric effect, monochromatic light is incident on a metal surface. Which of the following changes will result in an increase in the maximum kinetic energy of the emitted photoelectrons?
Oxford AQA International A-level · Physics (9630)
Photoelectric effect:练习题
5 道选择题即时批改,另有 5 道文字题附完整解题步骤,全部围绕「Photoelectric effect」。
A metal has a threshold wavelength of \( 540 \text{ nm} \). Calculate the work function of this metal in electron volts (eV).
Use: \( h = 6.63 \times 10^{-34} \text{ J s} \), \( c = 3.00 \times 10^8 \text{ m s}^{-1} \), \( 1 \text{ eV} = 1.60 \times 10^{-19} \text{ J} \).
When light of frequency \( f \) is incident on a metal, the stopping potential is \( V_s \). When light of frequency \( 1.5f \) is incident on the same metal, the stopping potential becomes \( 2V_s \). What is the work function of the metal in terms of \( hf \)?
A metal surface has a work function \(\phi = 2.50 \text{ eV}\). Light of frequency \(1.20 \times 10^{15} \text{ Hz}\) is incident upon the surface. Calculate the maximum speed of the photoelectrons emitted.
Planck constant, \(h = 6.63 \times 10^{-34} \text{ J s}\). Elementary charge, \(e = 1.60 \times 10^{-19} \text{ C}\). Mass of electron, \(m_e = 9.11 \times 10^{-31} \text{ kg}\).
Two different metals, X and Y, are illuminated with the same monochromatic light of frequency \( f \). The work function of metal X is \( \phi_X \) and the work function of metal Y is \( \phi_Y \). If \( \phi_X = 2\phi_Y \), which of the following expressions correctly relates the maximum kinetic energies \( E_X \) and \( E_Y \) of the emitted photoelectrons?
Light of frequency \(7.5 \times 10^{14} \text{ Hz}\) is incident on a metal surface. If the metal has a work function \(\phi = 2.0 \text{ eV}\), calculate the maximum kinetic energy, in joules, of the emitted photoelectrons. (Use \(h = 6.63 \times 10^{-34} \text{ J s}\) and \(1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}\)).
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The classical wave theory of light predicted that the maximum kinetic energy of photoelectrons should increase with the intensity of the incident light. State two key experimental observations from the photoelectric effect that contradicted this prediction and support the photon model.
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A photoelectric experiment uses monochromatic light of a fixed frequency. Explain why increasing the intensity of the incident light does not change the stopping potential required to halt the emitted photoelectrons.
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A student performs an experiment on the photoelectric effect using a metallic surface. When light of wavelength \(450 \text{ nm}\) is incident on the surface, the maximum kinetic energy of the emitted photoelectrons is found to be \(0.85 \times 10^{-19} \text{ J}\).
(a) Calculate the energy of a single photon of this incident light.
(b) Determine the work function \(\phi\) of the metallic surface, expressing your answer in electron volts (eV).
(c) Calculate the threshold frequency for this metal.
(Use: Planck constant, \(h = 6.63 \times 10^{-34} \text{ J s}\); Speed of light, \(c = 3.00 \times 10^8 \text{ m/s}\); Elementary charge, \(e = 1.60 \times 10^{-19} \text{ C}\))
先自己写一遍答案,再对照解题步骤。
A photoelectric experiment is conducted to investigate the properties of a specific metal surface. Monochromatic light is shone onto the metal, and the stopping potential is measured for two different wavelengths.
(a) Define the term stopping potential and explain its relationship with the maximum kinetic energy of the emitted photoelectrons.
(b) When light of wavelength \(\lambda_1 = 300 \text{ nm}\) is used, the stopping potential is \(V_{s1} = 2.14 \text{ V}\). When the wavelength is increased to \(\lambda_2 = 500 \text{ nm}\), the stopping potential falls to \(V_{s2} = 0.49 \text{ V}\). Use these data to calculate the Planck constant \(h\).
(c) Determine the work function of the metal in electron-volts (\(\text{eV}\)).
(d) Explain why the existence of a threshold frequency provides evidence for the particle nature of light, contrasting this with the predictions of classical wave theory.
(Use: Speed of light, \(c = 3.00 \times 10^8 \text{ m/s}\); Elementary charge, \(e = 1.60 \times 10^{-19} \text{ C}\))
先自己写一遍答案,再对照解题步骤。
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