Photoelectric effect：練習問題

In a photoelectric effect experiment, monochromatic light is shone onto a metal surface with a threshold frequency of \(f_0\). If the light source is replaced by another one with a frequency of \(0.8f_0\) and its intensity is tripled, what will be the result?

Monochromatic light with a frequency of \(8.50 \times 10^{14} \text{ Hz}\) is incident on a metal plate with a work function of \(2.30 \text{ eV}\). Calculate the maximum speed of the emitted photoelectrons.
(Given: Mass of electron \(m_e = 9.11 \times 10^{-31} \text{ kg}\), \(h = 6.63 \times 10^{-34} \text{ J s}\))

When monochromatic light of frequency \( f_1 = 2f_0 \) (where \( f_0 \) is the threshold frequency of the metal) is incident on a surface, the maximum speed of the emitted photoelectrons is \( v_1 \). If the frequency of the incident light is changed to \( f_2 = 5f_0 \), what is the new maximum speed \( v_2 \) of the photoelectrons in terms of \( v_1 \)?

According to Einstein's photoelectric theory, if the frequency of the incident light on a metal surface is increased while keeping the intensity constant, which of the following physical quantities will increase?

A metal surface is illuminated by monochromatic light of frequency \( f \) and intensity \( I \). The saturation photoelectric current is measured to be \( i \). If the frequency of the incident light is kept at \( f \) but its intensity is increased to \( 3I \), what will be the new saturation photoelectric current?

The threshold frequency of a specific metal is \(f_0\). If the metal is illuminated by monochromatic light of frequency \(1.5f_0\), express the maximum kinetic energy of the emitted photoelectrons in terms of the work function \(\phi\) of the metal.

Monochromatic light of wavelength \(\lambda\) is incident on a metal surface, resulting in a measured stopping potential of \(V\). If the wavelength of the incident light is reduced to \(0.5\lambda\), show mathematically using Einstein's equation why the new stopping potential \(V'\) must be greater than \(2V\).

A metal surface is illuminated by monochromatic light of wavelength \(\lambda\), and the maximum kinetic energy of the emitted photoelectrons is found to be \(K\). If the light source is replaced by one with double the wavelength \(2\lambda\), the new maximum kinetic energy is \(K'\). Use Einstein's photoelectric equation to explain why \(K > 2K'\).

A student investigates the photoelectric effect using a phototube with a clean metallic cathode. When monochromatic ultraviolet (UV) light of wavelength \(\lambda = 250 \times 10^{-9} \text{ m}\) is incident on the cathode, the maximum kinetic energy of the emitted photoelectrons is determined by measuring the stopping potential \(V_s\). The measured stopping potential is \(1.50 \text{ V}\).

Use the following physical constants where necessary:

• Speed of light, \(c = 3.00 \times 10^8 \text{ m s}^{-1}\)
• Planck constant, \(h = 6.63 \times 10^{-34} \text{ J s}\)
• Elementary charge, \(e = 1.60 \times 10^{-19} \text{ C}\)

(a) Calculate the energy of a single photon of the incident UV light, expressing your answer in electron volts ( \(\text{eV}\)\).

(b) Determine the work function ( \(\Phi\)) of the metallic cathode in electron volts ( \(\text{eV}\)\).

(c) Calculate the threshold frequency ( \(f_0\)) for this metallic cathode.

(d) The UV light source is replaced by a red laser source emitting light of frequency \(f = 4.00 \times 10^{14} \text{ Hz}\). Explain whether photoelectrons will be emitted from the cathode, and state how increasing the intensity of the red laser source would affect the emission process.

A monochromatic light source of wavelength \( 420 \text{ nm} \) is incident on a metal surface with a work function of \( 2.10 \text{ eV} \).

(a) Calculate the maximum kinetic energy of the emitted photoelectrons in Joules.
(b) Determine the stopping potential required to reduce the photoelectric current to zero.
(c) Calculate the maximum speed of the emitted photoelectrons.

(Given: \( h = 6.63 \times 10^{-34} \text{ J s} \), \( c = 3.00 \times 10^8 \text{ m/s} \), \( e = 1.60 \times 10^{-19} \text{ C} \), mass of electron \( m_e = 9.11 \times 10^{-31} \text{ kg} \))