In an experiment with the photoelectric effect,the stopping potential:

  • A
    increases with an increase in the wavelength of the incident light
  • B
    increases with an increase in the intensity of the incident light
  • C
    is $\left(\frac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons
  • D
    decreases with an increase in the intensity of the incident light

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Similar Questions

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $3\lambda$,the stopping potential is $\frac{V}{6}$. The threshold wavelength for the surface is:

When a metal surface is illuminated by light of wavelength $\lambda$,the stopping potential is $8 \ V$. When the same surface is illuminated by light of wavelength $3 \lambda$,the stopping potential is $2 \ V$. The threshold wavelength for this surface is:

The maximum kinetic energy of the emitted photoelectrons from a photosensitive material of work function $\phi$,when light of frequency $\nu$ is incident on it,is $E$. If the frequency of the incident light is $3\nu$,the maximum kinetic energy of the emitted photoelectrons is:

From the figure describing the photoelectric effect,we may infer correctly that:

When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

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