In a photoemission experiment,the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_P, E_Q$ and $E_R$,respectively,and they are related by $E_P = 2E_Q = 2E_R$. In this experiment,the same source of monochromatic light is used for metals $P$ and $Q$,while a different source of monochromatic light is used for metal $R$. The work functions for metals $P, Q$ and $R$ are $4.0 \ eV$,$4.5 \ eV$ and $5.5 \ eV$,respectively. The energy of the incident photon used for metal $R$,in $eV$,is:

  • A
    $6$
  • B
    $8$
  • C
    $9$
  • D
    $10$

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Similar Questions

$A$ photon of energy $5.5 \ eV$ strikes a surface that emits photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping potential for these electrons is ............ $V$.

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda / 3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case,the work function of the surface of the material is ($h=$ Planck's constant,$c=$ speed of light).

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

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