Energy of a photoelectron depends on which factors?

  • A
    Intensity of incident light
  • B
    Frequency of incident light
  • C
    Material of the photosensitive surface
  • D
    Both $(B)$ and $(C)$

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Similar Questions

The threshold wavelength for photoelectric emission from a material is $5500\,\mathring A$. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a:
$A.$ $75\,W$ infra-red lamp
$B.$ $10\,W$ infra-red lamp
$C.$ $75\,W$ ultra-violet lamp
$D.$ $10\,W$ ultra-violet lamp
Choose the correct answer from the options given below:

Light of two different frequencies with photon energies of $1 \ eV$ and $2.5 \ eV$ are incident one after another on a metal surface having a work function of $0.5 \ eV$. The ratio of the maximum kinetic energy of the emitted photoelectrons will be:

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

The threshold wavelength for photoelectric emission from a metal surface is $5200 \ \mathring A$. Which of the following sources will cause photoelectric emission from this surface?

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

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