Light of frequency $v$ falls on a material of threshold frequency $v_0$. Maximum kinetic energy of emitted electron is proportional to

  • A
    $v-v_0$
  • B
    $v$
  • C
    $\sqrt{v-v_0}$
  • D
    $v_0$

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Similar Questions

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

Assertion : In the process of photoelectric emission, all emitted electrons do not have the same kinetic energy.
Reason : If radiation falling on the photosensitive surface of a metal consists of different wavelengths, then the energy acquired by electrons absorbing photons of different wavelengths shall be different.

Light rays consisting of photons with energy $1.8 \ eV$ are incident on a metal surface with a work function of $1.2 \ eV$. What is the stopping potential required to stop the emitted electrons in $eV$?

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?

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