Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4v$ is incident on the metal plate,the maximum velocity of the emitted photoelectrons will be ($m=$ mass of photoelectron,$h=$ Planck's constant).

  • A
    $\sqrt{\frac{6hv}{m}}$
  • B
    $\sqrt{\frac{3hv}{m}}$
  • C
    $\sqrt{\frac{hv}{m}}$
  • D
    $\sqrt{\frac{5hv}{m}}$

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Similar Questions

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

An isolated metallic sphere is illuminated by light of $4\,eV$ photon energy. If the work function of the metal is $2\,eV$,then the minimum potential of the sphere so that no photoelectrons are ejected is ................ $V$.

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 $e = 1.6 \times 10^{-19} \, C$):

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Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

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