# For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) Option 1) $\sqrt{\frac{hv}{m}}$ Option 2) $\sqrt{\frac{2hv}{m}}$ Option 3) $2\sqrt{\frac{hv}{m}}$ Option 4) $\sqrt{\frac{hv}{(2m)}}$

P Prateek Shrivastava

As we discussed in concept

Conservation of energy -

$h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}$

$h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}$

$h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}$

$where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function$

- wherein

$\frac{1}{2}mv^{2}_{max}=h(\nu-\nu_{o})$

here $\nu\:=\:2\nu$

$\nu_{o}=\nu$

=> $\\ \frac{1}{2}mv^{2}_{max}=h.(2\nu-\nu)=h\nu \\ \\ \therefore\:V_{max}=\sqrt{\frac{2h\nu}{m}}$

Option 1)

$\sqrt{\frac{hv}{m}}$

This option is incorrect.

Option 2)

$\sqrt{\frac{2hv}{m}}$

This option is correct.

Option 3)

$2\sqrt{\frac{hv}{m}}$

This option is incorrect.

Option 4)

$\sqrt{\frac{hv}{(2m)}}$

This option is incorrect.

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