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)}}

 

Answers (1)
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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