# The threshold frequency for a photosensitive metal is $3.3\times 10^{14}$ Hz. If light of frequency $8.2\times 10^{14}$ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly Option 1) 2V Option 2) 3V Option 3) 5V Option 4) 1V

D Divya Saini

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

&

Stopping Potential /Cut-off Potential -

It is defined as the potential necessary to stop any electron from reaching the other side.

-

$ev_{s}=h \nu-h \nu_{o}$

= $6.62\times 10^{-34}[8.2\times 10^{14}-3.3\times 10^{14}]$

= $6.62\times 10^{-34}\times 4.9\times 10^{14}$

$v_{s}=\frac{32.438\times 10^{20}}{1.6\times 10^{19}}=2V$

Option 1)

2V

This option is correct.

Option 2)

3V

This option is incorrect.

Option 3)

5V

This option is incorrect.

Option 4)

1V

This option is incorrect.

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