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

 

Answers (1)
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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