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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)

best_answer

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.

Posted by

divya.saini

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