The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

  • Option 1)

    2.4 V

  • Option 2)

    -1.2 V

  • Option 3)

    -2.4 V

  • Option 4)

    1.2 V

 

Answers (1)
V Vakul

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- \theta\:=\:\frac{1242}{200}\:ev-5.01\:ev

ev_{s}=6.21\:ev-5.01\:ev=1.2\:ev

\therefore\:v_{s}=1.2V

 

 

 


Option 1)

2.4 V

This option is incorrect.

Option 2)

-1.2 V

This option is incorrect.

Option 3)

-2.4 V

This option is incorrect.

Option 4)

1.2 V

This option is correct.

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