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

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.

Exams
Articles
Questions