Question

The threshold frequency for a photosensitive metal is $3.3\times {10}^4$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

• A. 2V
• B. 3V
• C. 5V
• D. 1V

Answer 1

The correct option is A 2V

Given : ${\nu }_{th}=3.3\times {10}^{14}Hz$ $\nu =8.2\times {10}^{14}Hz$
Cut-off voltage $e{V}_o=h\nu -h{\nu }_{th}$
Or $V_o=\frac{h}{e}(\nu -{\nu }_{th})$
$⟹V_o=\frac{6.6\times {10}^{-34}}{1.6\times {10}^{-19}}(8.2-3.3)\times {10}^{14}=2V$