Question

Radiation of $\lambda =155nm$ was irradiated on Li (work function - 5 e V) plate.The stopping potential (in V ) is:

• A. 3 V
• B. 8 V
• C. 9 V
• D. 5 V

Answer 1

The correct option is A 3 V

$\lambda =155nm$

Energy of radiation=$\frac{hc}{\lambda }$

$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{155\times {10}^{-9}\times 1.6\times {10}^{-19}}$

$=\frac{19.878}{248}\times {10}^2$

$=8.015ev$

Energy required to stop emmision of electron

$=(8.015-5)eV$

Stopping potential =$3.015v\simeq 3v$