Question

Radiation of wavelength 155 nm is incident on Li metal with work function 5.0 eV. The stopping potential will be

• A. 3 V
• B. 8 V
• C. 9 V
• D. 5 V

Answer 1

The correct option is A 3 V

Given : λ=155nm
Energy of radiation $=\frac{hc}{\lambda }$​
$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{155\times {10}^{-9}\times 1.6\times {10}^{-19}}$
=8.015 eV
Energy required to stop emission of electron
=(8.015−5) eV
$\therefore Stoppingpotential=3.015V\simeq 3V$