Question

On shining light of wavelength $6.2\times {10}^{-6}$ m on a metal surface photo-electrons are emitted. The work
function of the metal is 0.1 eV. Find the kinetic energy of a photo-electron (in eV)

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Answer 1

The correct option is A 0.1

Here, $\lambda =6.2\times {10}^{-6}m,{\varphi }_0=0.1eV$
Energy of the incident photon,$E=hv=\frac{hc}{\lambda }orE=\frac{(6.6\times {10}^{-34})(3\times {10}^8)}{6.2\times {10}^{-6}}J=\frac{(6.6\times 3\times {10}^{-26})}{(6.2\times {10}^{-6})(1.6\times {10}^{-19})}eV=0.2eVAsE=K+{\varphi }_0,K=E-{\varphi }_0=0.2eV-0.1eV=0.1eV$