Question

A photon of wavelength $0.1\stackrel{o}{A}$ is emitted by a helium atom as a consequence of the emission of photon. The K.E. gained by helium atom is :

• A. 0.05eV
• B. 1.05 eV
• C. 2.05 eV
• D. 3.05 eV

Answer 1

The correct option is C 2.05 eV

Given : $\lambda =0.1$ $A^o$

Momentum gained by atom $P=\frac{h}{\lambda }=\frac{6.626\times {10}^{-34}}{0.1\times {10}^{-10}}=6.626\times {10}^{-23}$ $kg.m/s$

Kinetic energy gained $K.E=\frac{P^2}{2m}=\frac{(6.626\times {10}^{-23}{)}^2}{2(4\times 1.667\times {10}^{-27})}=3.29\times {10}^{-19}$ J

$⟹$ $K.E=\frac{3.29\times {10}^{-19}}{1.6\times {10}^{-19}}=2.05$ eV