Q. An atom of mass M which is in the state of rest emits a photon of wavelength $\lambda.$ As a result, the atom will deflect with the kinetic energy equal to ($h$ is Planck’s constant)
A) $\frac{h^{2}}{M\lambda^{2}}$
B) $\frac{1}{2}\frac{h^{2}}{M\lambda^{2}}$
C) $\frac{h}{M\lambda}\ $
D) $\frac{1}{2}\frac{h}{M\lambda}$
Explanation
(b) Momentum of photon, $p = \frac{h}{\lambda}$
Kinetic energy of photon of mass M,
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$K = \frac{p^{2}}{2M}$
or $K = \frac{h^{2}}{2M\lambda^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.