Question

A nucleus of mass $20u$ emits a $\gamma$ photon of energy $6MeV$. If the emission assume to occur when nucleus is free and rest, then the nucleus will have kinetic energy nearest to ($1u=1.6\times {10}^{-27}kg$)

• A. $10KeV$
• B. $1KeV$
• C. $0.1KeV$
• D. $100KeV$

Answer 1

The correct option is B $1KeV$

Given : $E=6MeV=6\times 1.6\times {10}^{-13}J$ $m=20u=20\times 1.6\times {10}^{-27}$ kg

Momentum of the gamma photon $p=\frac{E}{c}$

According to conservation of momentum, momentum of nucleus is equal to that of photon.

$⟹$ Momentum of Nucleus $P=p=\frac{E}{c}$

Kinetic energy of nucleus $K=\frac{P^2}{2m}=\frac{E^2}{2m{c}^2}$

$\therefore$ $K=\frac{(6\times 1.6\times {10}^{-13}{)}^2}{2\times (20\times 1.6\times {10}^{-27})\times (3\times {10}^8{)}^2}$ $J=1.6\times {10}^{-16}J$

$⟹$ $K=\frac{1.6\times {10}^{-16}}{1.6\times {10}^{-19}}eV=1000eV=1keV$