Q. A nucleus of mass 214 amu in free state decays to emit an $\alpha$-particle. Kinetic energy of the $\alpha$-particle emitted is $6.7\ MeV$. The recoil energy (in $MeV$) of the daughter nucleus is
A) 1.0
B) 0.5
C) 0.25
D) 0.125
Explanation
(d) Using conservation of momentum $P_{daughter} = P_{\alpha}$
$\Rightarrow \frac{E_{d}}{E_{\alpha}} = \frac{m_{\alpha}}{m_{d}} \Rightarrow E_{d} = \frac{E_{\alpha} \times m_{\alpha}}{m_{d}} = \frac{6.7 \times 4}{214} = 0.125MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.