Question

A nucleus of mass 218 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
• B. 0.5
• C. 0.25
• D. 0.125

Answer 1

Answer: (D)

0.125

Conservation of total momentum of system implies:

$P_1$(Momentum of alpha particle) + $P_2$ (Momentum of daughter nucleus) $=0$ (Initial momentum)

Thus, $P_1=-P_2$

$M=$ mass of daughter nucleus = 214 amu

$m=$ mass of alpha particle = 4 amu

$E$ is the energy of daughter nucleus in MeV

Now , $6.7MeV=\frac{P_1^2}{2m}$, for alpha particle

$E=\frac{P_2^2}{2M}$, for daughter nucleus

Dividing the above two equations, we get

$\frac{E}{6.7}=\frac{m}{M}$ (as $P_1=-P_2$ or $P_1^2=P_2^2$)

Therefore, $E=6.7\times \frac{4}{214}=0.125$