Question

A nucleus $220X$ at rest decays emitting an $\alpha particle$. If energy of daughter nucleus is $0.2MeV,Q$ value of the reaction is

• A. $10.8MeV$
• B. $10.9MeV$
• C. $11MeV$
• D. $11.1MeV$

Answer 1

Answer: (C)

$11MeV$

${}_z{X}^{220}\to {}_{z-2}{Y}^{216}+{}_2{He}^4$

$\begin{array}{c}\text{From energy and momentum conservation}\\ \text{we can derive the equation for K.E. of alpha}\\ \text{particle as}\end{array}$ $\begin{array}{c}\text{K.E. of}\alpha \text{particle}=\left[\right(A-4\left)/4\right]\times \text{K.E. of daughter nucles}\end{array}$

$\begin{array}{c}=\left[\right(220-4\left)/4\right]\times 0.2MeV\\ =10.8MeV\end{array}$

$\begin{array}{c}\text{value of the reaction is given by}\\ \text{Q=K.E. of alpha particle +K .E of daughter nucleus}\end{array}$

$\begin{array}{c}=10.8MeV+0.2MeV\\ =11MeV\end{array}$