Question

In an $\alpha$-decay, the kinetic energy of $\alpha$-particle is $48MeV$ and $Q$-value of the reaction is $50MeV$. The mass number of the mother nucleus is $X$. Find the value of $X/25$. (Assume that daughter nucleus is in ground state)

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

The $\alpha -$decay nuclear reaction: ${}_Z^{X}B{\to }_{Z-2}^{X-4}C+\alpha +Q$

where, $B$ and $C$ are the mother and daughter nuclei respectively.

Given: $Q=50MeV,E_{alpha}=48MeV$

Using: $Q=E_{\alpha }(\frac{M_{\alpha }+M_C}{M_C})$

we get, $Q=E_{\alpha }(\frac{4+X-4}{X-4})=E_{\alpha }(\frac{X}{X-4})$

$\therefore 50=48(\frac{X}{X-4})⟹X=100$

Hence, $\frac{X}{25}=\frac{100}{25}=4$