Question

A nucleus with mass number $220$ initially at rest emits an $\alpha $ -particle. If the $Q$ value of the reaction is $5.5 \, MeV$ , then calculate the kinetic energy of the $\alpha $ -particle.

• A. $4.4 \, MeV$
• B. $5.4 \, MeV$
• C. $5.6MeV$
• D. $6.5MeV$

Answer 1

Answer: (B)

Given that $K_{1}+K_{2}=\text{5.5} \, MeV$ ... (i)
From conservation of linear momentum,
$p_{1}=p_{2}$
$\sqrt{2 K_{1} \left(\right. 216 m \left.\right)} = \sqrt{2 K_{2} \left(\right. 4 m \left.\right)}$ (As $p = \sqrt{2 K m}$ )
$K_{2}=54K_{1}$ ... (ii)
Solving Eqs. (i) and (ii), we get
$K_{2}=$ K.E. of $\alpha $ -particle $\text{5.4} \, MeV$ .