Question

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

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

Answer 1

The correct option is C $5.4MeV$

Let $K_2$ be kinetic energy of $\alpha$-particle and
$K_1$ be kinetic energy of daughter nucleus.

Given that, $K_1+K_2=5.5$ MeV

From Conservation of Linear Momentum,

$P_1=P_2$

As $P=\sqrt{2Km}$

$\Rightarrow \sqrt{2K_1\left(216m\right)}=\sqrt{\left(2K_2\right(4m)}$

$\Rightarrow K_2=54K_1\dots \left(2\right)$

Solving equations (1) and (2),

$K_2=KE$ of $\alpha$ - particle = $5.4MeV$.