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 the kinetic energy of the $\alpha$ -particle is

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

Answer 1

The correct option is C

5.4 MeV

Given that $K_1+K_2=5.5$ MeV

From Conservation of Linear Momentum,

$P_1=P_2$

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

As $P=\sqrt{2Km}$

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

Solving equations (1) and (2), we get $K_2=KE$ of $\alpha$ - particle = 5.4 MeV.