Question

A nucleus of mass $M+△m$ is at rest and it decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. The speed of the light is $c$. Then what is the speed of the daughter nuclei?

• A. $c\frac{△m}{M+△m}$
• B. $c\sqrt{\frac{2△m}{M}}$
• C. $c\sqrt{\frac{△m}{M}}$
• D. $c\sqrt{\frac{△m}{M}}$

Answer 1

The correct option is B

$c\sqrt{\frac{2△m}{M}}$

The momentum is conserved!

Earlier the parent nucleus was at rest,

So,

$\therefore \frac{M}{2}{V}_1=\frac{M}{2}{V}_2\Rightarrow V_1=V_2=V\left(say\right)$
now. Kinetic energy = change in rest energy

$\Rightarrow \frac{1}{2}(\frac{M}{2}{)}^{V^2}+\frac{1}{2}(\frac{M}{2}{)}^{V^2}=(M+△m-[\frac{M}{2}+\frac{M}{2}]{)}^{c^2}$
$\Rightarrow \frac{1}{2}M{V}^2=△m{c}^2$

$\Rightarrow thespeedofthedaughternucleiV=\sqrt{\frac{2△m}{M}}c$