Question

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

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

Answer 1

The correct option is B $\text{c}\sqrt{\frac{2\Delta \text{m}}{\text{M}}}$

Momentum is always conserved! Earlier the parent nucleus was at rest. So,

$\therefore \frac{\text{M}}{2}{\text{V}}_1=\frac{\text{M}}{2}{\text{V}}_2$

$\Rightarrow {\text{V}}_1={\text{V}}_2=\text{V}$(say)

Now, kinetic energy = Change in the rest energy

$\Rightarrow \frac{1}{2}\left(\frac{\text{M}}{2}\right){\text{V}}^2+\frac{1}{2}\left(\frac{\text{M}}{2}\right){\text{V}}^2$

$=(\text{M}+△\text{m}-[\frac{\text{M}}{2}+\frac{\text{M}}{2}\left]\right){\text{c}}^2$

$\Rightarrow \frac{1}{2}{\text{MV}}^2=△{\text{mc}}^2$

$\Rightarrow$ The speed of the daughter nuclei

$\text{V}=\text{c}\sqrt{\frac{2△\text{m}}{\text{M}}}$