$E_{P}+E_{Q}=c^{2} \delta$
$E_{P} = \left( \frac{M_{P}}{M_{P}+M_{Q}} \right) c^{2} \delta$
$\frac{v_{P}}{v_{Q}} = \frac{M_{Q}}{M_{p}}$
The magnitude of momentum for $P$ as well as $Q$ is $c \sqrt{2 \mu \delta}$ , where $\mu = \frac{M_{P} M_{Q}}{(M_{P}+M_{Q})}$

Solution:

For nuclear fission reaction,

N \rightarrow P+Q

Energy released

= \Delta M c^{2} = (M_{N} - M_{P} - M_{Q}) c^{2} = \delta_{=} c^{2}

This will be distributed as kinetic energy of

P

and

Q

\therefore \;\; E_{p} + E_{q} = \delta c^{2}

.....(i)

By conservation of momentum

V_{P} M_{P} = V_{Q} M_{Q}

\therefore \;\; \frac{V_{P}}{V_{q}} = \frac{M_{q}}{M_{p}}

.....(ii)

Kinetic energy be written as

\text{KE} = \frac{P^{2}}{2m}

Hence divided in inverse ratio of masses

E_{P} = \frac{M_{q}}{M_{P}+M_{q}} c^{2} \delta

....(iii)

By equation (i)

\Rightarrow \frac{P^{2}}{2 M_{P}} + \frac{P^{2}}{2 M_{q}} = \delta c^{2}

\Rightarrow \frac{P^{2}}{2 \mu} = \delta c^{2} \Rightarrow P = c \sqrt{2 \mu \delta} \quad \left[\because \mu = \frac{M_{P} M_{Q}}{M_{P}+M_{Q}}\right]

Go to Question:

Prev Question

Next Question