Question

A heavy nucleus $N$, at rest, undergoes fission $N \rightarrow P + Q$, where $P$ and $Q$ are two lighter nuclei. Let $\delta= M _{ N }$ - $M_{P}-M_{Q}$, where $M_{P}, M_{Q}$ and $M_{N}$ are the masses of $P, Q$ and $N$, respectively. $E _{P}$ and $E_{Q}$ are the kinetic energies of $P$ and $Q$, respectively. The speeds of $P$ and $Q$ are $v_{P}$ and $v_{Q}$, respectively. If $c$ is the speed of light, which of the following statement(s) is(are) correct?

• A. $E_{p}+E_{Q}=c^{2} \delta$
• B. $E_{p}=\left(\frac{M_{p}}{M_{p}+M_{Q}}\right) c^{2} \delta$
• C. $\frac{ v _{ P }}{ v _{ Q }}=\frac{ M _{ Q }}{ M _{ P }}$
• D. The magnitude of momentum for $P$ as well as $Q$ is $c \sqrt{2 \mu \delta}$, where $\mu=\frac{ M _{ p } M _{ Q }}{\left( M _{ p }+ M _{ Q }\right)}$

Answer 1

Answer: (A), (C), (D)

Energy released during process
$Q =\delta c ^{2}$ .... (1)
$\because$ momentum is conserved in process.
$\Rightarrow 0= m _{ p } v _{ p }- m _{ Q } . v _{ Q }$
$\Rightarrow \frac{ v _{ P }}{ v _{ Q }}=\frac{ m _{ Q }}{ m _{ P }}$ ....(2)
$E _{ P }=\frac{1}{2} m _{ p } v _{ P }^{2}$
$E _{ Q }=\frac{1}{2} m _{ Q } v _{ Q }^{2}$
$\Rightarrow \frac{ E _{ p }}{ E _{ Q }}=\frac{ m _{ Q }}{ m _{ p }}$ .... (3)
Solving (1) and (3),
$E_{P}=\frac{m_{Q}}{m_{P}+m_{Q}} \delta c^{2}$
$E_{Q}=\frac{m_{p}}{m_{p}+m_{Q}} \delta c^{2}$
Momentum $P=\sqrt{2 m_{P} E_{p}}=\sqrt{2 m_{Q} E_{Q}}$
$=\sqrt{2 m _{ p } \frac{ m _{ Q }}{ m _{ p }+ m _{ Q }} \cdot \delta c ^{2}}= c \sqrt{\frac{2 m _{ p } m _{ Q }}{ m _{ p }+ m _{ Q }} \cdot \delta}$