Question

A heavy nucleus $N$, at rest, undergoes fission $N\to P+Q$, 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 $P=\sqrt{2m_P{E}_p}=\sqrt{2m_Q{E}_Q}$
$=\sqrt{2m_p\frac{m_Q}{m_p+m_Q}\cdot \delta c^2}=c\sqrt{\frac{2m_p{m}_Q}{m_p+m_Q}\cdot \delta }$