A heavy nucleus N at rest, undergoes fission $N \to P + Q$ , where P and Q are two lighter nuclei let $δ = m_{N} - m_{P} - m_{Q}$ where $m_{N}, m_{P}$ and $m_{Q}$ are masses of N, P and Q respectively $E_{P}$ and $E_{Q}$ are the kinetic energy of P and Q. The speeds of P and Q are $V_{P}$ and $V_{Q}$ . If C is the speed of light, which of the following statement(s) is (are) correct?

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$E_{P} + E_{Q} = C^{2} δ$

$E_{P} = (\frac{m_{P}}{m_{P} + m_{Q}}) C^{2} δ$

$\frac{V_{P}}{V_{Q}} = \frac{m_{Q}}{m_{P}}$

The magnitude of momentum for P as well as Q is $C \sqrt{2 μ δ}$ where $μ = \frac{m_{P} m_{Q}}{(m_{P} + m_{Q})}$

answer is A, C, D.

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Detailed Solution

$\begin{array}{l} P_{i} = P_{f} O = P_{P} + P_{Q} = 0 P_{P} = - P_{Q} \\ Energy released \\ Δ m c^{2} = δ c^{2} = k \cdot E_{p} + K \cdot E_{Q} δ C^{2} = E_{P} + E_{Q} \\ K . E = \frac{P^{2}}{2 m} \frac{K \cdot E_{P}}{K \cdot E_{Q}} = \frac{m_{Q}}{m_{P}} \\ \to m_{P} V_{P} = m_{Q} V_{Q} \frac{V_{P}}{V_{Q}} = \frac{m_{Q}}{m_{P}} \\ \to K . E_{P} = (\frac{m_{Q}}{m_{P} + m_{Q}}) δ c^{2} \\ \to \frac{P^{2}}{2 m_{P}} + \frac{P^{2}}{2 m_{Q}} = δ c^{2} \\ P = C \sqrt{2 [\frac{m_{P} m_{Q}}{m_{P} + m_{Q}}] δ} \end{array}$