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 $M_{P}, M_{Q} and M_{N}$ 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?

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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 $C \sqrt{2 μδ}$ , where $μ = \frac{M_{P} M_{Q}}{(M_{P} + M_{Q})}$

answer is A.

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Detailed Solution

$N \to P + Q$

$\begin{array}{l} δ = M_{N} - M_{P} - M_{Q} Q = δ c^{2} \\ ∴ E_{P} + E_{Q} = δ C^{2} \end{array}$

Since there are no external forces acting on the system. Momentum has to be conserved.

$\begin{array}{l} P_{P} = P_{Q} = P \\ M_{P} V_{P} = M_{Q} V_{Q} \\ \frac{V_{P}}{V_{Q}} = \frac{M_{Q}}{M_{P}} \\ K . E = \frac{P^{2}}{2 M} ∴ \frac{P^{2}}{2 M_{P}} + \frac{P^{2}}{2 M_{Q}} = δ c^{2} \\ P^{2} = \frac{2 M_{P} M_{Q} \cdot {Sc}^{2}}{M_{P} + M_{Q}} \\ P = \sqrt{2 μ δ c^{2}} \\ P = c \sqrt{2 μ δ} \end{array}$