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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The magnitude of momentum for P as well as Q is $c \sqrt{2 μδ}, where μ = \frac{M_{P} M_{Q}}{(M_{P} + M_{Q})}$

$E_{P} + E_{Q} = c^{2} δ$

$E_{P} = (\frac{M_{P}}{M_{P} + M_{Q}}) c^{2} δ$

$\frac{v_{P}}{v_{Q}} = \frac{M_{Q}}{M_{P}}$

answer is A, C, D.

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Detailed Solution

For nuclear fission reaction $N \to P + Q$
$Energyreleased = Δ {Mc}^{2} = (M_{N} - M_{P} - M_{Q}) c^{2} = δ c^{2}$
This will be distributed as kinetic energy of P
$and E_{P} + E_{P} = δ c^{2}$ ……….(i)

By Conservation of momentum $V_{P} M_{P} = V_{Q} M_{Q}$

$∴ \frac{V_{p}}{V_{q}} = \frac{M_{p}}{M_{q}}$ ………(ii)
From (i) and (ii) , $E_{P} = \frac{M_{Q}}{(M_{P} + M_{Q})} c^{2} δ$

By equation (i) and (ii) , $\Rightarrow \frac{P^{2}}{2 M_{P}} + \frac{P^{2}}{2 M_{P}} = δ c^{2}$
$\Rightarrow \frac{P^{2}}{2 μ} = δ c^{2} \Rightarrow P = c \sqrt{2 μ δ} [∴ μ = \frac{M_{P} M_{Q}}{M_{P} + M_{Q}}]$