A heavy nucleus N at rest, undergoes a reaction N $\to$ P + Q, where P and Q are two lighter nuclei. Let $δ = M_{N} - M_{P} - M_{Q}$ where M_P, M_Qand 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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$\frac{V_{P}}{V_{Q}} = \frac{M_{Q}}{M_{P}}$

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

$E_{P} + E_{Q} = C^{2} δ$

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

answer is A, B, C.

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Detailed Solution

Given $N \to P + Q$
Mass defect $δ = M_{N} - M_{P} - M_{Q}$
$\Rightarrow$ Energy released= $δ C^{2}$
$∴ E_{P} {+E}_{Q} {=C}^{2} δ \to (1)$
From conservation of momentum (linear)
M_PV_P=M_QV_Q
$∴ \frac{V_{P}}{V_{Q}} = \frac{M_{P}}{M_{Q}} \to (2)$
Eq (1) can become
$\frac{p^{2}}{2 M_{P}} + \frac{p^{2}}{2 M_{Q}} = C^{2} δ \Rightarrow \frac{p^{2}}{2} [\frac{1}{M_{P}} + \frac{1}{M_{Q}}] = C^{2} δ \frac{p^{2}}{2} . \frac{M_{P} + M_{Q}}{M_{P} . M_{Q}} = C^{2} δ \frac{p^{2}}{2} . μ = C^{2} δ where μ = \frac{M_{P} +M {}_{Q}}{M_{P} . M_{Q}} = Re d u c e d m a s s p^{2} = 2 μ C^{2} δ p = C \sqrt{2 μ δ}$