
Answer & explanation
Correct answer: option 2
$\text{Charge on proton } q_p = e \text{Charge on }\alpha \text{particle is }q_{\alpha} = 2e$
$ \text{From Energy Conservation }E_i = E_f$
$\text{Linear Momentum of proton and }\alpha \text{particle are equal to p}$
$ K_p = \frac{p^2}{2m_p}$
$K_{\alpha} = \frac{p^2}{2m_\alpha} = \frac{p^2}{8m_p} = \frac{K_p}{4}$
$ K_f = K_p + K_{\alpha} = \frac{5}{4} K_p$
$ U_i + K_i = U_f + K_f $
$ \frac{ke.2e}{r} + 0 = 0 + \frac{5}{4} K_p$
$ K_f = \frac{8ke^2}{5r}$