A proton is bombarded on a stationary lithium nucleus. As a result of the collision two  $\alpha$ – particles are produced. If the direction of motion of the $\alpha$ – particles with the initial direction of motion makes an angle $co{s}^{-1}(1/4),$ then the kinetic energy of the striking proton is [Given, binding energies per nucleon of $L{i}^7=5.60MeV$  and  $H{e}^4=7.06MeV,$ $m_{proton}\approx m_{neutron}$]

(1) Q value of reaction is,
Q = (2 x 4 x 7.06 – 7 x 5.6)MeV = 17.28 MeV

Applying conversation of energy for collision,
Kp + Q = $K_{\alpha }$   ……(i)
(Here, K_p and $K_{\alpha }$ are the kinetic energies of proton and – particle respectively)
From the conservation of linear momentum (As there is no external force) ….(ii)
$[HereP=\sqrt{2mk}]$
$\Rightarrow K_p=16K_{\alpha }{\cos }^2\theta =\left(16K_{\alpha }\right){\left(\frac{1}{4}\right)}^2(as{m}_{\alpha }=4m_p)$
$\therefore K_{\alpha }=K_p\mathrm{....}\left(iii\right)$
solving equations (i) and (iii) with Q = 17.28MeV
we get       K_p = 17.28MeV