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# A proton is bombarded on a stationary lithium nucleus. As a result of collision two $\;\alpha\;$ particles are produced . If direction of motion of $\;\alpha \;$ particle with initial direction of motion makes an angle $\;cos^{-1}(\large\frac{1}{4})\;$ find kinetic energy of striking proton . Given binding energies per nucleon of $\;Li^{7}\;$ & $\;He^{4}\;$ are $5.60 MeV$ and $7.06 MeV$ respectively

$(a)\;19.23\;MeV\qquad(b)\;17.28\;MeV\qquad(c)\;20.16\;MeV\qquad(d)\;12.23\;MeV$

Answer : (b) $\;17.28\;MeV$
$Q=2K_{\alpha}-K_{p}$
Therefore , $Q=(2 \times 4 \times 7.06 -7 \times 5.6)$
$=17.28 MeV$
Where $\;K_{p}\;$ & $\;K_{\alpha}\;$ are kinetic energies of photon and $\;\alpha \;$ particle respectively . By conservation of linear momentum
$\sqrt{2m_{p} K_{p}}=2 \sqrt{2 m_{\alpha}K_{\alpha}} cos \theta$
$K_{p}=16 K_{\alpha} cos^{2} \theta =K_{\alpha}$
$K_{p}=17.28 \;MeV$