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

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Answer : (b) $\;17.28\;MeV$

Q value of reaction is

$Q=2K_{\alpha}-K_{p}$

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

$\sqrt{2m_{p} K_{p}}=2 \sqrt{2 m_{\alpha}K_{\alpha}} cos \theta$

$K_{p}=16 K_{\alpha} cos^{2} \theta =K_{\alpha}$

$K_{p}=17.28 \;MeV$

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