$(a)\;1.42 MeV\qquad(b)\;3 MeV\qquad(c)\;2 MeV\qquad(d)\;2.82 MeV$

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Answer : (d) $\;2.82 MeV$

Explanation :

We are given $\; K_{\alpha}=K_{\lambda}=0$

So $\;Q=K_{Li}+K_{He}$

Where $\;K_{He}=\large\frac{1}{2}\;M_{He}v_{He}^{2}=\large\frac{1}{2}\;(4.0026 u)\;(1.66\times10^{-27} Kg/u)\times(9.30\times10^{6} m/s)^2$

$=2.87 \times 10^{-13} J =1.8 MeV$

Hence $\;Q=1.02 MeV + 1.8 MeV = 2.82 MeV\;.$

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