$(a)\;29.3 MeV\qquad(b)\;24.4 MeV\qquad(c)\;21.6 MeV\qquad(d)\;18.6 MeV$

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Answer : (c) $\;21.6 MeV$

Explanation :

$_{1}T^{3}+_{1}D^{2} \to _{2}He^{4}+_{0}n^{1}+\varepsilon_{2}$

$\bigtriangleup m=[(3.016+2.014)-(4.0026+1.0086)]$

$=0.0188 a.m.u$

Therefore , energy released

$\varepsilon_{2}=0.01888 \times931 MeV=17.58 MeV$

Total energy released in both processess

$=17.58+4.03=21.611 MeV\;.$

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