$(a)\;23.6 MeV\qquad(b)\;22.8 MeV\qquad(c)\;19 MeV\qquad(d)\;23 MeV$

Answer : $\;23.6 \;MeV$

If two deuteron nuclei react to form a single helium nucleus ,$^2_1 H + ^2_1 H \rightarrow ^4_2 H + Q$. We need to calculate Q.

Energy released = total binding energy of product - total binding energy of reactants

$\Rightarrow Q = 4\times 7 - 2 \times ( 2\times 1.1 ) = 28 - 4.4=23.6 \;MeV \;.$

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