JEE MAIN - Physics (2004 - No. 6)
The binding energy per nucleon of deuteron $$\left( {{}_1^2\,H} \right)$$ and helium nucleus $$\left( {{}_2^4\,He} \right)$$ is $$1.1$$ $$MeV$$ and $$7$$ $$MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
$$23.6\,\,MeV$$
$$26.9\,\,MeV$$
$$13.9\,\,MeV$$
$$19.2\,\,MeV$$
Explicació
The nuclear reaction of process is $$2_1^2H \to {4 \over 2}$$ He
Energy released $$ = 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$$
Energy released $$ = 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$$