The binding energy per nucleon of deuteron $\left({ }_1^2 \mathrm{H}\right)$ and helium nucleus $\left({…

The binding energy per nucleon of deuteron $\left({ }_1^2 \mathrm{H}\right)$ and helium nucleus $\left({ }_2^4 \mathrm{He}\right)$ is $1.1 \mathrm{MeV}$ and $7 \mathrm{MeV}$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
  1. $13.9 \mathrm{MeV}$
  2. $26.9 \mathrm{MeV}$
  3. $23.6 \mathrm{MeV}$
  4. $19.2 \mathrm{MeV}$

Solution

Energy released $=$ total binding energy of product $-$ total binding energy of reactants $\Rightarrow 28-(2 \times 2.2)=28-4.4=236 \mathrm{MeV}$

Asked in: JEE Main 2004

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