Binding energy per nucleon of ${}_1^{2}H$ and ${}_2^{4}He$ are $1.1MeV$ and $7MeV$ respectively. Calculate the amount of energy released in the following process:
${}_1^{2}H{+}_1^{2}H⟶$ ${}_2^{4}He$

Q. The binding energy per nucleon for ${}_1^{2}H$ and ${}_2^{4}He$ respectively are $1.1MeV$ and $7.1MeV$. The energy released in $MeV$ when two ${}_1^{2}H$ nuclei fuse to form ${}_2^{4}He$ is :

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

Q. In the reaction ${}_1^{2}H{+}_1^{3}H⟶$ ${}_2^{4}He{+}_0^{1}n$, if the binding energies of ${}_1^{2}H$, ${}_1^{3}H$ and ${}_2^{4}He$ are $a$, $b$ and $c$ (in MeV) respectively, then energy (in MeV) released in this reaction is:

Q. Binding energy per nucleon of ${}_1{H}^2$ and ${}_{2}H{e}^4$ are 1.1 MeV and 7.0 MeV respectively. Energy released in the process ${}_1{H}^2{+}_1{H}^2{=}_{2}H{e}^4$ is

Q. Binding energy per nucleon of ${}_1{H}^2$ and ${}_{2}H{e}^4$ are $1.1MeV$ and $7.0MeV$, respectively. Energy released in the process ${}_1{H}^2{+}_1{H}^2{⟶}_{2}H{e}^4$ is