Question

The binding energies per nucleon for deuteron ${(}_1{H}^2)$ and helium ${(}_{2}H{e}^4)$ are 1.1 MeV and 7.0 MeV, respectively. Calculate the energy released when two deuterons fuse to form a helium nucleus ${(}_{2}H{e}^4)$

Answer 1

The equation of fusion is represented as:
${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^4+Energy$

The binding energy per nucleon of deuteron is $1.1MeV$.

Therefore, the net initial binding energy $=4\times 1.1=4.4MeV$

The binding energy per nucleon of ${}_{2}H{e}^4$ is $7.0MeV$.

Therefore, the net final binding energy is $=4\times 7.0=28.0MeV$
Hence, energy released $=28.0MeV-4.4MeV=23.6MeV$