Question

In the fusion reaction ${}_1{H}^2{+}_1{H}^2{⟶}_2{He}^4{+}_0{n}^1$, the masses of deuteron helium and neutron expresses in amu are $2.05$,$3.017$ and $1.009$, respectively. If $1Kg$ of deuterium undergoes complete fusion, then find the amount of total energy released.
$1amu=931.5MeV/c^2$

Answer 1

${}_1^{2}H+{}_1^{2}H\to {}_2^{4}He+{}_0^{1}n$

Mass of deuteron ${}_1^{2}H=2.05amu$

Mass of helium ${}_2^{4}He=3.017amu$

Mass of neutron ${}_0^{1}n=1.009amu$

Energy released = total binding energy of products – total binding energy of products

$Q=\left(\right(4\times 3.017+1\times 1.009)-2(2\times 2.05)$

$=13.077-8.2$

$=4.877amu$

$=4.877\times 931.5$

$=4542.9MeV$