${ }_{92}^{238} \mathrm{~A}|\rightarrow|_{90}^{234} \mathrm{~B}+{ }_2^4 \mathrm{D}+\mathrm{Q}$ In the given nuclear reaction, the approximate amount of energy released will be : [Given, mass of ${ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 \mathrm{MeV} / \mathrm{c}_{\text {mat }}^2$ mass of ${ }_{90}^{234} \mathrm{~B}=234.04363 \times 931.5 \mathrm{MeV} / \mathrm{c}^2$ mass of $\left.{ }_2^4 D=4.00260 \times 931.5 \mathrm{MeV} / \mathrm{c}^2\right]$