_92^238 A →_90^234 B+_2^4 D+Q In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of _92^238 A=238.05079 ×…

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