${ }_{92}^{238}\mathrm{A}{\to }_{90}^{234}\mathrm{B}{+}_2^4\mathrm{D}+\mathrm{Q}$
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of ${ }_{92}^{238}\mathrm{A}=238.05079\times 931.5{\text{\hspace{0.17em}MeV/c}}^2,$
Mass of ${ }_{90}^{234}\mathrm{B}=234.04363\times 931.5{\text{\hspace{0.17em}MeV/c}}^2,$
Mass of ${ }_2^4\mathrm{D}=4.00260\times 931.5{\text{\hspace{0.17em}MeV/c}}^2]$