Atoms And Nuclei

\(_{92}^{238}A \to _{90}^{234}B + _2^4D + Q\)

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of \({ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},\)

mass of \({ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},\)

mass of \(\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]\)