\(_{92}^{238}A\to\,_{90}^{234}B+\,_{2}^{4}D+Q\)
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of \(_{92}^{238}A=238.05079\times931.5\ \mathrm{MeV/c^2}\), mass of \(_{90}^{234}B=234.04363\times931.5\ \mathrm{MeV/c^2}\), mass of \(_{2}^{4}D=4.00260\times931.5\ \mathrm{MeV/c^2}\)]