JEE · NEET Previous Year Question

${}_{92}^{238}A\to {}_{90}^{234}B+{}_2^{4}D+Q$ In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of ${}_{92}^{238}A=238.05079\times 931.5\text{MeV}{c}^{-2}$, mass of ${}_{90}^{234}B=234.04363\times$ 931. $5\text{MeVc}{}^{-2}$, mass of ${}_2^{4}D=4.00260\times 931.5\text{MeV}{c}^{-2}$ ]

Topic: JEE PYQ

Answer: D. Full step-by-step solution below.