Q 12-13-077JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
$^{238}_{92}A\to{}^{234}_{90}B+{}^4_2D+Q$
In the given nuclear reaction, the approximate amount of energy released will be [Given, mass of $^{238}_{92}A=238.05079\times931.5\ \text{MeV}\,c^{-2}$, mass of $^{234}_{90}B=234.04363\times931.5\ \text{MeV}\,c^{-2}$, mass of $^4_2D=4.00260\times931.5\ \text{MeV}\,c^{-2}$]
Answer: (D) $4.25\ \text{MeV}$
$\Delta m=238.05079-234.04363-4.00260=0.00456\ \text{u}$; $Q=0.00456\times931.5\approx4.25\ \text{MeV}$.
Solution by Sreeraj P, M.Sc Physics