Q 12-13-076JEE MainJEE Main 2023 (12 Apr, Shift 1)Medium
A common example of alpha decay is
$$^{238}_{92}\text{U}\to{}^{234}_{90}\text{Th}+{}^4_2\text{He}+Q$$
Given: $^{238}_{92}\text{U}=238.05060$ u, $^{234}_{90}\text{Th}=234.04360$ u, $^4_2\text{He}=4.00260$ u and $1\ \text{u}=931.5\ \text{MeV}/c^2$.
The energy released ($Q$) during the alpha decay of $^{238}_{92}\text{U}$ is ______ MeV.
Numerical value type. Enter your answer.
Answer: 4
$\Delta m=238.05060-234.04360-4.00260=0.00440\ \text{u}$.
$Q=0.0044\times931.5\approx4.1\ \text{MeV}\approx4\ \text{MeV}$.
Solution by Sreeraj P, M.Sc Physics