Q 12-13-001NEETNEET 2026Top questionEasy
Consider the following nuclear reaction :
$$^{238}\text{U} \rightarrow {}^{234}\text{Th} + {}^{4}\text{He}$$
Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^{4}\text{He}$ as $238.050$ u, $234.043$ u and $4.003$ u, respectively. The Q value for the reaction, in keV, is :
[Given : $1\ \text{u} = 931.5\ \text{MeV}\,c^{-2}$]
Answer: (A) $3726$
Mass defect:
$$\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}$$
$$Q = \Delta m \times 931.5\ \text{MeV} = 0.004 \times 931.5 = 3.726\ \text{MeV} = 3726\ \text{keV}$$
Solution by Sreeraj P, M.Sc Physics