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Physics · PYQ NEET UG 2026
हिन्दीConsider the following nuclear reaction : $^{238}U \rightarrow ^{234}Th + ^4He Take masses of $^{238}U,^{234}Thand^4Heas238.050 u,234.043 uand4.003 u,respectively.TheQvalueforthereaction,inkeV$, is : [Given : $1 u = 931.5 MeV c^{-2}$]
Correct answer · Option 1
3726
Explanation
Mass defect $\Delta m = 238.050 - (234.043 + 4.003) = 0.004\text{ u}$. $Q = 0.004 \times 931.5\text{ MeV} = 3.726\text{ MeV} = 3726\text{ keV}$.
