Consider the following nuclear reaction ²³⁸U ²³⁴Th + ⁴He Take masses of ²³⁸U, ²³⁴Th and ⁴He as 238.050u, 234.043u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: 1u = 931.5 MeV c⁻² ] (a) 3740 (b) 3726 (c) 3730 (d) 3736

Answer: (b). Δ m = 238.050 - (234.043 + 4.003) = 0.004 u Q = 0.004 × 931.5 = 3.726 MeV So the answer is (b) 3726 keV.

Consider the following nuclear reaction ²³⁸U ²³⁴Th + ⁴He Take masses… | NEET 2026 (Re-neet)

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Nuclear Reactions & Q-value

Consider the following nuclear reaction

${}^{238}\text{U} \longrightarrow {}^{234}\text{Th} + {}^{4}\text{He}$

Take masses of ${}^{238}\text{U}, {}^{234}\text{Th}$ and ${}^{4}\text{He}$ as $238.050\text{u}, 234.043\text{u}$ and 4.003 u, respectively. The $Q$ value for the reaction, in keV, is:

[Given: $1\text{u} = 931.5\text{ MeV c}^{-2}$ ]

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Answer: (b)

$\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}$
$Q = 0.004 \times 931.5 = 3.726\ \text{MeV}$
So the answer is (b) $3726\ \text{keV}$ .

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