Question
Consider the following nuclear reaction
$$\ ^{238}U \longrightarrow \ ^{234}Th + \ ^{4}He$$
Take masses of $\ ^{238}U,\ ^{234}Th$ and $\ ^{4}He$ as $238.050u,234.043u$ and 4.003 u , respectively. The $Q$ value for the reaction, in keV, is:
[Given: $1u = 931.5MeVc^{- 2}$ ]
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Solution
Sol. [Key Concept: Nuclear reaction — conservation of mass and charge number]
$Q$-value for the nuclear reaction is given by, $Q = \Delta mc^{2}$
$$\ ^{238}U \longrightarrow \ ^{234}Th + \ ^{4}He\text{~}$$
$$\text{has mass defect}\text{~}\Delta m = m_{\text{reactants}\text{~}} - m_{\text{products}\text{~}}$$
$$\begin{matrix} & \ \Rightarrow \ \Delta m = 238.050 - (234.043 + 4.003) = 0.004u \\ & \ \therefore\ Q = 0.004 \times 931.5MeV = 3.726MeV = 3726keV \end{matrix}$$
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