A common example of alpha decay is 23892U → ^ 234 90Th + _ 2 He ^ 4 + Q Given: ^ 238 92U = 238 . 05060u ^ 234 90Th = 234 . 04360u ^ 4 2He = 4 . 00260u and 1u = 931 . 5 MeV c ^ 2 The energy released (Q) during the alpha

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The formula to calculate the Q-value of the given nuclear reaction can be written as

$Q = \left({m}_{U} - {m}_{Th} - {m}_{He}\right){c}^{2} . . . \left(1\right)$

Substitute the values of the known parameters into equation (1) to calculate the required Q-value for the reaction.

$\begin{cases}Q & = & \left(238 . 05060u - 234 . 04360u - 4 . 00260u\right){c}^{2} \\ & = & 0 . 0044{c}^{2} \times 931 . 5MeV / {c}^{2} \\ & \approx & 4MeV\end{cases}$

A common example of alpha decay is 23892U → ^ 234 90Th + _ 2 He ^ 4...

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