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A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $\mathrm{M}(\mathrm{A}, \mathrm{Z})$ denotes the mass of a single neutral atom of an element with mass number A and atomic number Z, then the minimal condition that the $\beta$ decay
$$
\mathrm{X}_{\mathrm{Z}}^{\mathrm{A}} \rightarrow \mathrm{Y}_{\mathrm{Z}+1}^{\mathrm{A}}+\beta^{-}+\overline{\mathrm{v}}_{\mathrm{e}}
$$
will occur is $\left(\mathrm{m}_{\mathrm{e}}\right.$ denotes the mass of the $\beta$ particle and the neutrino mass $\mathrm{m}_{\mathrm{v}}$ can be neglected):
Options:
$$
\mathrm{X}_{\mathrm{Z}}^{\mathrm{A}} \rightarrow \mathrm{Y}_{\mathrm{Z}+1}^{\mathrm{A}}+\beta^{-}+\overline{\mathrm{v}}_{\mathrm{e}}
$$
will occur is $\left(\mathrm{m}_{\mathrm{e}}\right.$ denotes the mass of the $\beta$ particle and the neutrino mass $\mathrm{m}_{\mathrm{v}}$ can be neglected):
Solution:
1753 Upvotes
Verified Answer
The correct answer is:
$M(A, Z)>\mathbb{M}(A, Z+1)+m_{e}$
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