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Question

Natural aluminum is all 1327Al_{13}^{27} \mathrm{Al}. If it absorbs a neutron, what does it become? Does it decay by β+\beta^{+} or β?\beta^{-} ? What will be the product nucleus?

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If the aluminum isotope absorbs a neutron, the reaction must still follow the conservation of charge and mass number. Hence, the reaction can be written as,

1327Al+01n1328Al\begin{aligned} ^{27}_{13}Al+^1_0n\rightarrow\boxed{^{28}_{13}Al} \end{aligned}

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