Chapter 31, Problem 1P

To determine

The condition of the aluminiumif it absorbs the neutron and what does it becomes. Also determine if it decay by ${\mathrm{\beta }}^{+}$ for ${\mathrm{\beta }}^{-}$ and find the product nucleus.

Answer to Problem 1P

The new element is ${}_{14}^{28}\text{Si}$ and the product is silicon after the ${\mathrm{\beta }}^{-}$ emission.

Explanation of Solution

Given Information:

The natural aluminium is all ${}_{13}^{27}\text{Al}$ .

Calculation:

Consider the absorption of the neutron means increase in the atomic weight by the one unit and the symbol of the neutron is ${}_0^1\text{n}$ .

Thus, the absorption of the neutron ${}_{13}^{27}\text{Al}$ is,

  ${}_{13}^{27}{\text{Al}}_0^1\text{n}{\to }_{13}^{28}\text{Al}$

Thus, the atom decay by the ${\mathrm{\beta }}^{-}$ emission as the nucleus has the extra neutron as,

  ${}_{13}^{28}\text{Al}{\to }_{-1}^0\mathrm{\beta }{+}_{14}^{28}\text{Si}+\text{<mover accent="true"><mo>v</mo><mo>¯</mo></mover>}$

Conclusion:

Thus, the new element is ${}_{14}^{28}\text{Si}$ and the product is silicon after the ${\mathrm{\beta }}^{-}$ emission.