^{27}_{13}A is a stable element. ^{29}_{13}A | vedclass

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(B) The stable isotope of Aluminum is $^{27}_{13}A$,which has $13$ protons and $14$ neutrons. The ratio of neutrons to protons $(n/p)$ is $14/13 \appr

Vedclass · Accepted Answer

$\beta$-emission

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(B) The stable isotope of Aluminum is $^{27}_{13}A$,which has $13$ protons and $14$ neutrons. The ratio of neutrons to protons $(n/p)$ is $14/13 \approx 1.077$. For the isotope $^{29}_{13}A$,the number of neutrons is $16$. The $n/p$ ratio is $16/13 \approx 1.23$. Since this isotope has an excess of neutrons compared to the stable isotope,it will undergo $\beta$-decay to convert a neutron into a proton,thereby increasing the atomic number and moving towards a more stable $n/p$ ratio. The reaction is: $^{29}_{13}A \rightarrow ^{29}_{14}Si + ^{0}_{-1}e$.

$^{27}_{13}A$ is a stable element. $^{29}_{13}A$ is expected to disintegrate by:

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