Question

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

• A. $\alpha$-emission
• B. $\beta$-emission
• C. positron emission
• D. proton emission

Answer 1

Answer: (B)

$\beta$-emission

$\to$Stable isotope : ${}_{13}^{27}Al$

n/p ratio = 14/13

$\to$Unstable isotope = ${}_{13}^{29}Al$

n/p ratio = 16/13

Nuclides having n/p ratio greater than 1 undergoes $\beta$-emission to decrease n/p ratio in order to attain belt of stability.