Question

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

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

Answer 1

Answer: (B)

$\beta$-emission

In any atom, protons and neutrons are packed together in a very small space in the nucleus.

• There must be some force of attraction between these particles to maintain the existence of nucleus.
• The columbic force of attraction is not enough to hold many particles in such a small space, as the repulsion force between protons would render the nucleus highly unstable and tend to disintegration.
• Hence, the force of attraction between nucleons which is stronger than the repulsive force between protons is called as nuclear forces.
• Nuclear forces operate within a small distance around 0.10 fermi only.

Those nucleus with n/p ratio one is highly stable. Elements with high n/p ratio are unstable an radioactive in nature. Elements till atomic number 20 are stable with n/p =1, as atomic number increases, n/p ratio increases and instability increases.
Nucleides having $n/p>1$ undergoes $\beta$-emission to decrease $n/p$ ratio in order to attain belt of stability.