Answer
a) $\underline{\color{red}{\text{ Gamma decay }}}$
b) $\underline{\color{red}{\text{ $\beta^-$ decay }}}$
c) $\underline{\color{red}{\text{ None}}}$
Work Step by Step
Let's analyze each of the nuclei in the given figure and determine the possible decay processes based on the arrangement of protons and neutrons.
$$\color{blue}{\bf [a]}$$
In the first diagram (a), we can observe that one of the protons is in an excited state, occupying a higher energy level ($n = 3$) than the ground state ($n = 2$). Since this proton is in an excited state, gamma decay is possible. $\underline{\color{red}{\text{ Gamma decay }}}$ occurs when an excited nucleon (proton or neutron) drops to a lower energy level by emitting a gamma photon. In this case, the excited proton can undergo gamma decay and drop from the $n = 3$ level to the $n = 2$ level, emitting a gamma photon in the process.
$$\color{blue}{\bf [b]}$$
In the second diagram (b), there is an imbalance between the number of neutrons and protons. There are more neutrons than protons, which means the nucleus is neutron-rich. $\underline{\color{red}{\text{ $\beta^-$ decay }}}$ is a process that occurs in neutron-rich nuclei, where a neutron transforms into a proton, emitting an electron (beta particle) and an antineutrino. In this case, beta-minus decay can occur to change one of the neutrons into a protonز
$$\color{blue}{\bf [c]}$$
In the third diagram (c), all nucleons are in their ground-state energy levels. Since there are no excited nucleons. So, $\underline{\color{red}{\text{ there is no decay allowed}}}$. This configuration is stable, meaning there is no need for the nucleus to undergo any decay process.
