Animated Solution for Physics - Atoms and Nuclei: Which of the following processes represent a $\gamma$-decay ?

When a nucleus undergoes radioactive decay, it often ends up in an excited energy state. Just like an electron in an atom transitions to a lower energy level by emitting a photon, an excited nucleus transitions to its ground state (or a lower energy state) by emitting a high-energy photon. This photon is what we call a gamma ray ($\gamma$-ray).

The most crucial property of a gamma ray is that it is a quantum of electromagnetic radiation. It has no rest mass and no electric charge.

Because of this, when a nucleus emits a gamma ray, it does not lose any protons or neutrons. Therefore, both the atomic number ($Z$) and the mass number ($A$) of the nucleus remain completely unchanged. The general equation for gamma decay is written as:

Here, the asterisk ($\ast$) denotes the excited state of the nucleus.

Let's evaluate each option to see which one perfectly describes a gamma decay process, keeping in mind that $A$ and $Z$ must remain constant.

Option (a): In this reaction, the atomic number changes from $Z$ to $Z-1$. This violates the condition for gamma decay.

Option (b): Here, both the mass number and the atomic number are changing. The mass number decreases by 3, and the atomic number decreases by 2. This is clearly not gamma decay.

Option (c): In this process, the parent nucleus ${}_Z^{A}X$ transforms into the exact same nucleus ${}_Z^{A}X$, while emitting a particle $f$. Since neither $A$ nor $Z$ has changed, the emitted particle $f$ must be a gamma photon. This perfectly represents gamma decay.

Option (d): This reaction shows an electron capture process, where the atomic number decreases from $Z$ to $Z-1$. Again, this is not gamma decay.

By systematically checking the conservation of mass and atomic numbers, we can confidently conclude that option (c) is the only process that represents a gamma decay.