Question

When will $\gamma \text{- decay}$ take place?
A. Prior to alpha decay.
B. Prior to beta decay.
C. Prior to positron decay.
D. Due to the-excitement of nuclear levels.

Answer 1

We must know that radioactive elements are unstable and emit radiation to achieve states of greater stability. So, the parent nucleus emits $\alpha, \beta \text{ and }\gamma$ particles to release the excess energy. So, the emission of $\gamma$- particles to achieve stability by a nucleus is known as $\gamma \text{- decay}$.

Complete answer:
Basically, the emission photons by means of $\gamma \text{- rays}$ from a radioactive nucleus is known as $\gamma \text{- decay}$. During a $\gamma \text{- decay}$, an excited daughter nucleus releases high energy photons in the range of MeV and de-excite to a stable energy level.
That means, after $\alpha \text{ and }\beta$ decay, the daughter nucleus will be still in an excited state. So, for achieving its greater stability, it emits one or more $\gamma \text{- ray}$ photons.
Now, the general expression for $\gamma \text{- decay}$ is,
$_{Z}{{X}^{A}}{{\to }_{Z}}{{Y}^{A}}+\gamma$
Where, X is the daughter nuclei formed by other decays and Y is the stable product after $\gamma \text{- decay}$.
As an example, the $\beta \text{- decay}$of $_{27}C{{o}^{60}}$ transforms into an exciting $_{28}N{{i}^{60}}$ nucleus. It reaches the ground state by emitting $\gamma \text{- rays}$. The representation for this reaction will be as follows,

& _{27}C{{o}^{60}}{{\to }_{28}}N{{i}^{{{60}^{**}}}}{{+}_{-1}}{{e}^{0}} \\\ & _{27}C{{o}^{60}}{{\to }_{28}}N{{i}^{{{60}^{*}}}}+{{E}_{\gamma }}\text{ (=1}\text{.17MeV)} \\\ & _{27}C{{o}^{60}}{{\to }_{28}}Ni+{{E}_{\gamma }}\text{ (=1}\text{.33MeV)} \\\ \end{aligned}$$ Here, $${{E}_{\gamma }}$$ is the energy emitted by $\gamma \text{- decay}$. Now, we can conclude that $\gamma \text{- decay}$ will only take place after $\alpha \text{ and }\beta $ decay. So, $\gamma \text{- decay}$ occurs when an excited nucleus makes a transition from a higher energy level to a lower state of energy. That means due to the-excitement of nuclear levels. **So, the correct answer is “Option D”.** **Note:** We must know that as gamma rays are electromagnetic waves, they travel with the speed of light. Gamma rays are used in radiotherapy to treat tumors and cancers, sterilizing medical equipment, etc. Also, they are used to look for distant gamma rays sources in astronomy and to develop nuclear reactors and bombs.