Question

If \[_{{\text{43}}}{\text{T}}{{\text{c}}^{98}}\] undergoes \[\gamma \] decay. How would you determine the resulting nucleus?

Answer 1

Hint: \[\gamma \] rays are a form of energy. They are electromagnetic radiation with short wavelengths. They do not have any mass and charge. Their emission will not affect atomic number and atomic mass of the species undergoing decay.

Complete step by step answer:
Some radioactive nuclei release excess of energy in the form of \[\gamma \] rays. They are emitted due to energy rearrangements in the nucleus. These rays are a form of pure energy with no mass and no charge. Therefore, their emission can not change the number of protons and neutrons in the nucleus.
The mentioned \[\gamma \] decay can be represented in the form of equation below:
$_{43}Tc^{98}\to_{43}Tc^{98}+\gamma$
In this equation, \[*\] represents an excited state of isotope of atom ‘technetium’ \[_{{\text{43}}}{\text{T}}{{\text{c}}^{98}}\].
Hence, the resulting nucleus will remain as \[_{{\text{43}}}{\text{T}}{{\text{c}}^{98}}\].

Additional Information:
Generally, all other forms of decay (alpha and beta) accompanied by gamma decay. This represents release of energy. A radioactive nucleus can change to other species by emitting \[\alpha \] or \[\beta \] rays with release of \[\gamma \] rays. The resulting ‘daughter nucleus’ is usually in an excited state. Then this excited state can convert to a more stable form by releasing \[\gamma \] rays.
In this question, the given species must be in its excited state. An excited state of the nucleus can decay gamma rays to release excess energy. In this way, it reaches a more stable form. This is because we know that any species is more stable in its lower energy form than that in its higher energy form.

Note: We should not confuse the given question with ‘bombardment’ of gamma rays. In this case, an isotope of radioactive element can disintegrate into a different isotope and other subatomic particles.