Question

An isotope of the element lead ($\text{Pb}$) undergoes a nuclear reaction that results in the release of a β-particle, an anti-neutrino, and an element $\text{X}$. Find $\text{X}$.

• A. $20680\mathrm{Hg}$
• B. $21083\mathrm{Bi}$
• C. $21082\mathrm{Pb}$
• D. $21481\mathrm{Tl}$

Answer 1

The correct option is B $21083\mathrm{Bi}$

Given:
The parent nucleus is $21082\mathrm{Pb}$.
The given reaction is an example of β - emission or β - decay because a beta particle, and an anti-neutrino are released along with the daughter nucleus, $AZX$.
In a beta emission, the atomic number of the daughter nucleus will be one more than the atomic number of the parent nucleus, and there is no change in the mass number of the parent nucleus.
$⟹$ The daughter nucleus has $\text{Z}=82+1=83$ and $\text{A}=210$.
So, among the given options, $21083\mathrm{Bi}$ is the daughter nucleus.