Question

If a nucleus P converts into a nucleus Q by the decay of one alpha particle and two $\beta^-$ particles, then the nuclei P and Q are

• A. Isotopes
• B. Isobars
• C. Isotones
• D. Isomers

Hint:

Remember the changes in A (mass number) and Z (atomic number) for different types of radioactive decay: Alpha decay ($\alpha$): A decreases by 4, Z decreases by 2. Beta-minus decay ($\beta^-$): A is unchanged, Z increases by 1. Beta-plus decay ($\beta^+$): A is unchanged, Z decreases by 1. Gamma decay ($\gamma$): A and Z are unchanged.

Answer 1

The Correct Option is A

Let the parent nucleus P have mass number A and atomic number Z. We can represent it as $^A_Z P$.
Step 1: Alpha decay.
An alpha particle is a helium nucleus, represented as $^4_2 \alpha$.
When P undergoes alpha decay, its mass number decreases by 4 and its atomic number decreases by 2. Let the intermediate nucleus be X.
$^A_Z P \rightarrow ^{A-4}_{Z-2} X + ^4_2 \alpha$.
Step 2: Two beta-minus decays.
A beta-minus ($\beta^-$) particle is an electron, represented as $^0_{-1} e$.
In a $\beta^-$ decay, a neutron in the nucleus converts into a proton and an electron. This increases the atomic number by 1 and leaves the mass number unchanged.
The intermediate nucleus X undergoes two successive $\beta^-$ decays to become the final nucleus Q.
First $\beta^-$ decay: $^{A-4}_{Z-2} X \rightarrow ^{A-4}_{Z-2+1} Y + ^0_{-1} e \implies ^{A-4}_{Z-1} Y$.
Second $\beta^-$ decay: $^{A-4}_{Z-1} Y \rightarrow ^{A-4}_{Z-1+1} Q + ^0_{-1} e \implies ^{A-4}_{Z} Q$.
Step 3: Compare the parent nucleus P and the final nucleus Q.
Parent nucleus: $^A_Z P$.
Final nucleus: $^{A-4}_{Z} Q$.
Nuclei that have the same atomic number (Z) but different mass numbers (A) are called isotopes.
Since both P and Q have the same atomic number Z, they are isotopes.
% Quick tip