Question

A radioactive nucleus \({}_{Z}^{A}X\) undergoes spontaneous decay in the sequence \({}_{Z}^{A}X\rightarrow {}_{Z-1}^{\square }B\rightarrow {}_{Z-3}^{\square }C\rightarrow {}_{Z-2}^{\square }D\) where \(Z\) is the atomic number of element. The possible decay particles in the sequence are

• A. \(\alpha ,\beta ^{+},\beta\)
• B. \(\beta ^{+},\alpha ,\beta\)
• C. \(\beta ,\alpha ,\beta ^{+}\)
• D. \(\alpha ,\beta ,\beta ^{+}\)

Hint:

Track the change in atomic number at each step: positron emission lowers Z by 1, alpha lowers it by 2 and beta raises it by 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Alpha emission reduces Z by 2 and A by 4. Beta-minus emission increases Z by 1. Positron (beta-plus) emission decreases Z by 1. A stays the same for beta decays.

Step 2: Step 1: Z to Z-1
Z falls by 1 and the mass number is unchanged, so this is \(\beta^{+}\) emission.

Step 3: Step 2: Z-1 to Z-3
Z falls by 2, so this is an alpha particle emission.

Step 4: Step 3: Z-3 to Z-2
Z rises by 1, so this is \(\beta\) (beta-minus) emission.

Step 5: Result
The sequence is \(\beta^{+},\alpha,\beta\), option (B).

Final Answer:
The decay particles are beta-plus, alpha and beta, in that order. \[ \boxed{\text{(B) }\beta^{+},\alpha,\beta} \]