Question

A radioactive element $^{242}_{92}X$ emits two $\alpha$ particles, one electron and two positrons. The product nucleus is represented by $^{234}_P Y$. The value of P is

• A. 87
• B. 85
• C. 92
• D. 96

Hint:

$\alpha$ emission reduces Z by 2; $\beta^-$ (electron) increases Z by 1; $\beta^+$ (positron) reduces Z by 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to find the atomic number $Z$ (represented by $P$) of the daughter nucleus after a series of radioactive decays. Conservation of mass number $A$ and charge number $Z$ is used.
Step 2: Key Formula or Approach:
1. $\alpha$-decay: $A \rightarrow A-4, Z \rightarrow Z-2$.
2. $\beta^-$ (electron) decay: $A \rightarrow A, Z \rightarrow Z+1$.
3. $\beta^+$ (positron) decay: $A \rightarrow A, Z \rightarrow Z-1$.
Step 3: Detailed Explanation:
Initial nucleus: $^{242}_{92}X$.
Change in Atomic Number ($Z$):
- Two $\alpha$ particles: $2 \times (-2) = -4$
- One electron ($e^-$): $1 \times (+1) = +1$
- Two positrons ($e^+$): $2 \times (-1) = -2$
Net change in $Z$:
\[ \Delta Z = -4 + 1 - 2 = -5 \]
Final atomic number $P$:
\[ P = 92 - 5 = 87 \]
(Note: Net change in $A = 2 \times (-4) = -8$. Final $A = 242 - 8 = 234$, which matches the given daughter nucleus).
Step 4: Final Answer:
The value of P is 87.