Question

A radioactive element \( {}^{242}_{92}\text{X} \) emits two \( \alpha \)-particles, one electron and two positrons. The transformed nucleus is represented by \( {}^{234}_{P}\text{Y} \). The value of \( P \) is

• A. 85
• B. 87
• C. 92
• D. 96

Hint:

To solve nuclear decay problems quickly, remember that charge is strictly conserved. You can write a simple algebraic equation of the form \( Z_{\text{initial}} = Z_{\text{final}} + \sum Z_{\text{emitted}} \) to directly find the unknown atomic number.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Nuclear transformations involve changes in atomic number (\(Z\)) and mass number (\(A\)).
- \(\alpha\)-particle (\(^4_2He\)): Decreases \(A\) by 4 and \(Z\) by 2.
- \(\beta^-\)-particle (electron, \(^0_{-1}e\)): Increases \(Z\) by 1, \(A\) remains unchanged.
- \(\beta^+\)-particle (positron, \(^0_{+1}e\)): Decreases \(Z\) by 1, \(A\) remains unchanged.
Total change in \(Z\) and \(A\) is the sum of changes from all emitted particles.
Step 2: Key Formula or Approach:
Atomic number of final nucleus \(P = Z_{initial} + \sum \Delta Z\).
Mass number check: \(A_{final} = A_{initial} + \sum \Delta A\).
Step 3: Detailed Explanation:
Initial state: \(Z = 92, A = 242\).
Step 1: Two \(\alpha\)-particles (\(2 \times ^4_2He\)).
Change in \(Z = 2 \times (-2) = -4\).
Change in \(A = 2 \times (-4) = -8\).
New mass number = \(242 - 8 = 234\) (matches the question).
Step 2: One electron (\(1 \times \beta^-\)).
Change in \(Z = +1\).
Step 3: Two positrons (\(2 \times \beta^+\)).
Change in \(Z = 2 \times (-1) = -2\).
Total change in \(Z\):
\[\Delta Z_{total} = -4 + 1 - 2 = -5\]
Final atomic number \(P\):
\[P = 92 + (-5) = 87\]
This matches option (B).
Step 4: Final Answer:
The final atomic number \(P\) is 87, calculated by accounting for the shifts caused by two alphas (-4), one electron (+1), and two positrons (-2).