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 Question:
The problem requires us to find the atomic number \( P \) of the transformed nucleus \( {}^{234}_{P}\text{Y} \) after the parent nucleus \( {}^{242}_{92}\text{X} \) undergoes radioactive decay by emitting two \( \alpha \)-particles, one electron, and two positrons.

Step 2: Key Formula or Approach:
We will use the laws of conservation of mass number (\( A \)) and charge/atomic number (\( Z \)) in nuclear reactions.
- Emission of an \( \alpha \)-particle (\( {}^4_2\text{He} \)) decreases the mass number by 4 and the atomic number by 2.
- Emission of an electron (\( \beta^- \)-decay, \( {}^0_{-1}e \)) leaves the mass number unchanged and increases the atomic number by 1.
- Emission of a positron (\( \beta^+ \)-decay, \( {}^0_{+1}e \)) leaves the mass number unchanged and decreases the atomic number by 1.

Step 3: Detailed Explanation:
Let the reaction be represented as:
\[ {}^{242}_{92}\text{X} \rightarrow {}^{234}_{P}\text{Y} + 2 \left( {}^{4}_{2}\text{He} \right) + 1 \left( {}^{0}_{-1}e \right) + 2 \left( {}^{0}_{+1}e \right) \]
By applying the law of conservation of atomic number (total charge on both sides must be equal):
\[ 92 = P + 2(2) + 1(-1) + 2(+1) \]
\[ 92 = P + 4 - 1 + 2 \]
\[ 92 = P + 5 \]
\[ P = 92 - 5 = 87 \]

Step 4: Final Answer:
The atomic number \( P \) of the transformed nucleus is 87.