Question

A radioactive nucleus \( {}^{290}_{92}X \) decays to \( {}^{278}_{87}Y \). The number of \( \alpha \) and \( \beta \) particles emitted during this decay are

• A. 12\( \alpha \) and 1\( \beta^+ \)
• B. 6\( \alpha \) and 1\( \beta^- \)
• C. 3\( \alpha \) and 1\( \beta^+ \)
• D. 3\( \alpha \) and 1\( \beta^- \)

Hint:

For decay problems, remember the effects of \( \alpha \)- and \( \beta \)-decays: - \( \alpha \)-decay reduces both the atomic number by 2 and the mass number by 4. - \( \beta^- \)-decay increases the atomic number by 1 with no change in the mass number.

Answer 1

The Correct Option is D

The decay process can be understood by considering the changes in both the atomic number (Z) and the mass number (A) of the nucleus.
1. The initial nucleus \( {}^{290}_{92}X \) has:
- Atomic number (Z) = 92
- Mass number (A) = 290
2. The final nucleus \( {}^{278}_{87}Y \) has:
- Atomic number (Z) = 87
- Mass number (A) = 278
The change in atomic number is \( 92 - 87 = 5 \), and the change in mass number is \( 290 - 278 = 12 \).
Understanding \( \alpha \) and \( \beta \) emissions:
- An \( \alpha \)-particle emission results in a decrease of 2 in the atomic number (Z) and 4 in the mass number (A).
- A \( \beta^- \)-particle emission results in an increase of 1 in the atomic number (Z) but no change in the mass number (A).
Step-by-step breakdown:
- To decrease the atomic number by 5 and the mass number by 12, we can emit:
- 3 \( \alpha \)-particles, each reducing Z by 2 and A by 4. This accounts for a decrease of 6 in Z and 12 in A.
- 1 \( \beta^- \)-particle, which increases Z by 1, compensating for the 5 total decrease in Z after the 3 \( \alpha \)-particles.
Thus, the correct answer is 3 \( \alpha \)-particles and 1 \( \beta^- \)-particle.