Question

A \(_{92}^{238}U\) nucleus decays to a \(_{82}^{206}Pb\) nucleus. The number of \( \alpha \) and \( \beta^- \) particles emitted are?

• A. \( 6 \) and \( 2 \)
• B. \( 3 \) and \( 3 \)
• C. \( 2 \) and \( 6 \)
• D. \( 3 \) and \( 4 \)

Hint:

In radioactive decay, \( \alpha \)-decay reduces the atomic number by 2 and mass number by 4 per particle, while \( \beta^- \)-decay increases the atomic number by 1 without changing the mass number.

Answer 1

The Correct Option is A

Step 1: Understanding the Alpha (\( \alpha \)) Decay Contribution 
Alpha (\( \alpha \)) decay reduces the atomic number (\( Z \)) by 2 and the mass number (\( A \)) by 4 per emitted \( \alpha \) particle. The given reaction is: \[ _{92}^{238}U \rightarrow _{82}^{206}Pb + x \alpha + y \beta^- \] Since the mass number decreases from 238 to 206, the total mass loss is: \[ 238 - 206 = 32 \] Since each \( \alpha \) particle reduces \( A \) by 4, the number of emitted \( \alpha \) particles is: \[ x = \frac{32}{4} = 8 \] Step 2: Understanding the Beta (\( \beta^- \)) Decay Contribution 
Beta-minus (\( \beta^- \)) decay increases the atomic number (\( Z \)) by 1 without changing the mass number. The total decrease in atomic number is: \[ 92 - 82 = 10 \] Each \( \alpha \) decay reduces \( Z \) by 2, contributing: \[ 8 \times 2 = 16 \] Since the final change in \( Z \) should be 10, the number of \( \beta^- \) emissions must compensate: \[ y = 16 - 10 = 6 \] Thus, the number of \( \alpha \) and \( \beta^- \) particles emitted are 6 and 2.