Question

A radioactive isotope \(U^{238}\) decays into \(Pb^{206}\) in a series by emission of \(n_\alpha\) alpha particles and \(n_\beta\) beta particles. Find \(n_\alpha\) and \(n_\beta\):

• A. \(n_\alpha = 8, n_\beta = 8\)
• B. \(n_\alpha = 6, n_\beta = 6\)
• C. \(n_\alpha = 8, n_\beta = 6\)
• D. \(n_\alpha = 6, n_\beta = 8\)

Hint:

Alpha decay decreases mass number by 4 and atomic number by 2; beta decay increases atomic number by 1.

Answer 1

The Correct Option is C

Step 1: Mass number reduction.
Each α decay lowers A by 4: 238 - 206 = 32 → n_α = 32/4 = 8.

Step 2: Atomic number change.
Z drops from 92 to 82, a net decrease of 10.

Step 3: Alpha contribution to Z.
8 α decays lower Z by 16, overshooting by 6; each β⁻ decay raises Z by 1.

Step 4: Beta count.
n_β = 6 compensates the excess drop.

Step 5: Consistency verification.
Final mass and atomic numbers match ²⁰⁶Pb.

Step 6: Conclusion.
n_α = 8, n_β = 6.