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 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 \(n_\beta\):
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Alpha decay decreases mass number by 4 and atomic number by 2; beta decay increases atomic number by 1.
Updated On: Jun 19, 2026
- \(n_\alpha = 8, n_\beta = 8\)
- \(n_\alpha = 6, n_\beta = 6\)
- \(n_\alpha = 8, n_\beta = 6\)
- \(n_\alpha = 6, n_\beta = 8\)
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The Correct Option is C
Solution and Explanation
Step 1: Use mass number change.
Each alpha decay reduces mass number by 4: \[ 238 - 206 = 32 \] \[ n_\alpha = \frac{32}{4} = 8 \]
Step 2: Use atomic number change.
Atomic numbers: \[ U = 92,\quad Pb = 82 \] Change: \[ 92 - 82 = 10 \]
Step 3: Account for alpha effect on Z.
Each alpha reduces Z by 2: \[ 8 \times 2 = 16 \] So Z becomes too low by 6, hence beta decays increase Z by 1 each.
Step 4: Determine beta decays.
\[ n_\beta = 6 \]
Step 5: Final consistency check.
Mass and atomic numbers match final nucleus Pb-206.
Step 6: Final conclusion.
\[ \boxed{n_\alpha = 8,\; n_\beta = 6} \]
Each alpha decay reduces mass number by 4: \[ 238 - 206 = 32 \] \[ n_\alpha = \frac{32}{4} = 8 \]
Step 2: Use atomic number change.
Atomic numbers: \[ U = 92,\quad Pb = 82 \] Change: \[ 92 - 82 = 10 \]
Step 3: Account for alpha effect on Z.
Each alpha reduces Z by 2: \[ 8 \times 2 = 16 \] So Z becomes too low by 6, hence beta decays increase Z by 1 each.
Step 4: Determine beta decays.
\[ n_\beta = 6 \]
Step 5: Final consistency check.
Mass and atomic numbers match final nucleus Pb-206.
Step 6: Final conclusion.
\[ \boxed{n_\alpha = 8,\; n_\beta = 6} \]
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