Question

The number of alpha and beta decays occurred when \(_{92}^{238}U\) changes to \(_{82}^{206}Pb\) are respectively:

• A. \(8,3\)
• B. \(8,6\)
• C. \(4,6\)
• D. \(4,3\)

Hint:

Each \(\alpha\)-decay decreases atomic number by 2 and mass number by 4, whereas each \(\beta^{-}\)-decay increases atomic number by 1 without changing mass number.

Answer 1

The Correct Option is B

Concept: In radioactive decay: \[ \alpha \text{-decay} : \quad A \rightarrow A-4, \quad Z \rightarrow Z-2 \] \[ \beta^- \text{-decay} : \quad A \rightarrow A, \quad Z \rightarrow Z+1 \] Mass number changes only during \(\alpha\)-decay.

Step 1: Find the number of alpha decays.
Let the number of \(\alpha\)-decays be \(n\). Mass number changes from \[ 238 \rightarrow 206. \] Therefore, \[ 238-4n=206 \] \[ 4n=32 \] \[ n=8. \] Hence, \[ \boxed{\text{Number of }\alpha\text{-decays}=8} \]

Step 2: Find atomic number after eight alpha decays.
Initially, \[ Z=92. \] After \(8\) alpha decays, \[ Z=92-16=76. \]

Step 3: Determine beta decays.
Final atomic number is \[ 82. \] Let the number of beta decays be \(m\). \[ 76+m=82 \] \[ m=6. \] \[ \boxed{\text{Number of }\beta\text{-decays}=6} \] Hence, \[ \boxed{(8,6)} \]