Question

$_{88}\text{R}_{\text{a}}^{226}$ is converted into $_{82}\text{P}_{\text{b}}^{206}$ by emission of alpha ( $\alpha$ ) and beta ( $\beta$ ) particles. The number of alpha and beta particles emitted are respectively}

• A. 5, 4
• B. 4, 5
• C. 6, 4
• D. 4, 6

Hint:

For nuclear decay: \[ \alpha : A-4,\ Z-2 \] \[ \beta^- : A\text{ same},\ Z+1 \] Always use mass number first, then atomic number.

Answer 1

The Correct Option is A

Concept:
In radioactive decay:
• each \(\alpha\)-particle decreases mass number by \(4\) and atomic number by \(2\),
• each \(\beta^-\)-particle does not change mass number but increases atomic number by \(1\). e} ip

Step 1: Find number of alpha particles from mass number change.
Initial mass number: \[ 226 \] Final mass number: \[ 206 \] So decrease in mass number: \[ 226-206=20 \] Each \(\alpha\)-particle reduces mass number by \(4\), so: \[ \text{Number of } \alpha = \frac{20}{4}=5 \] ip

Step 2: Find number of beta particles from atomic number change.
Initial atomic number: \[ 88 \] After emission of \(5\alpha\)-particles, atomic number becomes: \[ 88-5\times 2=88-10=78 \] Final atomic number is: \[ 82 \] So atomic number must increase by: \[ 82-78=4 \] Each \(\beta^-\)-particle increases atomic number by \(1\), so: \[ \text{Number of } \beta =4 \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 5,\ 4} \]