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

Step 1: Understanding the Concept:
Alpha decay reduces the mass number ($A$) by 4 and the atomic number ($Z$) by 2. Beta minus decay ($\beta^-$) does not change the mass number but increases the atomic number by 1.
Step 2: Key Formula or Approach:
Change in mass number: $\Delta A = 4 \times n_{\alpha}$
Change in atomic number: $\Delta Z = (2 \times n_{\alpha}) - (1 \times n_{\beta})$
Step 3: Detailed Explanation:
1. Find the number of alpha particles ($n_{\alpha}$):
\[ 226 - 206 = 20 \]
\[ n_{\alpha} = \frac{20}{4} = 5 \]
2. Find the number of beta particles ($n_{\beta}$):
Initial $Z = 88$. Final $Z = 82$.
After 5 alpha emissions, the intermediate atomic number would be:
\[ Z' = 88 - (2 \times 5) = 78 \]
Since the final atomic number is 82, beta emissions must have increased it by:
\[ n_{\beta} = 82 - 78 = 4 \]
Step 4: Final Answer:
The number of alpha and beta particles are 5 and 4 respectively.