Question 424: Atoms & Nuclei

Question

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

Accepted Answer

Ra-226 ($ ^{226}_{88}Ra $) decays to Pb-206 ($ ^{206}_{82}Pb $) by emission of $\alpha$ and $\beta$ particles. We need to find how many $\alpha$'s and $\beta$'s are emitted.
Step 1: Write data
Initial: $ A=226, Z=88$.
Final: $ A=206, Z=82$.
Step 2: Effect of alpha and beta emissions
An $\alpha$-particle emission: decreases mass number by 4, atomic number by 2. ($A 	o A-4$, $Z 	o Z-2$)
A $\beta^-$-particle emission: mass number unchanged, atomic number increases by 1 ($Z 	o Z+1$).
Step 3: Compare initial and final mass numbers
Mass number difference: $226 - 206 = 20$.
Each $\alpha$ reduces mass number by 4.
To reduce by 20, we need $ \frac{20}{4} = 5$ $\alpha$ emissions.
So number of $\alpha = 5$.
Step 4: Effect on atomic number
After 5 $\alpha$ decays, atomic number decreases by $5 	imes 2 = 10$.
Initial $Z=88$. After 5 $\alpha$, $Z=88-10=78$.
But final $Z$ must be 82.
So we must increase atomic number from 78 to 82 Ã¢ÂÂ increase by 4.
Each $\beta^-$ increases $Z$ by 1. Thus, 4 $\beta^-$ emissions.
Step 5: Answer
Number of $\alpha$: 5
Number of $\beta^-$: 4
So the correct answer =
Option A: 5,4
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Answer

5, 4

Answer

4, 5

Answer

6, 4

Answer

4, 6

Atoms & Nuclei

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

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