Question

What are the respective numbers of $\alpha$ and $\beta$ particles emitted respectively in the following radioactive decay?
$^{200}_{90}\text{X} \rightarrow ^{168}_{80}\text{Y}$

• A. 8 and 8
• B. 8 and 6
• C. 6 and 8
• D. 6 and 6

Hint:

Always calculate the change in mass number ($A$) first! Since $\beta$ particles have zero mass, the entire drop in mass is strictly due to alpha particles. This gives you $n_{\alpha}$ immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In radioactive nuclear decay, the emission of an alpha particle ($^{4}_{2}\text{He}$ or $\alpha$) reduces the mass number by 4 and the atomic number by 2. The emission of a beta particle ($^{0}_{-1}\text{e}$ or $\beta^{-}$) leaves the mass number unchanged but increases the atomic number by 1. By conserving total mass and charge balances across the reaction, we can solve for both counts.

Step 2: Key Formula or Approach:
Let $n_{\alpha}$ be the number of $\alpha$ particles and $n_{\beta}$ be the number of $\beta$ particles emitted. The nuclear decay balance equation is written as: $$^{200}_{90}\text{X} \rightarrow ^{168}_{80}\text{Y} + n_{\alpha}\left(^{4}_{2}\alpha\right) + n_{\beta}\left(^{0}_{-1}\beta\right)$$ 1. Conserve Mass Number ($A$): $200 = 168 + 4n_{\alpha} + 0n_{\beta}$ 2. Conserve Atomic Number ($Z$): $90 = 80 + 2n_{\alpha} - 1n_{\beta}$

Step 3: Detailed Explanation:
First, solve for the number of $\alpha$ particles using the mass conservation balance: \[ 200 = 168 + 4n_{\alpha} \] \[ 4n_{\alpha} = 200 - 168 \] \[ 4n_{\alpha} = 32 \implies n_{\alpha} = 8 \] Now, substitute $n_{\alpha} = 8$ into the charge/atomic number conservation balance equation to evaluate $n_{\beta}$: \[ 90 = 80 + 2(8) - n_{\beta} \] \[ 90 = 80 + 16 - n_{\beta} \] \[ 90 = 96 - n_{\beta} \] \[ n_{\beta} = 96 - 90 = 6 \] Therefore, $8$ alpha particles and $6$ beta particles are emitted during the transformation path.

Step 4: Final Answer:
The number of $\alpha$ and $\beta$ particles emitted are 8 and 6 respectively.