Question

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

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

Hint:

Always compute the $\alpha$ particles first, since beta decay has zero mass effect ($A=0$). Once you know $n_{\alpha} = 6$, you can eliminate options (A) and (B) instantly. Then, check the atomic numbers: $6$ alpha particles would drop the atomic number from $90$ down to $78$. Since the product nucleus has an atomic number of $80$, you need exactly $8$ beta emissions to bring it back up by $+8$.

Answer 1

The Correct Option is C

Concept: In a nuclear decay series, the total mass number ($A$) and total atomic number ($Z$) must be conserved on both sides of the reaction equation. Let $n_{\alpha}$ be the number of alpha particles emitted and $n_{\beta}$ be the number of beta ($\beta^{-}$) particles emitted:
• An alpha particle ($\alpha$) is a helium nucleus: ${}_{2}^{4}\text{He}$. It reduces the mass number by $4$ and the atomic number by $2$.
• A beta particle ($\beta^{-}$) is an electron: ${}_{-1}^{0}\text{e}$. It leaves the mass number unchanged and increases the atomic number by $1$. The balanced nuclear equation is represented as: \[ {}_{90}^{200}X \longrightarrow {}_{80}^{168}Y + n_{\alpha}\left({}_{2}^{4}\text{He}\right) + n_{\beta}\left({}_{-1}^{0}\text{e}\right) \]

Step 1: Conserve the total Mass Number ($A$) to find the number of alpha particles.
Equating the mass numbers from both sides of the equation: \[ 200 = 168 + n_{\alpha}(4) + n_{\beta}(0) \] \[ 200 = 168 + 4n_{\alpha} \] Subtracting $168$ from both sides: \[ 4n_{\alpha} = 200 - 168 = 32 \] \[ n_{\alpha} = \frac{32}{4} = 6 \] Thus, the total number of emitted $\alpha$ particles is $6$.

Step 2: Conserve the total Atomic Number ($Z$) to find the number of beta particles.
Equating the atomic numbers from both sides of the equation and substituting $n_{\alpha} = 6$: \[ 90 = 80 + n_{\alpha}(2) + n_{\beta}(-1) \] \[ 90 = 80 + 6(2) - n_{\beta} \] \[ 90 = 80 + 12 - n_{\beta} \] \[ 90 = 92 - n_{\beta} \] Rearranging the terms to solve for $n_{\beta}$: \[ n_{\beta} = 92 - 90 = 8 \] Thus, the total number of emitted $\beta$ particles is $8$.