Question

When an element $_{90}^{232}\text{Th}$ decays into $_{82}^{208}\text{Pb}$, the number of $\alpha$ and $\beta^{-}$ particles emitted respectively are

• A. 4, 8
• B. 8, 2
• C. 6, 2
• D. 6, 4

Hint:

In nuclear decay, balance the mass numbers (top numbers) first to find the number of alpha particles ($_{2}^{4}\text{He}$), then balance the atomic numbers (bottom numbers) to find the number of beta particles ($_{-1}^{0}e$). Remember that alpha decay reduces the mass number by 4 and atomic number by 2, while beta decay leaves the mass number unchanged and increases the atomic number by 1.

Answer 1

The Correct Option is D

Step 1: Set up the nuclear reaction equation.
A parent nucleus $P$ (Thorium) decays into a daughter nucleus $D$ (Lead) by emitting $n_{\alpha}$ alpha ($\alpha$) particles and $n_{\beta}$ beta ($\beta^-$) particles. \[ _{90}^{232}\text{Th} \to _{82}^{208}\text{Pb} + n_{\alpha} \left(_{2}^{4}\text{He}\right) + n_{\beta} \left(_{-1}^{0}e\right). \]

Step 2: Balance the mass number (superscripts) to find the number of alpha particles.
The mass number must be conserved: \[ 232 = 208 + n_{\alpha}(4) + n_{\beta}(0). \] \[ 232 - 208 = 4n_{\alpha}. \] \[ 24 = 4n_{\alpha} \implies n_{\alpha} = 6. \] So, 6 alpha particles are emitted.

Step 3: Balance the atomic number (subscripts) to find the number of beta particles.
The atomic number (charge) must be conserved: \[ 90 = 82 + n_{\alpha}(2) + n_{\beta}(-1). \] Substitute the value of $n_{\alpha}=6$ from Step 2: \[ 90 = 82 + 6(2) - n_{\beta}. \] \[ 90 = 82 + 12 - n_{\beta}. \] \[ 90 = 94 - n_{\beta}. \]

Step 4: Solve for $n_{\beta$.}
\[ n_{\beta} = 94 - 90 = 4. \] So, 4 beta-minus particles are emitted.
The number of $\alpha$ and $\beta^{-}$ particles emitted respectively are 6 and 4.