The nuclide ^ 212 _ 83 Bi is capable of disintegrating by emitting...

The initial mass of ^{212}_{83}\text{Bi}$is$1 \text{ g}$.<br><br>The overall half-life of ^{212}_{83}\text{Bi}$ is $1 \text{ hour}$. Since the time elapsed from $12.00$ noon to $1$ p.m. is exactly $1 \text{ hour}$ (one half-life), the amount of ^{212}_{83}\text{Bi}$remaining is:<br><br>$W_{\text{Bi}} = 1 \times \left(\dfrac{1}{2}\right)^1 = 0.50 \text{ g}$<br><br>The mass of ^{212}_{83}\text{Bi}$ that has decayed is:

$W_{\text{decayed}} = 1 - 0.50 = 0.50 \text{ g}$

The decay occurs via two parallel pathways:
$\alpha$-decay: ^{212}_{83}\text{Bi} \rightarrow ^{208}_{81}\text{Tl} + ^{4}_{2}\text{He}$<br>$\beta$-decay: ^{212}_{83}\text{Bi} \rightarrow ^{212}_{84}\text{Po} + \beta^{-}$

The probabilities for $\alpha$- and $\beta$-decays are in the ratio $7/13$. Therefore, the fractions of the decayed ^{212}_{83}\text{Bi}$that undergo$\alpha$- and$\beta$-decay are$\dfrac{7}{20}$and$\dfrac{13}{20}$, respectively.<br><br>Assuming the mass of the products is approximately equal to the mass of the decayed parent nuclei (as indicated by the options), we get:<br><br>Mass of Tl formed$= 0.50 \times \dfrac{7}{20} = 0.175 \text{ g}$<br><br>Mass of Po formed$= 0.50 \times \dfrac{13}{20} = 0.325 \text{ g}$<br><br>The composition of the sample at$1$p.m. is$0.50 \text{ g}$Bi,$0.175 \text{ g}$Tl, and$0.325 \text{ g} Po.