Question

${}_{83}{}^{212}\mathrm{Bi}$ can disintegrate either by emitting an α-particle of by emitting a β⁻-particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration α-and β-decays are in the ratio 7/13. The overall half-life of ²¹²Bi is one hour. If 1 g of pure ²¹²Bi is taken at 12.00 noon, what will be the composition of this sample at 1 P.m. the same day?

Answer 1

Given:
Half-life of ²¹²Bi, T_1/2 = 1 h^$-1$
When ${}_{83}{}^{212}\mathrm{Bi}$ disintegrates by emitting an α-particle

${}_{83}{}^{212}\mathrm{Bi}\to {}_{81}{}^{208}\mathrm{T}1+{}_2{}^4\mathrm{He}\left(\mathrm{\alpha }\right)$

When ${}_{83}{}^{212}\mathrm{Bi}$ disintegrates by emitting a β⁻particle

${}_{83}{}^{212}\mathrm{Bi}\to {}_{84}{}^{212}\mathrm{P}_0+{\mathrm{\beta }}^{-}+\overline{\mathrm{v}}$

Half-life period of ²¹²Bi, $T_{\frac{1}{2}}$= 1 h^$-1$
At t = 0, the amount of ²¹²Bi present = 1 g

At t = 1 = One half-life,
Amount of ²¹²Bi present = 0.5 g
Probability of disintegration of α-decay and β-decay are in the ratio $\frac{7}{13}$.
In 20 g of ²¹²Bi, the amount of ²⁰⁸Ti formed = 7 g
In 1 g of ²¹²Bi, the amount of ²⁰⁸Ti formed = 7/20 g
∴ Amount of ²⁰⁸Ti present in 0.5 g = $\frac{7}{20}\times 0.5=0.175\mathrm{g}$

In 20 g of ²¹²Bi, the amount of ²¹²Po formed = 13 g
In 1 g of ²¹²Bi, the amount of ²¹²Po formed = 13/20 g
∴ Amount of ²¹²Po present in 0.5 g = $\frac{13}{20}\times 0.5=0.325\mathrm{g}$