#### Question

`""_83^212"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 ^{212}Bi is one hour. If 1 g of pure ^{212}Bi is taken at 12.00 noon, what will be the composition of this sample at 1 P.m. the same day?

#### Solution

Given:-

Half-life of `""^212"Bi" , T_"1/2" = 1 "h"^-1`

When `""_83^212"Bi"` disintegrates by emitting an α-particle

`""_83^212"Bi" → ""_81^208"T1" + ""_2^4"He"(alpha)`

When `""_83^212"Bi"` disintegrates by emitting a β^{−}particle

`""_83^212"Bi" → ""_84^212P_0 + β^(-)+ bar"v"`

Half-life period of ^{212}Bi, `T_(1/2)`= 1 `"h"^-1`

At t = 0, the amount of ^{212}Bi present = 1 g

At t = 1 = One half-life,

Amount of ^{212}Bi present = 0.5 g

Probability of disintegration of α-decay and β-decay are in the ratio `7/13`.

In 20 g of ^{212}Bi, the amount of ^{208}Ti formed = 7 g

In 1 g of ^{212}Bi, the amount of ^{208}Ti formed = 7/20 g

∴ Amount of `""^208"Ti"` present in `0.5 "g" = 7/20 xx 0.5 = 0.175 "g"`

In 20 g of ^{212}Bi, the amount of ^{212}Po formed = 13 g

In 1 g of ^{212}Bi, the amount of ^{212}Po formed = 13/20 g

∴ Amount of `""^212"Po"` present in `0.5 "g" = 13/20 xx 0.5 = 0.325 "g"`