2 g of pure CsCl gives 200 counts `s^(-1)`. Calculate the relative abundance of `Cs^(137)` in natural caesium, if the half-life of `Cs^(137)` is 30 years.

Explanation

Number of counts by 2g of CsCl = 200
Number of counts by 1g of CsCl = 100
`T_(1//2) = 30` years
`lambda = 0.693/T_(1//2) = 0.693/(30 xx 365 xx 24 xx 3600)`
As activity `A = lambda N`
`therefore N = A/lambda`
N is number of atoms
`N = (100 xx 30 xx 365 xx 24 xx 3600)/0.693`
`=13.65 xx 10^(10)`
`6.023 xx 10^(23)` atoms are present in 137 g
Mass of 1 atom is `=137/(6.023 xx 10^(23))`g
`13.65 xx 10^(10)` atoms are present in
`=137/(6.023 xx 10^(23)) xx 13.65 xx 10^(10) = 310.48 xx 10^(-13)g`
Hence, `310.48 xx 10^(-13)`g of `Cs^(137)` is present in 1g of natural Cs.