A mixture of `Pu""^(239)` and `Pu""^(240)` has specific activity of `6xx10""^(9)`dps. The half lves of isotopes are `2.44xx10""^(4)` and `6.58xx10""^(3)`years respectively. The percent of `Pu""^(239)` in the mixture is x/10% (nearly). Then 'x' is

Explanation

Total activity of a sample is the sum of the individual activities of all its components. Let the total mass of the sample be 1 gm and the mass of 239Pu be x gm
`therefore x/239xx6.023xx10^(23) xx (0.693)/(2.44xx10^(4))+(1-x)/(240)xx`
`6.023xx 10^(23) xx (0.693)/(6.58xx10^(3))`
`= 6 xx10^(9) xx 365 xx 24 xx 60 xx60 `
On calculating, `x = 0.3896`
`therefore ""^(239)Pu = 38.96% and ""^(240)Pu = 61.04%`