The nuclide ratio, `._(1)^(3) H` to `._(1)^(1) H` in a sample of water is `8.0xx10^(-18) : 1` Tritium undergoes decay with a half-life period of `12.3yr` How much tritium atoms would `10.0g` of such a sample contains `40` year after the original sample is collected?

Explanation

The ratio of tritium atom to that of H-atom will be same as the ratio of moles of T-atoms to that of H-atoms, since 1 mole of `T_(2)O equiv 2` mole of T atoms and 1 mole of `H_(2)O equiv 2` mole of H atoms
Calculate the mass ratio
`("mass of "T_(2)O)/("mass of " H_(2)O)=((8xx10^(-18)xx22)/(1xx18))`
`rArr ` mass of `T_(2)O ` atoms in 10 gm sample
`=((8xx10^(-18)xx22)/(22xx8xx10^(-18)+1 xx18))xx10=9.78xx10^(-17)` gm
`rArr` no. of `T_(2)O` atoms `= (9.78xx1969-17)/22xx6xx10^(23)`
`= 2.67 xx 10^(6)` atoms
Now using Ist order kinetics rate expression
`lambda t=2.303 log_(10)frac(N_(0))(N_(t)) or (0.693)/(12.3)xx40 = 2.303 log_(10)`
`(2.67xx10^(6))/N_(t)or N_(t)=2.81xx10^(5)`
`rArr` number of `T_(2)O` atoms after 40 years
`= 2.81 xx 10^(5)`
1 mole `T_(2)O equiv 2` moles of T-atoms
`rArr` number of tritium atom after 40 years
`= 2 xx 2.81 xx 10^(5) = 5.62 xx 10^(5)`