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

Explanation

We know,
`18g H_(2)O` has `2N H` atoms in it
and `H^(3) : H^(1) : : 8xx10^(-18) : 1`
`:. 18g H_(2)O` has `._(1)H^(3)` atoms `= 8xx10^(-18) x6.023xx10^(23) xx2`
`:. 10g H_(2)O` has `._(1)H^(3)` atoms
`= (8xx10^(-18)xx6.023xx10^(23)xx2xx10)/(18)`
i.e., `N_(o) pf ._(1)H^(3) = 5.354 xx 10^(6)` atoms
Now `t = (2.303)/(lamda) log_(10) (N_(o))/(N)`
`40 = (2.303xx12.3)/(0.693) log_(10) (5.354xx10^(6))/(N)`
`N = 5.624xx10^(5)` atoms