Question

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?

Answer 1

`"Ratio of """_(1)""^(3)H:H=8.0xx10""^(-18):1`
`"Tritium decays as below:"`
`"Half life period "t""_(1//2)=12.3years`
`"Disintegration constant" =lambda=(0.693)/t""_(1//2)=(0.693)/(12.3)`
`"Consider one mole of water"`
`"mol.wt.of "H""_(2)O=18`
`"1 mole of "H""_(2)O=18g=wt.of" "2xx 6.023xx10""^(23)`
`"atom of hydrogen"`
`18" "g" ""of"" "H""_(2)O" ""contain hydrogen atoms"`
`=2xx6.023xx10""^(23)`
`10g" of "H""_(2)O" contain hydrogen atoms"`
`=(2xx6.023xx10""^(23)xx10)/18`
`"Since, the ratio of tritium "(.""_(1)H""^(3)),H=8xx10""^(-18):1`
`"No. of tritium atoms of 10g of "H""_(2)O`
`=(2xx6.023xx10""^(23)xx10)/(18)xx8xx10""^(-18)`
`=5.3532xx10""^(6)" atom"`
`"Initial conc. of """_(1)H""^(3)" atoms in 10g of "H""_(2)O`
`=5.3532xx10""^(6)`
`"Let the conc. left after 40yr = N"`
`lambda=(2.303)/(t)log.(N""_(0))/(N)`
`(0.693)/(12.3)=(2.303)/(40)log.(5.3532xx10""^(6))/(N)`
`log.(5.3523xx10""^(6))/(N)=(0.693xx40)/(12.3xx2.303)`
`N=5.6223xx10""^(5)" atoms"`