1.

The half-Life for radioactive decay of ""^(14)C is 5739 years.An archaeological artifact containing wood had only 80% of the ""^(14)C found in a living tree.Estimate the age of the sample.

Answer»

Solution :The half LIFE of `""^(14)Ct_((1)/(2))=5730 year`
After…….. Time it is 80 `""^(14)C`
Soif initially `""^(14)C_(0)=[R]_(0)`
So after t time `""^(4)C=[R]_(t)80%[R]_(0)=0.8[R_(0)]`
All RADIOACTIVE dcay are first ORDER reaction.
So,`t_((1)/(2))=(0.693)/(k)` (But `t_((1)/(2))`=5730 years)
`THEREFORE k=(0.693)/(5730)years`
For ,first order reaction,
`t=(2.303)/(k)` log `([R]_(0))/([R]_(t))=(2.303xx5730)/(0.693)` log `([R]_(0))/(0.78[R]_(0))`
`=(2.303xx5730)/(0.693)` log 1.25
`(2.303xx5730)/(0.693)xx00969=1845.2 years`


Discussion

No Comment Found

Related InterviewSolutions