The half life a radioactive element is `30` days, if the mass of the specimen reduces to `1//10^(th)` then the time taken is :-
A. `100` days
B. `50` days
C. `33` days
D. `16` days

4 views

1 Answers

Correct Answer - C
`N = N_(0)e^(-lambdat) rArr (N_(0))/(10) = N_(0)e^(-lambdat) rArr ln(1) - ln(10) = - lambdat`
`rArr t = ((ln10)/(ln2)) T_(1//2) rArr t = 33` days

4 views

Related Questions