A radioactive substance decays to `((1)/(16))^(th)` of its initial activity in 40 days. The half-life of the radioacctive substance expressed in days is

Options

A. 2.5
B. 5
C. 10
D. 20

Answer

C

Explanation

We known that the activity of radioactive substance, `A=A_(0)e^(-lamdat)`
`rArr A/(A_0) = e^(-lamdat)=e^(-(0.693)/(T^(1//2))t) [ therefore T_(1//2)=(0.693)/(lamda)]`
`rArr 1/16 = e^(-(0.693xx40)/(T^(1//2))) rArr (0.693xx40)/(T_(1//2))=ln 16`
`rArr T_(1//2)=(0.693xx40)/(ln16) = 9.997 ~~ 10` days