If a radioactive substance reduces to `(1)/(16)` of its original mass in `40` days, what is its half-life ?

Options

A. 10 days
B. 20 days
C. 40 days
D. None of these

Answer

A

Explanation

`(N)/(N_(0))= ((1)/(2))^((t)/(T_(1//2))) rArr (1)/(16) = ((1)/(2))^((40)/(T_(1//2)))`
`rArr T_(1//2)= (40)/(4)= 10` days