A radioactive material has half-life of `10` days. What fraction of the material would remain after `30` days ?

Options

A. `0.5`
B. `0.25`
C. `0.125`
D. `0.33`

Answer

C

Explanation

Fraction of material remain `(N)/(N_(0)) = ((1)/(2))^((t)/(T_(1//2))) = ((1)/(2))^((30)/(10)) = (1)/(8)`
`rArr (N)/(N_(0))= 0.125`