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

(a) `(N)/(N_0) = ((1)/(2))^n rArr (1)/(16) = ((1)/(2))^4 = ((1)/(2))^n rArr n = 4`
Also `n = (t)/(T_(1//2)) rArr T_(1//2) = (40)/(4) = 10 days`.