The ratio activity of an element becomes `1//64 th` of its original value in `60 sec`. Then the half-life period is

Options

A. 5 sec
B. 10 sec
C. 20 sec
D. 30 sec

Answer

B

Explanation

(b) `A = A_0 ((1)/(2))^((t)/(T_(1//2))) rArr (1)/(64) = ((1)/(2))^((60)/(T_1//2))`
`rArr ((1)/(2))^6 = ((1)/(2))^((60)/(T_1//2)) rArr T_(1//2) = 10 sec`.