The acticity of a sample is `64 xx 10^-5 Ci`. Its half-life is `3` days. The activity will become `5 xx 10^-6 C i` after.

Options

A. 21 days
B. 18 days
C. 12 days
D. 7 days

Answer

A

Explanation

`A = A_0 (1/2)^n =A_0 (1/2)^(t//T)`
Where t = total time and T = half life time
`therefore A/A_0 = (5xx10^(-6))/(64xx10^(-5)) = (1/2)^(t//3)`
`therefore (0.5xx10^(-5))/(64xx10^(-5)) = (1)/(128) = (1/2)^(t//3)`
`therefore (1/2)^7 = (1/2)^(t//3)`
`therefore t= 7xx3 = 21` days