The activity of a radioactive sample is `1.6` curie, and its half-life is `2.5 days`. Its activity after `10 days` will be

Options

A. `0.8 curie`
B. `0.4 curie`
C. `0.1 curie`
D. `0.16curie`

Answer

c

Explanation

Since four half-lives have elapesed
`A=(A_(0))/(2^(4)) =(A_(0))/(16)=(1.6)/(16) curie=0.1 curie`.