Half life of a radioactive substance is 15 seconds. After 15 seconds what percentage of activity will be left ?

Options

A. `15%`
B. `18%`
C. `12.5 %`
D. `10%`

Answer

C

Explanation

Suppose `N_(0)=100`
`N_(t)=N_(0)((1)/(2))^(n)`
[Given `t_(1//2)=15` seconds, T=45 seconds]
`N_(t)=100xx((1)/(2))^(n) " "[because n=(T)/(t_(1//2))=(45)/(15)=3]`
`=100xx((1)/(2))^(3)=(100)/(8)=12.5%`