At time `t=0`, activity of a radioactive substance is 1600 Bq, at t=8 s activity remains 100 Bq. Find the activity at t=2 s.

Answer

D

Explanation

`R=R_0(1/2)^n`
Here , n is the number of half-lives.
Given, `R=(R_0)/(16)`
:. `R_0/16 = R_0(1/2)^n` or `n=4`
Four half-lives are equivalent to 8 s. Hence, 2 s is equal to one half-life. So, in one half-life activity will remain half of `1600 Bq`, i.e. `800 Bq`.