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`.

Explanation

Activity of the sample after n half n half life is `R=R_(0)((1)/(2))^(n)`
Given , `R=(R_(0))/(16)`
`(R_(0))/(16)=R_(0)((1)/(2))^(n)`
or number of half - lives , n=4
four half are equivlent to 8 s. Hence ,2 s is equal to one half - life , so in one half - life activity will ramain half of 16000 Bq. i.e., 800 Bq.