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`.

Options

A. 400 Bq
B. 800 Bq
C. 200 Bq
D. 600 Bq

Answer

B

Explanation

`R=R_(0)((1)/(2))^(n)` where n is the number of half - lives.
Given : `R=(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 fall half of 1600 Bq, i.e., 800 Bq.