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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. |
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Answer» Correct Answer - D `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`. |
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