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A radioactive substance has half-life T_(1/2) and number of radioactive nuclides present initially was N_(0). What is the number of radioactive nuclides at time t ? |
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Answer» Solution :We know that number of radioactive nuclides left intact after a time `T_(1/2)` is `N = N_(0)/2` , Thus, after ANOTHER time equal to `T_(1/2)` (i.e., after 2 half-lives) VALUE `N =1/2(N_(0)/2)=N_(0)(1/2)^(2)`. It FOLLOWS that after a time equal to n half-lives, the number of radioactive nuclides left intact will be `N=N_(0)(1/2)^(n)` For time t, `n=t/T_(1/2)` and hence, `N=N_(0)(1/2)^(t/T_(1/2)`. |
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