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Calculate the half life periode of a radioactive substance if its activity drops to `1/16` th of its initial value of 30 yr. |
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Answer» Correct Answer - `7.5 yr` `T=?, N/(N_(0))=1/16, t=30years` As `N/(N_(0))=(1/2)^(n)=1/16=(1/2)^(4)` `:. n=4=t/T,T=t/4=30/4=7.5 years` |
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