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The decay constant for a radio nuclide, has a value of `1.386 day^(-1)`. After how much time will a given sample of this nuclide get reduced to only 6.25% of its present number ? |
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Answer» Correct Answer - `2days` Here, `lambda=1.386 day^(-1), t=?` `N/(N_(0))=6.25%` Half life `T=0.693/lambda=0.693/1.386=1/2=0.5 day` `N/(N_(0))=(1/2)^(n)=(6.25/100)=1/16=(1/2)^(4)` `:. n=4=t/T, t=4T=4xx0.5=2 days` |
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