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A F^32 radio nuclide with half-life T=14...

A `F^32` radio nuclide with half-life `T=14.3` days is produced in a reactor at a constant rate `q=2xx10^9` nuclei per second. How soon after the beginning of production of that radio nuclide will its activity be equal to `R=10^9` disintegration per second?

Text Solution

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The correct Answer is:
A, C, D
`N=R/lambda=((10^9)/(0.693))/(14.3xx3600)=7.43xx10^13`
Now, `(dN)/(dt)=q-lambdaN` or `int_0^N(dN)/(q-lambdaN)=int_0^tdt`
`:.` `N=q/lambda(1-e^(-lambdat))`
Substituting the values,
`7.43xx10^13=((2xx10^9)/(0.693))/(14.3xx3600)[1-e^(-(0.693//14.3xx3600)t)]`
Solving this equation we get,
`t=14.3h`
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