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A particular type of nucleus whose decay...

A particular type of nucleus whose decay constant is `lambda` is produced at a steady rate of `p` nuclei per second. Show that the number of nuclei `N` present `t` second after the production starts is `N=(P)/(lambda)[1-e^(~lambdat)]`.

Text Solution

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The instantaneous rate of growth of the nuclei is
`(dN)/(dt)=p-lambdaN` or `(dB)/(p-lambdaN)=dt`
On integrating,
`-(1)/(lambda)In|p-lambdaN|=t+c`
At `t=0`, `N=0rArr c=(-In|p|)/(lambda)`
or `-(1)/(lambda)[In|p-lambdaN|-In|p|]=t` or `In|1-(lambdaN)/(p)|=-lambda t`
or `N=(p)/(lambda)[1-e^(-lambda t)]`
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