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A radioactive substance with decay constant of 0.5 s^(-1) is being produced at a constant rateof 50 nuclei per second. If there are no nuclei present initially, the time in second) after which 25 nuclei will be present is |
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Answer» 1 `(DN)/(DT)=50-lamdbaN rArr (dN)/(50-lambdaN)=dt` Integrate both sides, we get `int_(0)^(N)(dN)/(50-lambdaN) =int_(0)^(t) dt rArr -(1)/(lambda)[ln(50-lambdaN)]_(0)^(N)=t` `ln((50-lambdaN)/(50))=-lambdat` `(50-lambdaN)/(50)=e^(-lambdat), 1-(lambdaN)/(50)=e^(-lambdaN), N=(50)/(lambda)(1-e^(-lambdat))` As, `N=25 and lambda=0.5s^(-1)` `25=(50)/(0.5)(1-e^(-0.5t)) or e^(-0.5t)=(3)/(4) therefore t=2ln((4)/(3))` |
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