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Nuclei of a radioactive element A are be...

Nuclei of a radioactive element A are being produced at a constant rate `alpha`. The element has a decay constant `lambda` . At time t=0, there are `N_0` nuclei of the element . The number N of nuclei of A at time t is

A

`1/lambda[alpha + (alpha -N_0lambda)e^(-lambdat)]`

B

`1/lambda[alpha - (alpha -N_0lambda)e^(-lambdat)]`

C

`lambda[alpha - (alpha -N_0lambda)e^(-lambdat)]`

D

`[alpha-(N_0lambda-alpha ) e^(-lambdat)]`

Text Solution

Verified by Experts

The correct Answer is:
B
Nuclei of a radioactive element A are being produced at a constant rate `alpha` .
Decay constant of element =` lambda`
At t=0, nuclei of the element present =`N_0`
Number N of nuclei of A at time t :
Net rate of formation of nuclei of element A =`"dN"/"dt"`
`therefore "dN"/"dt"=alpha -lambdaN` or `(dN)/(alpha - lambdaN)=dt`
or `underset(N_0)oversetNint (dN)/(alpha - lambdaN) = underset0oversett int dt` or `-1/lambda [ In (alpha - lambdaN)]_(N_0)^N=t`
or In` ((alpha-lambdaN)/(alpha - lambdaN_0))=-lambdat` or `(alpha - lambdaN)/(alpha - N_0lambda)=e^(-lambdat)`
or `alpha -lambdaN=e^(-lambdat)(alpha - lambdaN_0)`
or `N=1/lambda[alpha - (alpha -N_0lambda)e^(-lambdat)]`
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