1.

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. (a) Calculate the number N of nuclei of A at time t. (b) If alpha=2N_0lambda, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as trarroo.

Answer»

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

SOLUTION :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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