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 isA. Number of nuclei of `A` at time `t` is `(1)/(lambda)[alpha-(alpha-lambdaN_(0))e^(-lambdat)]`B. Number of nuclei of `A` at time `t` is `(1)/(lambda)[(alpha-lambdaN_(0))e^(-lambdat)]`C. If `alpha=2N_(0)lambda`, then the limiting value of number of nuclei of `A (rarr prop)` will be `2N_(0)`D. If `alpha=2N_(0)lambda`, then the number of nuclei of `A` after one half-life of `A` will be `N_(0)//2`

Nuclei of a radioactive element A are being produced at a constant rat

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 . The number N of nuclei of A at time t isA. Number of nuclei of `A` at time `t` is `(1)/(lambda)[alpha-(alpha-lambdaN_(0))e^(-lambdat)]`B. Number of nuclei of `A` at time `t` is `(1)/(lambda)[(alpha-lambdaN_(0))e^(-lambdat)]`C. If `alpha=2N_(0)lambda`, then the limiting value of number of nuclei of `A (rarr prop)` will be `2N_(0)`D. If `alpha=2N_(0)lambda`, then the number of nuclei of `A` after one half-life of `A` will be `N_(0)//2`
Answer» Correct Answer - A::C Let at time `t`, number of radioactive nuclei are `N`. Net rate of formation of nuclei of `A` `(dN)/(dt)=alpha-lambdaN` or `(dN)/(alpha-lambdaN)=dt` `-(1)/(lambda)[l n(alpha-lambdaN)]_(No)^(N)=t` `l n(alpha-lambdaN)/(alpha-N_(0))= -lambdat` `N=(1)/(lambda)[alpha-(a-lambdaN_(0))e^(-lambdat)]`

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