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Nuclei of a radioactive element X are be...

Nuclei of a radioactive element `X` are being produced at a constant rate `K` and this element decays to a stable nucleus `Y` with a decay constant `lambda` and half-life `T_(1//3)`. At the time `t=0`, there are `N_(0) `nuclei of the element X.
The number `N_(Y)` of nuclei of `Y` at time `t` is .

A

`(K+lamdaN_(0))/(2lamda)`

B

`2(lamdaN_(0)-K)(1)/(lamda)`

C

`(lamdaN_(0)+(K)/(2))(1)/(lamda)`

D

`(lamdaN_(a)-K)/(2lamda)`

Text Solution

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The correct Answer is:
A
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Nuclei of a radioactive element X are being produced at a constant rate K and this element decays to a stable nucleus Y with a decay constant lambda and half-life T_(1//2) . At the time t=0 , there are N_(0) nuclei of the element X . The number N_(X) of nuclei of X at time t=T_(1//2) is .

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 .

Knowledge Check

  • 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)]`

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