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A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment. The isotope has a half-life t_{1}_{/2}. Show that after a time t >> t_{1}_{/2} the number of active nuclei will become constant. Find the value of this constant.

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#### Solution

Given:

Half life period of isotope = t_{1}_{/2}

Disintegration constant, `lambda = 0.693/t_"1/2"`

Rate of Radio active decay (R) is given by

`R = "dN"/"dt"`

We are to show that after time t >> `t_"1/2"`the number of active nuclei is constant.

`("dN"/"dt")_"present" = R = ("dN"/"dt")_"decay"`

`therefore R = ("dN"/"dt")_"decay"`

Rate of radioactive decay, `R = lambdaN`

Here, λ = Radioactive decay constant

N = Constant number

`R = 0.693/t_"1/2" xx N`

⇒`Rt_"1/2" = 0.693 N`

⇒`N = (Rt_"1/2")/0.693`

This value of N should be constant.

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