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Question
A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment. The isotope has a half-life t1/2. Show that after a time t >> t1/2 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 = t1/2
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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