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.

Answer 1

Explanation:

It is given:

An isotope’s half-life period = $t_{1 / 2}$

Constant disintegration, $\lambda=0.693 t 1 / 2$

Radioactive decay’s rate, R is shown as $R=d N d t$

After time $t>>t_{1 / 2}$  there is a consistency in active nuclei’s number.

$dNdtdecay = dNdtPresent =\mathrm{R}$

Therefore, R = dNdtdecay

Radioactive decay’s rate,

$\mathrm{R}=\lambda \mathrm{N}$

where, λ = Constant of radioactive decay

N = Constant number

$\mathrm{Rt} 1 / 2=0.693 \mathrm{N}$

$\Rightarrow \mathrm{N}=\mathrm{R} \mathrm{t} 1 / 20.693$

There should be consistency in the value of N.