The nuclei of a certain radioactive element are produced at a constant rate of α per second. This element has a decay constant λ . Initially $$N0$$ nuclei of the element are present.What is the number N of nuclei of this element at a time t?If in the above question α$$=2$$ λ $$N0$$ calculate the number of nuclei left in the element after one half life period of the element.

Question

The nuclei of a certain radioactive element are produced at a constant rate of  α per second. This element has a decay constant λ . Initially $$N0$$  nuclei of the element are present.What is the number N of nuclei of this element at a time t?If in the above question α$$=2$$ λ $$N0$$  calculate the number of nuclei left in the element after one half life period of the element.

Infinity Learn · Accepted Answer

$$dNdt=$$α−λN∫$$N0NdN$$α−λ$$N=$$∫$$0tdtln$$α−λNα−λ$$N0$$−λ$$=t$$λ$$t=ln$$α−λ$$N0$$α−λNeλtα−λ$$N=$$α−λ$$N0$$αeλt−λNeλ$$t=$$α−λ$$N0N=$$αeλ$$t+$$λ$$N0$$−αλeλ$$t=1$$λα$$+$$λ$$N0e$$−λt−αe−λ$$tN=1$$λα$$1$$−e−λ$$t+$$λ$$N0e$$−λt λ$$t=ln$$ $$2$$λ$$N0$$−λ$$N02$$λ$$N0$$−λNλln $$2$$λ$$=ln$$ λ$$N02$$λ$$N0$$−λ$$N2=$$λ$$N02$$λ$$N0$$−λ$$N4$$λ$$N0$$−$$2$$λ$$N=$$λ$$N0N=32N0$$

Kinetics of nuclear reactions or nuclear chemistry

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The nuclei of a certain radioactive element are produced at a constant rate of $α$ per second. This element has a decay constant $λ$ . Initially $N_{0}$ nuclei of the element are present.

What is the number N of nuclei of this element at a time t?

If in the above question $α = 2 λ N_{0}$ calculate the number of nuclei left in the element after one half life period of the element.

$λ t = l n (\frac{2 λ N_{0} - λ N_{0}}{2 λ N_{0} - λ N})$
$λ \frac{l n 2}{λ} = l n (\frac{λ N_{0}}{2 λ N_{0} - λ N})$
$2 = \frac{λ N_{0}}{2 λ N_{0} - λ N}$
$4 λ N_{0} - 2 λ N = λ N_{0}$
$N = \frac{3}{2} N_{0}$

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Kinetics of nuclear reactions or nuclear chemistry

Unlock the full solution & master the concept.

The nuclei of a certain radioactive element are produced at a constant rate of α per second. This element has a decay constant λ . Initially N0 nuclei of the element are present.