`._(84)Po^(218) (t_(1//2) = 3.05 min)` decays to `._(82)Pb^(214) (t_(1//2) = 2.68 min)` by `alpha` emisison while `Pb^(214)` is `beta`-emitter. In an experiment starting with `1 g` atom of pure `Po^(218)`, how much time would be required for the concentration of `Pb^(214)` to reach maximum?

Explanation

`._(84)Po^(218) overset(-alpha)(rarr) ._(82)Pb^(214) overset(-beta)(rarr) ._(83)Bi^(214)`
Initial amount `m_(0)=1 g` atom
Let us assume `[A]_(0)` to be the initial concentration of `[Po^(218)]` and `[B]` to be the concentration of `Pb^(214)`, thus for maximum concentration in a sequential reaction.
`(d[B])/(dt)=[A]_(0)(K_(1)/(K_(2)-K_(1))){-K_(1) e^(K_(1)t)+K_(2)e^(-K_(2)t)}`
For the number of nuclei to be maximum, `(d[B])/(dt), i.e., (d[Po^(214)])/(dt)` should be equated to zero.
`:. K_(1)e^(-K_(1)t)=K_(2)e^(-K_(2)t)`
or, `(K_(1)/K_(2))=e^((K_(1)-K_(2))t)`
or, `ln(K_(1)/K_(2))=(K_(1)-K_(2))t`
Now, `K_(1)=0.693/3.05=0.227`
`K_(2)=0.693/2.68=0.259`
`t=(1)/(K_(1)-K_(2))"ln" K_(1)/K_(2)=4.12 min`