Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive . An example is
`.^(38)"Sulphur"underset(=2.48h)overset("half -life")(to) .^(38)Cl underset(=0.62 h)overset("half-life")(to) .^(38)Ar(" stable")`
Assume that we start with `1000 .^(38)S` nuclei at time t=0. the number of `.^(38)Cl` of count zero at t=0 and will again be zero at `t= infty.`At what value of t, would the number of counts be a maximum?

Explanation

The given decay sequence is `.^(38)("Sulper") overset ("half-life") underset(=2.48) to .^(38)Cl overset ("half-life") underset(=0.62h) to .(38)Ar ("stable")`
At any time t, suppose `.^(38)S` has `N_(1) (t)` active nuclei and `.^(38)Cl` have `N_(2)(t)` active nuclei.
`:. (dN_(1))/(dt)=-lambda_(1)N_(1)`=rate of formation of `.^(38)Cl`
and `(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)`= net rate of decay of `.^(38)Cl=-lambda_(2)N_(2)+lambda_(1)N_(0)e^(-lambda_(1)t)`
Multiplying both sides by `e^(lambda_(2)t)dt` and rearranging, `e^(lambda_(2)t)dN_(2)+lambda_(2)N_(2)e^(lambda_(2)t)d t=lambda_(1)N_(0)e^((lambda_(2)-lambda_(1)t)dt`
Integrating both sides, we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^((lambda_(2)-lambda_(1)t)+C.......(i)`
Where C is constant of integration.
At `t=0, N_(2)=0 :. C=-(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))`
Putting in (i), we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^((lambda_(2)-lambda_(1)t))-1)]`
`N_(2)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^(-lambda_(1))-e^(-lambda_(2)t)]`
for maximum count , `N_(2)=max, (dN_(2))/(dt)=0`
`(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(1)t) (-lambda_(1))=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(2)t) (-lambda_(2))` or `(lambda_(1))/(lambda_(2))=(e^(-lambda_(2)t))/(e^(-lambda_(2)t))=e^((lambda_(1)-lambda_(2))t)`
or `log_(e) ((lambda_(1))/(lambda_(2)))=(lambda_(1)-lambda_(2))tlog_(e) e` or `t=(log_(e)(lambda_(1)//lambda_(2)))/((lambda_(1)-lambda_(2)))=(log_(e) T_(2)//T_(1))/(0.693(1/(T_(1))-1/(T_(2))))`
`t=(2.303log_(10) (0.62//2.48)xxT_(1)T_(2))/(0.693 (T_(2)-T_(1)))=(2.303(0-0.602)xx2.48xx0.62)/(0.693(0.62-2.48))=(2.303xx0.602xx2.48xx0.62)/(0.693xx1.86)=1.65sec`