Human body excretes certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in $24 \mathrm{Hrs}$. A patient is given an injection containing a radio isotope ${\mathrm{Tc}}^{99}.$ This isotope has half life of 6 hours. Find the time after which the activity of $\mathrm{Tc}$ inside body of patient reduces to half.   

We can consider two simultaneous processes happening together. One is excretion of technetium from body and second is radioactive decay.

Half life through first process is ${\mathrm{t}}_1=24 \mathrm{hrs}$

Half life  through second process is ${\mathrm{t}}_2= 6 \mathrm{hrs}$

So, average half life for simultaneous processes can be written as,

$\frac{1}{{\mathrm{T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{1}{{\mathrm{t}}_1}+\frac{1}{{\mathrm{t}}_2}$

$\Rightarrow {\mathrm{T}}_{1/2}=\frac{{\mathrm{t}}_1{\mathrm{t}}_2}{{\mathrm{t}}_1+{\mathrm{t}}_2}=4.8\mathrm{h}$

Now, activity is given by,

$\mathrm{A}={\mathrm{A}}_0{2}^{-\frac{\mathrm{t}}{{\mathrm{T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}$

$\Rightarrow \frac{{\mathrm{A}}_0}{2}={\mathrm{A}}_0{2}^{-\frac{\mathrm{t}}{{\mathrm{T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}$

$\Rightarrow 2^{-1}=2^{-\frac{\mathrm{t}}{{\mathrm{T}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}$

$\Rightarrow \mathrm{t}={\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=4.8 \mathrm{hr}$