A sample of radioactive material decays simultaneously by two processes A and B with half -lives 1/2 and 1/4h, respectively. For first half hour it decays with the process A, next one hour with the process B and for further half an hour with both A and B. If originally there were N₀ nuclei, find the number of nuclei after 2h of such decay

In process 'A' half life period ${\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{A}}=\frac{1}{2}\mathrm{hr}$; No.of half lives occured
$\left(\mathrm{n}\right) =\frac{\mathrm{t}}{{\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{A}}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$
No.of of nucleii remaining ${\mathrm{N}}_1=\frac{{\mathrm{N}}_0}{2}$
In process ‘B’ half period ${\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{B}}=\frac{1}{4}\mathrm{hr}$
No.of half lives $\left(\mathrm{n}\right)=\frac{\mathrm{t}}{{\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{B}}}=\frac{\frac{1}{1}}{4}=4$
No.of nucleii remaining ${\mathrm{N}}_2=\left(\frac{{\mathrm{N}}_0}{2}\right)\times \frac{1}{2^4}=\frac{{\mathrm{N}}_0}{2^5}$
In simultaneous decay, equivalent half life period
$\left({\mathrm{t}}_{\frac{1}{2}}\right)=\frac{{\left({\displaystyle {\mathrm{t}}_{\frac{{\displaystyle 1}}{{\displaystyle 2}}}}\right)}_{\mathrm{A}}\times {\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{B}}}{{\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{A}}+{\left({\mathrm{t}}_{\frac{1}{2}}\right)}_{\mathrm{B}}}=\frac{\frac{1}{2}\times \frac{1}{4}}{\frac{1}{2}+\frac{1}{4}}=\frac{1}{6}\mathrm{hr}$
No.of half lives $\left(\mathrm{n}\right) \frac{\mathrm{t}}{{\mathrm{t}}_{{\displaystyle \frac{1}{2}}}}=\frac{1}{\frac{2}{\frac{1}{6}}}=3$
final no.of nuclii remaining
${\mathrm{N}}_3=\left({\mathrm{N}}_2\right)\times \frac{1}{2^3}=\frac{{\mathrm{N}}_0}{2^5}\times \frac{1}{2^3}=\frac{{\mathrm{N}}_0}{2^8}$