Half lives of two substances A and B are 20 min and 40 min respectively.Initially the sample had equal no. of nuclei.Find the ratio of the remaining no. of nuclei of A and B after 80 min.

Hi,

Please find below the solution to the asked query: 

Let N₁ and N₂ are the remaining number of nuclei of A and B after 80 min. Let T₁ and T₂ be the half lives of A and B. Let N₀ be the initial number of nuclei.

$$\begin{gathered} N_1=N_0{\left(\frac{1}{2}\right)}^n \\ \mathrm{But}, n=\frac{t}{T_1}=\frac{80}{20}=4 \\ \Rightarrow N_1=N_0{\left(\frac{1}{2}\right)}^4 .....\left(1\right) \\ \mathrm{Similarly} \\ {N}_2=N_0{\left(\frac{1}{2}\right)}^2 .....\left(2\right) \\ \mathrm{Dividing} \mathrm{equation} \left(1\right) \mathrm{by} \left(2\right), \mathrm{we} \mathrm{get} \\ \frac{N_1}{N_2}=\frac{1}{16}\times \frac{4}{1}=\frac{1}{4} \end{gathered}$$
So, the required ratio is 1 : 4.