Question

The half life of a radio active substance is $20$ days. If $\frac{2}{3}$ part of the substance has decayed in time $t_2$ and $\frac{1}{3}$ part of it has decayed in time $t_1$ then the time interval between $t_2$ and $t_{1}is\left(t_2-t_1\right)=$

• A. $5$ days
• B. $10$ days
• C. $20$ days
• D. $40$ days

Answer 1

The correct option is C $20$ days

$\begin{array}{c}\text{Given-}\ast \text{Half life}=20\text{days}\\ \begin{array}{cc}& \ast \frac{2}{3}\text{part decays in time}{t}_2\\ & \ast \frac{1}{3}\text{part decays in time}{t}_1\end{array}\\ \text{Taking the time}t=t_1\\ \text{we have, Decayed Radio active Nuclei}=\frac{1}{3}\text{of the total}\\ \Rightarrow \text{Total Left}=\frac{2}{3}\text{of the TotalNuclei}\end{array}$

$\begin{array}{c}\text{Taking the time}t=t_2\\ \text{we have, Decayed Nuclei}=\frac{2}{3}\text{of the total}\\ \Rightarrow \text{total Left}=\frac{1}{3}\text{of the total.}\\ \text{Now from}t=t_1\text{to}t=t_2\\ \text{we see that the number of Nuclei gets halved.}\\ \text{Initially}\frac{2}{3}{N}_0\text{to finally}\frac{\text{No}}{3}\\ \Rightarrow \text{Then}{t}_2-t_1=\text{Half life}=20\text{days.}\end{array}$