Half-life of a radioactive substance is 18 minutes. The time interval between its $20 \%$ decay and $80 \%$…

1. 6
2. 9
3. 18
4. 36

Solution

After $n$ half-life, the numbers of atom left undecayed is given by, $\begin{aligned} & N=N_0\left(\frac{1}{2}\right)^n \\ & n=\frac{t}{T_{\frac{1}{2}}} \text { or } N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} \end{aligned}$ For $20 \%$ decay of radioactive substance, $N=N_0-\frac{20}{100} N_0=0.8 N_0$ For $80 \%$ decay of radioactive substance,
From Eqs. (i) and (ii), we get $\Rightarrow 4=\left(\frac{1}{2}\right)^{\frac{t_1-t_2}{18}}$ Taking log on the both sides, we get $\begin{aligned} \log 4 & =\log \left(\frac{1}{2}\right)^{\frac{t_1-t_2}{18}} \Rightarrow \log 2^2=\log 2^{\left(\frac{t_2-t_1}{18}\right)} \\ \Rightarrow \quad 2 & =\frac{t_2-t_1}{18} \Rightarrow t_2-t_1=36 \mathrm{~min} \end{aligned}$

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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