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(B) Let $N_0$ be the initial amount of the radioactive substance.The amount remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lamb

Vedclass · Accepted Answer

$\frac{20 \ln 3}{\ln 2} \, 	ext{min}$

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(B) Let $N_0$ be the initial amount of the radioactive substance.The amount remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.At time $t_1$,$\frac{1}{4}$ has decayed,so the remaining amount is $N(t_1) = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0$.At time $t_2$,$\frac{3}{4}$ has decayed,so the remaining amount is $N(t_2) = N_0 - \frac{3}{4}N_0 = \frac{1}{4}N_0$.Using the decay law: $\frac{N(t_2)}{N(t_1)} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} = e^{-\lambda(t_2 - t_1)}$.Substituting the values: $\frac{1/4 N_0}{3/4 N_0} = e^{-\lambda(t_2 - t_1)} \Rightarrow \frac{1}{3} = e^{-\lambda(t_2 - t_1)}$.Taking the natural logarithm on both sides: $\ln(1/3) = -\lambda(t_2 - t_1) \Rightarrow \ln 3 = \lambda(t_2 - t_1)$.Substituting $\lambda = \frac{\ln 2}{20}$: $\ln 3 = \frac{\ln 2}{20} (t_2 - t_1)$.Therefore,$t_2 - t_1 = 20 \frac{\ln 3}{\ln 2} \, 	ext{min}$.

The half-life of a radioactive substance is $20 \, \text{minutes}$. The approximate time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{3}{4}$ of it has decayed and time $t_1$ when $\frac{1}{4}$ of it has decayed is:

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