The half-life of a radioactive substance is 2 | vedclass

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(B) The decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the number of undecayed atoms at time $t$.At time $t_1$,$\frac{1}{3}$ of the

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(B) The decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the number of undecayed atoms at time $t$.At time $t_1$,$\frac{1}{3}$ of the substance has decayed,so the remaining amount is $N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0$.Thus,$\frac{2}{3}N_0 = N_0 e^{-\lambda t_1} \Rightarrow e^{-\lambda t_1} = \frac{2}{3}$.Taking the natural logarithm: $-\lambda t_1 = \ln(\frac{2}{3}) \Rightarrow \lambda t_1 = \ln(1.5)$.At time $t_2$,$\frac{2}{3}$ of the substance has decayed,so the remaining amount is $N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0$.Thus,$\frac{1}{3}N_0 = N_0 e^{-\lambda t_2} \Rightarrow e^{-\lambda t_2} = \frac{1}{3}$.Taking the natural logarithm: $-\lambda t_2 = \ln(\frac{1}{3}) \Rightarrow \lambda t_2 = \ln(3)$.The time interval is $t_2 - t_1 = \frac{\ln(3) - \ln(1.5)}{\lambda} = \frac{\ln(3/1.5)}{\lambda} = \frac{\ln(2)}{\lambda}$.Since the half-life $T_{1/2} = \frac{\ln(2)}{\lambda} = 20 \, min$,we have $t_2 - t_1 = 20 \, min$.

The half-life of a radioactive substance is $20 \, min$. The approximate time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it has decayed and the time $t_1$ when $\frac{1}{3}$ of it has decayed is .......... $min$.

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