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

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(B) The amount of substance remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.At time $t_2$,$3/4$ of the substance has decayed,so the rema

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

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(B) The amount of substance remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.At time $t_2$,$3/4$ of the substance has decayed,so the remaining amount is $N(t_2) = N_0 - \frac{3}{4}N_0 = \frac{1}{4}N_0$.Thus,$\frac{1}{4}N_0 = N_0 e^{-\lambda t_2} \implies e^{-\lambda t_2} = \frac{1}{4} \implies \lambda t_2 = \ln 4 = 2 \ln 2$.At time $t_1$,$1/4$ of the substance has decayed,so the remaining amount is $N(t_1) = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0$.Thus,$\frac{3}{4}N_0 = N_0 e^{-\lambda t_1} \implies e^{-\lambda t_1} = \frac{3}{4} \implies \lambda t_1 = \ln(4/3) = \ln 4 - \ln 3 = 2 \ln 2 - \ln 3$.The time interval is $t_2 - t_1 = \frac{1}{\lambda} (\lambda t_2 - \lambda t_1) = \frac{1}{\lambda} (2 \ln 2 - (2 \ln 2 - \ln 3)) = \frac{\ln 3}{\lambda}$.Given the half-life $T_{1/2} = 20 	ext{ min}$,we have $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.Substituting $\lambda$,we get $t_2 - t_1 = \frac{\ln 3}{(\ln 2 / 20)} = \frac{20 \ln 3}{\ln 2} 	ext{ min}$.

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 $3/4$ of it has decayed and time $t_1$ when $1/4$ of it has decayed is:

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