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(B) The amount of substance remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $33\%$ decay,the remaining amount is $N_1 = N_0(1 - 0

Vedclass · Accepted Answer

$20$

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(B) The amount of substance remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $33\%$ decay,the remaining amount is $N_1 = N_0(1 - 0.33) = 0.67 N_0$. Thus,$0.67 = e^{-\lambda t_1}$,which gives $t_1 = -\frac{1}{\lambda} \ln(0.67)$.For $67\%$ decay,the remaining amount is $N_2 = N_0(1 - 0.67) = 0.33 N_0$. Thus,$0.33 = e^{-\lambda t_2}$,which gives $t_2 = -\frac{1}{\lambda} \ln(0.33)$.The time interval $\Delta t = t_2 - t_1 = -\frac{1}{\lambda} (\ln(0.33) - \ln(0.67)) = \frac{1}{\lambda} \ln(\frac{0.67}{0.33})$.Since $\frac{0.67}{0.33} \approx 2.03 \approx 2$,we have $\Delta t \approx \frac{\ln 2}{\lambda} = t_{1/2}$.Given $t_{1/2} = 20\, minutes$,the time interval is approximately $20\, minutes$.

Calculate the time (in $minutes$) interval between $33\%$ decay and $67\%$ decay if the half-life of a substance is $20\, minutes$.

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