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(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$, where $N(t)$ is the amount remaining at time $t$.For $20 \%$ decay, the amount

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$, where $N(t)$ is the amount remaining at time $t$.For $20 \%$ decay, the amount remaining is $N_1 = 80 \% 	ext{ of } N_0 = 0.8 N_0$. Thus, $0.8 N_0 = N_0 e^{-\lambda t_1} \Rightarrow e^{-\lambda t_1} = 0.8$.For $80 \%$ decay, the amount remaining is $N_2 = 20 \% 	ext{ of } N_0 = 0.2 N_0$. Thus, $0.2 N_0 = N_0 e^{-\lambda t_2} \Rightarrow e^{-\lambda t_2} = 0.2$.Dividing the two equations: $\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = \frac{0.8}{0.2} = 4$.This simplifies to $e^{\lambda(t_2 - t_1)} = 4$.Taking the natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln(4) = 2 \ln(2)$.Since $\lambda = \frac{\ln 2}{T_{1/2}}$, we have $\frac{\ln 2}{T_{1/2}}(t_2 - t_1) = 2 \ln 2$.Therefore, $t_2 - t_1 = 2 	imes T_{1/2} = 2 	imes 20 	ext{ min} = 40 	ext{ min}$.

Half-life of a radioactive substance is $20 \text{ min}$. The time between $20 \%$ and $80 \%$ decay will be: (in $\text{ min}$)

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