A radioactive sample has an average life of 3 | vedclass

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(C) The charge on the capacitor at time $t$ is given by $q = q_0 e^{-\frac{t}{RC}}$,where $RC$ is the time constant of the $RC$ circuit.The activity $

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$150$

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(C) The charge on the capacitor at time $t$ is given by $q = q_0 e^{-\frac{t}{RC}}$,where $RC$ is the time constant of the $RC$ circuit.The activity $A$ of the radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$,where $A_0$ is the initial activity and $\lambda$ is the decay constant.Given that the ratio $\frac{q}{A}$ is constant with respect to time,we have:$\frac{q}{A} = \frac{q_0 e^{-\frac{t}{RC}}}{A_0 e^{-\lambda t}} = 	ext{constant}$For this ratio to be independent of time,the exponential terms must cancel out,which implies:$-\frac{t}{RC} = -\lambda t \implies \lambda = \frac{1}{RC}$We know that the average life $	au = \frac{1}{\lambda} = 30 \, ms = 30 	imes 10^{-3} \, s$.Therefore,$\lambda = \frac{1}{30 	imes 10^{-3}} \, s^{-1}$.Substituting $\lambda = \frac{1}{RC}$ into the equation for $R$:$R = \frac{1}{\lambda C} = 	au 	imes \frac{1}{C}$Given $C = 200 \, \mu F = 200 	imes 10^{-6} \, F$:$R = \frac{30 	imes 10^{-3}}{200 	imes 10^{-6}} = \frac{30000}{200} = 150 \, \Omega$.

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