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A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance $200 μ F$ is first charged and later connected with resistor $R .$ If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of $R$ should be ______ $Ω$

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answer is 150.

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Detailed Solution

The ratio of charge on capacitor q to the activity of radioactive sample is fixed So, $\frac{q}{N} = \frac{Q_{0} e^{- t / R C}}{N_{0} e^{- λ t}} = \frac{Q_{0}}{N_{0}} e^{t (λ - \frac{1}{R C})}$

So, $λ = \frac{1}{R C}, R = \frac{1}{λ C} = \frac{T_{m}}{C} = \frac{30 \times 10^{- 3}}{200 \times 10^{- 6}} = 150 Ω$

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