A radioactive sample has an average life of 30 ms and is decaying.

Question:

A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 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_____________$\Omega .$

Solution:

$\mathrm{T}_{\mathrm{m}}=30 \mathrm{~ms}$

$\mathrm{C}=200 \mu \mathrm{F}$

$\frac{\mathrm{q}}{\mathrm{N}}=\frac{\mathrm{Q}_{0} \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}=\frac{\mathrm{Q}_{0}}{\mathrm{~N}_{0}} \mathrm{e}^{\left(\lambda-\frac{1}{\mathrm{RC}}\right)}$

Since q/N is constant hence

$\lambda=\frac{1}{R C}$

$\mathrm{R}=\frac{1}{\lambda \mathrm{C}}=\frac{\mathrm{T}_{\mathrm{m}}}{\mathrm{C}}=\frac{30 \times 10^{-3}}{200 \times 10^{-6}}=150 \Omega$

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