A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $\mu$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 _____________ $\Omega$.
Answer (integer)
150
Solution
T<sub>m</sub> =30 ms<br><br>C = 200 $\mu$F<br><br>$${q \over N} = {{{Q_0}{e^{ - t/RC}}} \over {{N_0}{e^{ - \lambda t}}}} = {{{Q_0}} \over {{N_0}}}{e^{t\left( {\lambda - {1 \over {RC}}} \right)}}$$<br><br>Since q/N is constant hence<br><br>$\lambda ={1 \over {RC}}$<br><br>$$R = {1 \over {\lambda C}} = {{{T_m}} \over C} = {{30 \times {{10}^{ - 3}}} \over {200 \times {{10}^{ - 6}}}} = 150\Omega $$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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