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Question

A radioactive sample decays with an average-life of $$20ms$$. A capacitor of capacitance $$100\mu F$$ is charged to some potential and then the plates are connected through a resistance $$R$$. What should be the value of $$R$$ (in $$\times {10}^{2}\Omega$$) so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time?

A
$$2$$
Solution
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Correct option is A. $$2$$
The activity of the sample at time $$t$$ is given by $$A={A}_{0}{ e }^{ -{ \lambda }_{ }t }$$ where $$\lambda$$ is the decay constant and $${A}_{0}$$ is the activity at time $$t=0$$ when the capacitor plates are connected. The charge on the capacitor at time $$t$$ is given by
$$Q={ Q }_{ 0 }{ e }^{ -t/CR }\quad $$
where $${ Q }_{ 0 }$$ is the charge at $$t=0$$ and $$C=100\mu F$$ Thus
$$\cfrac { Q }{ A } =\cfrac { { Q }_{ 0 } }{ { A }_{ 0 } } \cfrac { { e }^{ -t/CR } }{ { e }^{ -\lambda t } } $$
It is independent of it if $$\lambda =\cfrac { 1 }{ CR }$$
$$R=\cfrac { 1 }{ \lambda C } =\cfrac { { t }_{ av } }{ C }$$
$$=\cfrac { 20\times { 10 }^{ -3 }s }{ 100\times { 10 }^{ -6 }F } =200\Omega \quad $$

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