InterviewSolution
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InterviewSolution
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A radioactive sample decays with an average life of `20 ms`. A capacitor of capacitance `100 muF` is charged to some potential and then the plates are connected through a resistance `R`. What should be the value of `R` so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time? |
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Answer» The activity of the sample at time `t` is given by `A=A_(0)e^(-lambdat)` 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 where `Q_(0)` is the charge at `t=0` and `C=100 muF` is the capacitance. Thus, `(Q)/(A)=(Q_(0))/(A_(0))(e^(-t//CR))/(e^(-lambdat))` It is independent of `t` if `lambda=(1)/(CR)` or `R=(1)/(lambda C)=(t_(av))/(C )=(20xx10^(-3)s)/(100xx10^(-6)F)=200 Omega` |
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