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  • A radioactive sample decays with a mean life of 20 ms. A capacitor of capacitance <img alt="\mathrm{100 \mu \mathrm{F}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B100%20

A radioactive sample decays with a mean life of 20 ms. A capacitor of capacitance \mathrm{100 \mu \mathrm{F}} is charged to some potential and then the plates are connected through a resistance R, The resistance R for which the ratio of the charge on the capacitance to the activity of the radioactive sample remains constant in time is found \mathrm{\text { y } \Omega} Then value of y is:

Option: 1

200


Option: 2

150


Option: 3

125


Option: 4

250


Answers (1)

best_answer

The activity of radioactive sample of decay constant\lambda at time t is given by

\mathrm{\mathrm{A}=\mathrm{A}_0 \mathrm{e}^{-\lambda \mathrm{t}} \ldots \ldots . .(1)}

Where \mathrm{\mathrm{A}_0} is initial activity The charge on capacitor in series RC-circuit after time t is

\mathrm{\mathrm{Q}=\mathrm{Q}_0 \mathrm{e}^{-\frac{1}{\mathrm{RC}}} \ldots(2)}

\mathrm{\text { Where } Q_0 \text { is initial charge }}

Dividing (2) by (1) , we get

\mathrm{\begin{aligned} & \frac{Q}{A}=\frac{Q_0}{A_0} \frac{e^{t / R C}}{e^{-\lambda t}} \\ & \Rightarrow \frac{Q}{A}=\frac{Q_0}{A} e^{\left(\lambda-\frac{1}{R C}\right) t} \end{aligned}}

Clearly this ratio will be independent of t if

\mathrm{\begin{aligned} &\lambda-\frac{1}{\mathrm{RC}}=0 \Rightarrow \lambda=\frac{1}{\mathrm{RC}} \Rightarrow \mathrm{R}=\frac{1}{\mathrm{C} \lambda}\\ &\text { or as } \lambda=\frac{1}{\tau}\\ &\Rightarrow \mathrm{R}=\frac{\tau}{\mathrm{C}}=\frac{20 \times 10^{-3} \mathrm{~s}}{100 \times 10^{-6} \mathrm{~F}}=200 \Omega \end{aligned}}

 

 

 

Posted by

rishi.raj

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