Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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A Charged Capacitor of Capacitance C is Discharged Through a Resistance R. a Radioactive Sample Decays with an Average-life τ. Find the Value of R for Which the Ratio - Physics

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Question

A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life τ. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

Solution

Discharging of a capacitor through a resistance R is given by

`Q = qe^-t"/"CR`

Here, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance

Energy , `E = 1/2 Q^2/C = (q^2e^(-2t"/"CR))/(2C)`

Activity , `A = A_0e^(-lambdat)`

Here , `A_0` = Initial activity

`lambda` = Disintegration constant

∴ Ratio of the energy to the activity = `E/A = (q^2 xx e^(-2t"/"CR))/(2CA_0e^(-lambdat)`

Since the terms are independent of time, their coefficients can be equated.

`(2t)/(CR) = lambdat`

⇒ `lambda = 2/(CR)`

⇒`1/tau = 2/(CR)`

⇒ `R = 2 tau/C`

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Solution A Charged Capacitor of Capacitance C is Discharged Through a Resistance R. a Radioactive Sample Decays with an Average-life τ. Find the Value of R for Which the Ratio Concept: Radioactivity - Introduction of Radioactivity.
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