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प्रश्न
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.
उत्तर
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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