A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 μF is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be _____________ Ω.

Solution

Physics sir · 4 steps

Both quantities fall exponentially, so their ratio stays fixed only when the two time constants match.

1. Discharging capacitor. Through a resistor, the charge falls as q = q₀ e^(-t/RC). Its time constant is RC.
2. Decaying sample. The activity falls as A = A₀ e^(-λ t) = A₀ e^(-t/τ), where τ = 1/λ is the average (mean) life.

   q/A = q₀/A₀ e^(-t(1/RC - 1/τ))

3. Match time constants. This ratio is independent of t only if the exponent is zero, so RC = τ. That gives R = τ/C = (30 × 10⁻³ s)/(200 × 10⁻⁶ F).
4. So. R = 150 Ω, so the answer is 150.

The idea to keep. Two exponential decays keep a constant ratio only when their time constants are equal. For a nucleus that time constant is the mean life τ = 1/λ, not the half-life.

Answer from the NTA final answer key. Solution by PrepWiser.