Question
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate $\frac{\text{dN}}{\text{dt}}=\text{R}.$ An inductor of inductance 100mH, a resistor of resistance $100\Omega$ and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that $\frac{\text{i}}{\text{N}}$ remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.

Answer

$\text{R}=100\Omega;\text{ L}=100\text{mH}$After time t, $\text{i = i}_0\Big(1-\text{e}^{\frac{-\text{t}}{\text{Lr}}}\Big)\text{ N = N}_0\big(\text{e}^{-\lambda\text{t}}\big)$
$\frac{\text{i}}{\text{N}}=\frac{\text{i}_0\big(1-\text{e}^{-\frac{\text{tR}}{\text{L}}}\big)}{\text{N}_0\text{e}^{-\lambda\text{t}}}\frac{\text{i}}{\text{N}}$ is constant i.e. independent of time.
Coefficients of t are equal $-\frac{\text{R}}{\text{L}}=-\lambda\Rightarrow\frac{\text{R}}{\text{L}}=\frac{0.693}{\text{t}_{\frac{1}{2}}}$
$=\text{t}_{\frac{1}{2}}=0.693\times10^{-3}=6.93\times10^{-4}\text{sec}.$

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