Question

Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt = R. An inductor of inductance 100 mH, a resistor of resistance 100 Ω 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 i/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 1

Given:
Resistance of resistor, R = 100 Ω
Inductance of an inductor, L = 100 mH
Current $\left(i\right)$ at any time $\left(t\right)$ is given by
$i=i_0\left(1-e^{\frac{-Rt}{L}}\right)$
Number of active nuclei $\left(N\right)$ at any time $\left(t\right)$ is given by
$N=N_0{e}^{-\lambda t}$
Where N₀ = Total number of nuclei
$\lambda$ = Disintegration constant
Now,
$\frac{i}{N}=\frac{i_0\left(1-e^{-tR/L}\right)}{N_0{e}^{-\lambda t}}$
As $\frac{i}{N}$ is independent of time, coefficients of t are equal.

Let $t_{\frac{1}{2}}$ be the half-life of the isotope.

$$\begin{gathered} \Rightarrow \frac{-R}{L}=-\lambda \\ \Rightarrow \frac{R}{L}=\frac{0.693}{t_{{\displaystyle \frac{1}{2}}}} \\ \Rightarrow t_{\frac{1}{2}}=0.693\times {10}^{-3}=6.93\times {10}^{-4}\mathrm{s} \end{gathered}$$