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 (i) at any time (t) is given by

`i = i_0 (1-e^((-Rt)/L))`

Number of active nuclei (N) at any time (t) is given by 

`N = N_0e^(-lambdat)`

Where N₀ = Total number of nuclei

`lambda` = Disintegration constant
Now,

`i/N = (i_0(1-e^(-tR"/"L)))/(N_0e^(-lambdat)`

As `i/N` is independent of time, coefficients of t are equal.

Let `t_1/2` be the half-life of the isotope.

⇒ `(-R)/L = -lambda`

⇒ `R/L = 0.693/t_"1/2"`

⇒ `t_(1/2) = 0.693 xx 10^-3 = 6.93 xx 10^-4  "s"`