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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.
Solution
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"`
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