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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.

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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=i01-e-RtL
Number of active nuclei N at any time t is given by
N=N0e-λt
Where N0 = Total number of nuclei
λ = Disintegration constant
Now,
iN=i01-e-tR/LN0e-λt
As iN is independent of time, coefficients of t are equal.

Let t12 be the half-life of the isotope.

-RL=-λRL=0.693t12t12=0.693×10-3=6.93×10-4 s

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