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

The half-life of the isotope

$=6.93 \times 10.4 \mathrm{s}$

Explanation:

It is given:

• Resistor’s resistance, R = 100 Ω

• Inductor’s inductance, L = 100 mH

• At any time t, current i is shown as i = i01-e-RtL

• At any time (t), active nuclei’s number (N) is shown as

• N = N0e-λt

• where λ = constant of disintegration, N0 = Nuclei’s total number  

• Now,

• iN = i01-e-tR/LN0e-λtAs

• iN is not dependent on time and the t coefficients are equal.

• Let the isotope’s half-life be t12.  

• ⇒-RL=-λ  

• ⇒$\mathrm{t} 12=0.693 \times 10-3$

• $=6.93 \times 10-4 s$