In a nuclear physics experiment, radioactive isotopes are produced at a constant rate dN/dt = R . A series circuit is formed by connecting an inductor of inductance 100 mH , a resistor of resistance 100 , and a battery.

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 6.93 × 10^ -4 s

The rate of change of the number of active nuclei is given by

\dfrac{dN}{dt} = R - \lambda N

, where

R

is the constant production rate and

\lambda

is the decay constant.

Integrating with the initial condition

N(0) = 0

, we get

N(t) = \dfrac{R}{\lambda}(1 - e^{-\lambda t})

.

For the series LR circuit, the current at time

t

is given by

i(t) = i_{0}(1 - e^{-\frac{R_{res}}{L}t})

, where

R_{res}

is the resistance and

L

is the inductance.

Given that the ratio

\dfrac{i(t)}{N(t)}

is constant over time, the exponential terms must be identical. Thus,

\lambda = \dfrac{R_{res}}{L}

.

Substituting the given values

R_{res} = 100 \Omega

and

L = 100 \text{ mH} = 0.1 \text{ H}

:

\lambda = \dfrac{100}{0.1} = 1000 \text{ s}^{-1}

The half-life of the isotope is

T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{1000} = 6.93 \times 10^{-4} \text{ s}

.

In a nuclear physics experiment, radioactive isotopes are produced...

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