Question

Some smoke detectors (see the figure below) use the radiation produced by the decay of 241 95 Am (half-life 432 yr) to ionize air molecules, which in turn produces a steady current across two electrodes. If smoke enters the region of ionization, the alpha particles are absorbed by the smoke particulates, reducing the current across the electrodes. The electronic circuitry then sounds the alarm when it senses any current reduction. How long does it take for the activity of 241 95 Am to drop to 3.4% of its initial value?

yr

Answer 1

activity of a sample is given by

$A =Ao e^{-\lambda t}$

$\lambda$ = 0.693/T(1/2) = 1.6 x 10^-3 per year

A= 3.4/100Ao

$t =ln(\frac{Ao/A}{\lambda })$

plugging in the values we get

t =2113.37 years