Some smoke detectors (see the figure below) use the radiation produced by the de

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 5.0% of its initial value

Explanation / Answer

we were given half life 432 years

Nuclead decays follows first ordr kinectis

hence t12 = 0.693 /k    where k = decay constant

432 = 0.693 / k

k = 0.001604167

Now they asked time for drop to 5 % of initial value.

Hence initial = 100 % , = a

final = 5 % = a-x

we have formula t = ( 1/k) ln ( a/a-x)

t = ( 1/ 0.001604167) ln ( 100/5)

= 1867 years

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