Question

Americium-241 is used in smoke detectors. It undergoes alpha decay with a half life of 432 years. The alpha particles ionize some of the air near the americium-241 source, and these charged ions are collected on a metal plate that has a small voltage applied to it. The collected particles cause a small electric current to flow in the detector. If smoke particles get in between the americium-241 source and the detector and reduce the current, the alarm is triggered. A typical smoke detector experiences 370000 alpha decays each second. How much time is needed for the number of decays to reduce to 93000 decays per second?

Answer 1

860.6 years.

Step-by-step explanation:

The parameters given are;

Initial detector activity = 370000 alpha decays per second

Final detector activity = 93000 alpha decays per second

Formula for time to change in activity is given by the following relation;

`t_(93000) = (-ln(A)/(A_0) )/(\lambda) = (-ln(93000)/(370000) )/(5.08 * 10^(-11)) = 2.72 * 10^(10) \, seconds`

t₉₃₀₀₀ = 2.72 × 10¹⁰ seconds = 860.6 years.