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

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?

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Mộc Miên 3 years 2021-09-03T10:50:48+00:00 1 Answers 31 views 0

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  1. King
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    2021-09-03T10:52:29+00:00
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    Answer:

    860.6 years.

    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} = \dfrac{-ln\dfrac{A}{A_0} }{\lambda} = \dfrac{-ln\dfrac{93000}{370000} }{5.08 \times 10^{-11}} = 2.72 \times 10^{10} \, seconds

    t₉₃₀₀₀ = 2.72 × 10¹⁰ seconds = 860.6 years.

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