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In a fire alarm, if we start with americium-241 which emits radiation at 370 kBq (as an example) what will have happened to the decay rate after 470 years (half life is 470 years). The amount of americium will be half, but does this mean that the radiation will also half, thus being 185 kBq?

In other words, does the rate at which radioactive particles are emitted also change with time?

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  • $\begingroup$ On a probability per atom basis, no. $\endgroup$
    – Jon Custer
    Sep 4, 2019 at 20:54
  • $\begingroup$ @JonCuster no, but say you have a hundred atoms, and then after one half life you have fifty, then the total radiation per second will also be cut in half? $\endgroup$
    – Melvin
    Sep 4, 2019 at 20:55
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    $\begingroup$ One should be careful about exactly what scenario you are dealing with. In the case of Am-241, it decays to Np-237 which itself is unstable with a half-life of about 2 million years. Such decay chains can get pretty messy if you are just asking about the total rad or rem emitted with time. $\endgroup$
    – Jon Custer
    Sep 4, 2019 at 20:59
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    $\begingroup$ After one half-life, the radiation from that isotope is cut in half, yes. $\endgroup$
    – Jon Custer
    Sep 4, 2019 at 21:00
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    $\begingroup$ @Melvin, Yes. Exactly right. The rate of disintigrations always is proportional to the number of atoms in your sample. $\endgroup$
    – Solomon Slow
    Sep 4, 2019 at 21:00

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In other words, does the rate at which radioactive particles are emitted also change with time?

The founding principle of the half-life Law is that:

$$\frac{\text{d}N(t)}{\text{d}t}=-kN(t)$$

where $k$ is a constant. In plain English:

The decay rate of a radioactive substance is proportional to the amount of radioactive substance.

In the case of $^{241}\text{Am}$, indeed after 470 years radioactivity will have dropped to half the initial value.

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