Smoke detectors are fascinating. They contain a tiny radioactive sample of Americium-241, the atoms of which are continuously decaying into Neptunium-237. Alpha decay is when the nucleus of an atom emits an alpha particle, which is made of 2 protons and 2 neutrons. The half-life of \(Am\) is approximately 432.2 years, which means that if you start off with \(N_0\) atoms at \(t=0\) then at \(t=1\) you'll have \(\frac{1}{2}N_0\) and at \(t=2\) you'll have \(\frac{1}{4}N_0\) left. Generally:
$$N_t = N_0(\frac{1}{2})^t\tag{1}$$
This is an exponential function, so if you wanted to, you could rewrite it as:
$$N_t = N_0e^{t\ln\frac{1}{2}}\tag{2}$$
\(\ln\frac{1}{2}\) is an interesting constant that will reappear later. It's roughly equivalent to \(-0.69314718\).
If you plotted the above graph, you might note that time is on the x-axis and the number of \(Am\) atoms is on the y-axis. This number is always decreasing as time increases. For a given time interval \(t=a, t=b\), the decrease in \(N\) from \(N_{a}\) to \(N_{b}\) is a direct consequence of the activity.
Activity, measured in Becquerels \(Bq\), is the number of decays per second, but as our graph's time axis is currently measured in half-lives we can't just read the activity directly from the slope.
Approach #1
$$432.2 years\times365.25\times24\times3600 = 13639194720 secs\tag{3}$$
If we already had \(N_0\) we could plug \(\frac{1}{13639194720}\) into our first equation and see \(N\) one second later. This would be the number of Becquerels.
Approach #2
Modify the original equation so that \(t\) is no longer measured in half-lives but instead in seconds:
$$N_t = N_0(\frac{1}{2})^\frac{t}{13639194720}\tag{4}$$
So actually, both approaches are identical, but the second approach is more reusable: we've taken the general equation \(1\) and altered it to be specific to the half-life of Americium-241.
Let's find a typical value of \(N_0\) for our smoke detector example, where a Google search reveals about 0.3μg is common. How do we find the number of Americium-241 atoms in 0.3μg? Avogadro's number: \(6.02214076\times10^{23}\) is the number of particles in one mole.
$$N_0 = 0.3\times10^{-6} / 241 \times 6.02214076\times10^{23} = 7.49644\times10^{14}\tag{5}$$
Let's plug this number into equation \(4\). Unsurprisingly, if you follow along in Excel, you'll see the same number for \(N_0\) as \(N_1\).
$$N_1 = 7.49644\times10^{14}\tag{6}$$
However, if you rearrange it, you'll find that Excel didn't throw away the bits, it just hid them away with some formatting that prioritises the most significant digits:
$$N_0 - N_1 = 38097.125\tag{7}$$
Here you have it. In a smoke detector with just 0.3μg of Americium 241, there are approximately 38k decays per second. Alpha particles are being emitted at 38kHz! Note that in equation \(7\), \(N_1 - N_0\) would give us a negative slope of \(-38097.125\) indicating that our sample contained fewer atoms at \(t = 1\).
