Decay

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 $t = 1$.

Bonus section:

The slope of the exponential function $1$ is always directly proportional to the value on the y-axis! In this case, the $N$ value is scaled by $ln\frac{1}{2}$ while exponential function $4$ is scaled by a different number $\frac{\ln\frac{1}{2}}{13639194720}$. Intuitively this should make sense: when the amount of radioactive material is halved, so is the activity.