Assume that there is a pulse on a scope with \$100\;mV\$ height, as shown in picture
This pulse is coming from a photomultiplier tube. How to determine the number of electrons from which this pulse is constructed?
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Assuming 1MOhm input resistance of the oscilloscope you can compute the current from the voltage and integrate it to get the charge. Divide by the charge of an electron and you get the number of electrons.
For the lower pulse that would give about -200nA * 15ns = -3e-15 C ~ 20000e.
Note that I'm not an expert w.r.t. photomultipliers, so I don't know whether this is of the right magnitude and whether there are some caveats of this measurement method.
PMT pulse heights generally go up to 1 or 2 volts. Here is an example of an FEU-84 at ~1400V, looking at an LED flasher. The oscilloscope is a 300 MHz Tek.
Historically, people have often used a triangular approximation for the total charge, which will give you a fairly good estimate. Looking at the image above, let's say we have a pulse width of 40 ns and a pulse height of 700 mV, which is a current 'height' of 700 mV / 50 \$\Omega\$= 14 mA. The area of a triangle is base times height over two, so the total charge is 280 pC.
One wrinkle: if you are interested in knowing the number of electrons that hit the anode, you need to look at the base circuit. There may or may not be a termination resistor on the PMT side of the output cable. If there is no resistor, the 'scope is seeing the entire signal from the anode. If there is a resistor, then the 'scope is only seeing half of the signal. This assumes that cables and terminations are consistent, e.g. 50 \$\Omega\$.