I've done a little bit of research and it seems Millikan was able to measure the ratio between the charge of the electron and its mass. But how can one measure one of the two constants to get the value of the other?
Experimental Physics – How to Measure the Mass of the Electron
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Because $e/m_e$ appears in macroscopic equations, but $e,m_e$ individually do not.
The force equation for a general electron in an electromagnetic field is $F=e(...)$. This makes acceleration proportional to $\frac{e}{m_e}$.
Acceleration is not that hard to measure, especially if you measure it in the form of deflection. Piping electrons from a beta emitter into a calibrated magnetic field will give you a deflection in their path, and this can be measured by having them strike a phosphor screen.
So, $\frac{e}{m_e}$ is easy to measure.
On the other hand, it is hard to measure $e$ and $m_e$. To do so we would need to create a system where the force is acting only on the electron, but it is attached to a larger, measurable mass. This replaces the $m_e$ in $F=m_ea=e(...)$ with a larger $m$, and now from experiment we can measure $e$.
That's exactly what Millikan did when he measured $e$. He took oil drops with charges that were small multiples of $e$. The mass of the drops could be calculated from their behavior in freefall1, and the electrostatic force required to balance them is measured.
1. While Galileo's ball drop experiment tells us that the behavior of a falling body doesn't depend on mass; this is no longer applicable when we consider viscosity, and thus by carefully monitoring the behavior we can determine the mass of a falling drop.
Yes, the mass of the electron can be measured by the energy of the gamma rays involved in electron-positron annihilation. The mass of an electron is equal to the energy (divided by the speed of light squared) of one of those gamma rays, as measured in the center of momentum frame.
The measurement of gamma ray energies is known as Gamma-ray spectroscopy. The general idea is that the gamma ray is absorbed by some material, and the resulting ionization is detected in some way (cheaper detectors tend to use a scintillating material, while more expensive detectors use semiconductor technologies). The detector response is calibrated with a source of known gamma energies, and then the energy of another gamma ray can be measured.
The detector has a finite resolution, but the measurement can be enhanced by looking at extra features at the spectrum. For instance, there's a "Compton edge" that appears from gamma rays that Compton scatter rather than being absorbed, and an extra peak at twice the electron mass (when both gamma rays from the annihilation are absorbed).
That said, the relative uncertainty on the electron mass is only around $10^{-8}$, which is much better than you can get with standard gamma ray spectroscopy.
Some physics programs (Rutger's, for instance) have their students actually run this experiment in a lab course.
Best Answer
The mass-to-charge ratio $m/e$ of the electron was first measured by J.J. Thomson, the discoverer of the electron, using cathode rays in 1897:
It should not be surprising that one may measure this ratio even without isolating "individual electrons" because the electric force acting on a charge may be written as $$ F = ma = Ee, \quad a = E\cdot \frac em $$ So what was left was just to measure the mass or charge separately. Millikan and Fletcher did the relevant oil drop experiment in 1909.
The electric force $F=Ee$ acting on a single drop with charge $e$, a single extra (or deficit) electron, may be calculated when it is set equal to the drag force from hydrodynamics, $6\pi e\eta v_1$. The viscosity $\eta$ is the most difficult thing to know but otherwise all quantities are known so $e$ may be calculated.
If one knows the charge and the ratio, one may calculate the mass as $m = e/ (e/m)$.