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Chemistry · Ch 2 — Structure of Atom

Charge on the Electron

2.1.3

Charge on the Electron

The Charge on the Electron: Millikan’s Oil Drop Experiment

The discovery of the electron by J.J. Thomson gave us the charge-to-mass ratio (e/mee/m_e), but it did not give us either quantity separately. To know the mass of the electron, we first needed to know its charge. That measurement came from Robert A. Millikan, who, between 1906 and 1914, devised the famous oil drop experiment.

Millikan’s method was elegant in its simplicity. He sprayed tiny droplets of oil into a chamber. As the droplets fell, they passed through a hole and entered a region between two charged metal plates. By observing the motion of a single oil drop under the influence of gravity and an adjustable electric field, Millikan could determine the charge on that drop.

The key insight was that the charge on any oil drop was always an integer multiple of a smallest, indivisible unit of charge. By measuring many drops, he found this fundamental unit to be −1.6×10−19 C-1.6 \times 10^{-19} \, \text{C}. The modern accepted value is slightly more precise: −1.602176×10−19 C-1.602176 \times 10^{-19} \, \text{C}.

Note

The negative sign is a convention. It tells us the electron carries a negative charge, but the magnitude of the charge — the number 1.6×10−191.6 \times 10^{-19} — is what matters for calculations.

Determining the Mass of the Electron

Once Millikan had measured the charge (ee), the mass of the electron (mem_e) could be calculated directly. Thomson had already given us the ratio e/mee/m_e. We now had the numerator (ee), so the denominator (mem_e) was just a division away.

The calculation is straightforward:

me=e(e/me)m_e = \frac{e}{(e/m_e)}

Using Millikan’s value for ee (−1.6×10−19 C-1.6 \times 10^{-19} \, \text{C}) and Thomson’s value for e/mee/m_e (−1.76×1011 C/kg-1.76 \times 10^{11} \, \text{C/kg}), we get: …