During Robert Milikan's time, he uses the charge-to-mass ratio from J.J. Thomson's research of the electron (which is $1.759\cdot10^{11}\ \mathrm{C/kg}$) to calculate the mass of electron.

Did he know the mass of a hydrogen atom at the time? Because J.J. Thomson already discovered that electron is a lot lighter than a hydrogen atom (about $1800$ I believe). Because if he does, I feel like maybe he could find the mass of electron by $$\frac{\text{hydrogen atom}}{1800}\cdot1.759 \cdot 10^{11}\ \mathrm{\frac C{kg}}$$ I'm still pretty new to chemistry so I'd appreciate if the explanation can go easy on me.

  • 1
    $\begingroup$ The mass of the electron is equal to the mass of the hydrogen atom divided by about 1800. Why are you multiplying by $1.759·10^{11}$ C/kg ? $\endgroup$
    – Maurice
    Aug 27 at 16:15
  • $\begingroup$ See also History of science and mathematics $\endgroup$
    – Poutnik
    Aug 27 at 18:23
  • $\begingroup$ Your argument is circular: the mass of the hydrogen atom and of the electron were determined separately. The ratio which you quote was computed from these separate masses. $\endgroup$
    – Buck Thorn
    Aug 28 at 9:08

Did he know the mass of a hydrogen atom at the time? Because J.J. Thomson already discovered that electron is a lot lighter than a hydrogen atom (about 1800 I believe).

The idea that the mass to charge ratio of cathode ray "particle" (modern day electron) is a lot lighter than hydrogen came about indirectly from electrochemical experiments and the concept of equivalents.

Very briefly, from a Chapter written by Sir J.J. Thomson in 1924


[Continued discussion] previously determined $e / m v^{2}$, so that when we know $v$ we can find $e / m$. This was found to be equal to $1.8 \times 10^{7}$.

Now if $\mathrm{E}$ is the charge of electricity carried by the hydrogen atom in the electrolysis of solutions, and M the mass of that atom, $\mathrm{E} / \mathrm{M}$ can be determined by measuring the quantity of hydrogen liberated when a known quantity of electricity passes through an aqueous solution. This was done long ago, and the result was that $\mathrm{E} / \mathrm{M}=10^{4}$. Special investigations have shown that $e$, the charge on the electron, is equal to $\mathrm{E}$, the charge on the hydrogen ion ; hence since $e=\mathrm{E}$ and $e / m=1.8 \times 10^{7}$, while $\mathrm{E} / \mathrm{M}=10^{4}, m=\mathrm{M} / 1800$ or the mass of an electron is only $1 / 1800$ of that of an atom of hydrogen.

By the way, the absolute mass of atoms in the periodic table is not known. All the atomic weights or atomic masses are relative, relative to an arbitrary element-chosen by humans rather than any fundamental reason. Historically, one of the elements has been given a fixed number as its mass. In the long history of atomic weights, the element with a "fixed" mass was hydrogen. It was given a fixed mass of 1 (exact), later oxygen was chosen as 16 (exact) and this choice moved back and forth. Recently, from 1960s, the fixed element was an isotope of carbon, called C-12. It was given an exact mass of 12. Recently this has been slightly changed.

Following the same lines of reasoning, finding the absolute mass of electrons by experiments was not easy. You had to know the exact magnetic field and electric field values. Finding $e/m$ was a scientific feat of its own. Absolute measurements are always difficult.


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