Weighing Electrons
Don Herbison-Evans
(donherbisonevans@outlook.com)

I have this silly notion that no-one has actually weighed electrons, and maybe their gravitational mass is not the same as their inertial mass. The Eotvos experiment certainly implies that electrons are subject to gravity, but I thought it would be nice to actually weigh them. Two methods come to mind.

1. Gravimetric Chemistry
For the general case, consider the conversion of molecule S with molecular mass 'S' Daltons, into a molecule T with molecular mass 'T' Daltons. Let 's' be the number of electrons in S, and 't' be the number electrons in T. Let 'e' be the mass of an electron in Daltons. Bearing in mind that, generally, atomic masses are determined by Mass Spectrometry, and so are inertial masses, then if electrons do have gravitational weight: 1 gram of S will be produce T/S grams of T. But if electrons have no gravitational weight: (S-se) grams of S will produce (T-te) grams of T. So then 1 gram of S will produce (T-te)/(S-se) grams of T if electrons have no weight

If x is the difference in weight of product T from 1 gram of S

x = (T-te)/(S-se) - T/S
   = [S(T-te) - T(S-se)]/[S.(S-se)]
   = (sTe - tSe)/[S.(S-se)]
   ≈ e.(sT - tS)/(S2)             : equation 1
   ≈ e.(s/S - t/T).(T/S)             : equation 2
where e = 548 micrograms Electrons constitute the highest proportion of the mass of molecules that are composed of atoms of the lowest atomic number, so it is tempting to use the lightest elements for this experiment. However measurements based on chemical reactions can be complicated by changes in isotopic ratios for molecules containing elements with more than one naturally ocurring stable isotope. The lightest elements that have only one stable isope, and so avoid this fractionation, are Be and F.

So let us consider the reaction:

Be + F2 -> BeF2

The principle is to find out what weight of BeF2 is produced from a weighed amount of Be.

Electrons only contribute about 548 micrograms per Gram Dalton, but balances are now commercially available that can measure 2.5 grams to an accuracy of 1 microgram, such as the Sartorius Cubis Ultramicro Balance MSA

With this reaction: we have

Be -> BeF2 :-
S = 9.012182 ≈ 9
s = 4
T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47
t = 22
sT ≈ 4 x 47 = 188
tS ≈ 22 x 9 = 198
S2 ≈ 81
x ≈ e.[188 - 198)]/81 = -e.[10/81] = -0.123e = -67.7 micrograms
So 1 gram of Be should produce approximately either 5.216167 or 5.216099 grams of BeF2 depending on whether electrons are heavy or weightless.

One would of course work with arbitrary but accurately weighed amounts of the reactants and scale the results accordingly. So if the values are scaled to consider the production of 1 gram of BeF2, the electrons would make a difference of -12.9 micrograms. Limiting the effects of impurities and interactions with the atmosphere and reaction vessels to values well below this would be a challenge, particularly as metallic Be in air becomes covered in a thin oxide layer, and F2 attacks nearly everything.

From equation 2, we appear to need to maximise the difference s/S - t/T, while minimising T/S. Hydrogen has s/S ≈ 1, but gases are hard to work with and weigh accurately. Other lighter elements have s/S ratios near 0.5, and the heavier elements have values approaching 92/238 ≈ 0.39. So we might try to drop the t/T ratio by using a reaction that converts between heavy elements to light elements, for example: reacting BeI2 with F2, and evaporating off the iodine and iodine fluorides produced, would give

BeI2 -> BeF2: S = 263, s = 110, T = 47, t = 22.
sT ≈ 110 x 47 = 5,170
tS ≈ 22 x 263 = 5,786
S2 ≈ 69,169
x ≈ e.(5170 - 5786)/69169 = -e.616/69169 ≈ -0.008906e
≈ -4.88 micrograms
This appears to be a worse result than before, but the electron weight difference for 1 gram of the product BeF2 -4.88 x (263/47) ≈ -27.3 micrograms which is an improvement over -12.9 micrograms from the reaction Be -> BeF2. However, the spontaneous inflammability of BeI2 in air might make this experiment even more difficult and hazardous than Be -> BeF2.

One might alternatively consider other reactions involving the other stable mono-isotopic elements:

  Element  
  Atomic  
Number
  Approx  
Atomic
Mass
Be
4
9
F
9
19
Na
11
23
Al
13
27
P
15
31
Sc
21
45
V
23
51
Mn
25
55
Co
27
58
As
33
75
Y
39
89
Nb
41
93
Rh
45
103
I
53
127
Cs
55
133
Pr
59
141
Tb
65
159
Ho
67
165
Tm
69
169
Au
79
197
Bi
83
209

As can be seen, only four of these are non-metals, which severely limits our options. However, it is also possible to use compounds with multi-isotopic elements although the weights would be complicated by the isotopic ratios, which would need to be measured and allowed for.

2. Beam sag
Consider a vacuum tube L metres long with an electron gun at one end accelerating an electron beam by some voltage V striking a phosphorescent screen at the other end. If electrons have a gravitational weight proportional to their inertial mass: the beam will sag due to the electrons falling by gravity as they fly from one end of the tube to the other.

The amount of sag can be calculated by noting that the electrons will have an energy of

Ve = (1/2)mv2 joules, where V is in volts
e is the electron charge ≈ 2 x 10-19 coulombs
m is the electron mass ≈ 10-30 kilograms
so that v2 = 2Ve/m
v = (2Ve/m)1/2 metres/second

So the time of flight travelling L metres will be

t = L/v seconds during which it will sag s = (1/2)gt2 metres where g ≈ 10 metres/second2
so that s = (1/2)gL2/v2 = (1/2)gL2/(2Ve/m) = gmL2/(4Ve) If V = 1 Volt, and L = 1 metre: s ≈ 10 x 10-30/(4 x 2 x 10-19) ≈ 10-11 metres Sadly this appears to be too small to detect by conventional methods.

The result varies as the square of the tube length and inversely with the accelerating voltage. The minimum voltage will be determined by the thermal distribution of electron velocities as they leave the cathode of the electron gun.

If the cathode temperature is T Kelvin, the electrons will have a range of velocities about a median which is the equivalent of a voltage Vc where

eVc = k.T where k = Boltzmann's constant = 1.3806488 x 10-23 J/K so that Vc = kT/e For a cathode temperature of say 500 K Vc ≈ 1.4 x 10-23 x 500/(2 x 10-19) ≈ 0.035 Volt So if we dropped the voltage to say 0.1 Volt and lengthened the tube to 1 Kilometre, the sag of the electron beam would become s ≈ 10-11.106/0.1 = 10-4 m = 0.1 mm The curvature of the electron beam by the magnetic field of the earth can be minimised by aligning the tube along the local field lines of force, and shielding it with mu-metal.

These considerations suggest that this way of determining the weight of electrons may just be on the limit of feasibility, but would require the building of extensive and expensive apparatus.

Conclusion

All the above considerations suggest that gravimetric chemistry is a better way of determining the weight of electrons than using beam sag. The gravimetric chemistry method is probably feasible with available equipment and materials, but would require very pure reactants, ideally 99.9999% pure, very careful handling, and possibly the use of inert atmospheres.

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written 26 August 2014, updated 17 November 2018

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