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. Sagging Electrons
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 = 1.3806488×10-23 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 = 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.

2. Quantitative 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. Then if electrons have gravitational weight: then 1 gram of T will be produced from S/T grams of S. If electrons have no gravitational weight: (S-se) grams of S will produce (T-te) grams of T. So S/T grams of S will produce either one gram of T if electrons have weight, or if electrons have no weight: then S/T grams of S will produce

(S/T).(T-te)/(S-se) grams of T.

If x is the difference in weight of product T from S/T grams of S

1+x = (S/T).(T-te)/(S-se)
   = (1-te/T)/(1-se/S)
   ≈ (1-te/T)(1+se/S)
So x ≈ e(s/S - t/T) grams 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 + 2HF -> BeF2 + H2

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
s/S ≈ 4/9 = 0.444
t/T ≈ 22/47 = 0.468
S/T = 0.19171167 x ≈ e.(4x47 - 9x22)/423 = e.(188 - 198)/423 = 10e/438 = -0.0236e = -12.9 micrograms
So 0.19171167 grams of Be should produce either 1.000000 or 0.999987 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 to values well below this would be a challenge, particularly as metallic Be in air becomes covered in a thin oxide layer.

So we might consider instead

BeH2 -> BeF2: S = 11, s = 6, T = 47, t = 22.
s/S = 6/11 = 0.5455
t/T = 22/47 = 0.4681
x ≈ e.(6x47 - 11x22)/517 = e.(282 - 242)/517 = 40e/517 = 0.0774e = 42.4 micrograms
This is a better result by a factor of over 3, and avoids the problem of the superficial oxide layer on metallic Be, but the 42.4 micrograms difference would be reduced by about 0.03% by the deuterium content of the BeH2.

We might try to drop the t/T ratio by trying

BeH2 -> BeI2: S = 11, s = 6, T = 263, t = 110.
s/S = 6/11 = 0.5455
t/T = 110/263 = 0.4183
x ≈ e.(6x263 - 11x110)/2630 = e.(1578 - 1210)/2630 = 368e/2630 = 0.1399e = 76.7 micrograms
Although this has the best result of the examples shown, the spontaneous inflammability of BeI2 might make the experiment even more difficult and hazardous than the other two.

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

  Element  
  Atomic  
Number
  Approx  
Atomic
Mass
Be
4
9
F
9
19
Na
11
23
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

As we have seen, the use of other compounds with multi-isotopic elements is also possible, the weights would be complicated by possible changes in isotopic ratios during the reaction, which would need to be measured and allowed for.

These considerations suggest that this way of determining the weight of electrons is probably feasible with readily available equipment and materials.

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written 26 August 2014, updated 9 May 2018

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