Weighing Electrons

Don Herbison-Evans

(donherbisonevans@outlook.com)

**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

= [S(T-te) - T(S-se)]/[S.(S-se)]

= (sTe - tSe)/[S.(S-se)]

≈ e.(sT - tS)/(S

≈ e.(s/S - t/T).(T/S) : equation 2

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.

**Be -> BeF _{2}**

So let us consider the reaction:

The principle is to find out what weight of BeF_{2}
is produced from a weighed amount of Be.

With this reaction: we have

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

S

x ≈ e.[188 - 198)]/81 = -e.[10/81] ≈ -0.123e ≈ -0.123x548 ≈ -68 micrograms per gram Be

either 47.009042/9.012182 = 5.216167

or (47.009042 + 22x0.000548)/(9.012182 + 4x0.000548)

= (47.009042 + 0.012056)/(9.012182 + 0.002192)

= 47.021098/9.014374 = 5.216236 grams of BeF

depending on whether electrons are weightless or heavy.

**BeH _{2} -> BeF_{2}**

Now let us consider the reaction:

S = 9.012182 + 2x1.007825 = 9.012182 + 2.015650 = 11.027832 = ≈ 11

s = 6

T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47

t = 22

sT ≈ 6 x 47 = 282

tS ≈ 22 x 11 = 242

S

x ≈ e.[282 - 242)]/121 = e.[40/121] ≈ 0.331e ≈ 0.331x548 ≈ 181 micrograms / gram BeH

either 47.009042/11.027832 = 4.262764

or (47.009042 + 22x0.000548)/(11.027832 + 6x0.000548)

= (47.009042 + 0.012056?)/(11.027832 + 0.003288?)

= 47.021098/11.031120 = 4.262586 grams of BeF

depending on whether electrons are weightless or heavy.

**BeD _{2} -> BeF_{2}**

Because natural Hydrogen contains some Deuterium, let us consider the reaction :

S = 9.012182 + 2x2.014102 = 9.012182 + 4.028204 = 13.040386 = ≈ 13

s = 6

T = 2 x 18.998403 + 9.012182 = 47.009042 ≈ 47

t = 22

sT ≈ 6 x 47 = 282

tS ≈ 22 x 13 = 286

S

x ≈ e.[282 - 286)]/169

= -e.[4/169] ≈ -0.024e ≈ -0.024x548 ≈ -13 micrograms / gram BeD

(47.009042 + 22x0.000548)/(11.027832 + 6x0.000548)

= (47.009042 + 0.012056)/(11.027832 + 0.003288)

= 47.021098/11.031120 = 4.262586 grams of BeF

depending on whether electrons are weightless or heavy.

Combining these two experiments, with BeH_{2} and BeD_{2},
assuming the proportion D/H in natural Hydrogen is about 1/5000,
if we did the experiment with natural Hydrogen we would get
a differnece of approximately

**General Considerations**

One would of course work with arbitrary
but accurately weighed amounts of the reactants
and scale the results accordingly.
Limiting the effects of impurities and interactions with the atmosphere
and reaction vessels to values well below a few micrograms would be a challenge,
particularly as metallic Be in air becomes covered in a thin oxide layer,
and F_{2} 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,
Fortunately hydrides are solid and could be used to advantage.
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 BeI_{2} with F_{2},
and evaporating off the iodine and iodine fluorides produced,
would give

sT ≈ 110 x 47 = 5,170

tS ≈ 22 x 263 = 5,786

S

x ≈ e.(5170 - 5786)/69169 = -e.616/69169 ≈ -0.008906e

≈ -4.88 micrograms per gram of BeI

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

Number | Atomic Mass | |

As can be seen, only four of these are non-metals, which severely limits our options. However, as we have seen: 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

e is the electron charge ≈ 2 x 10

m is the electron mass ≈ 10

v = (2Ve/m)

So the time of flight travelling L metres will be

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 V_{c} where

**Conclusion**

Both methods described here are at the limit of feasibility. 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. Determining the weight of electrons by the Beam Sag would require the building of extensive and expensive apparatus. The result of either experiment may be that Einstein's Principle of Equivalence is true. Then nothing would change. But it may be that the result would disprove the Principle, which might then lead to an understanding of how to reconcile the two main incompatible current theories of theoretical physics: Quantum Theory and Relativity Theory. That tantalising posibility may make it worthwhile to undertake one of these experiments.

written 26 August 2014, revised 1 May 2020