Weighing Electrons

Don Herbison-Evans

(donherbisonevans@outlook.com)

**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

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

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

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

= (1-te/T)/(1-se/S)

≈ (1-te/T)(1+se/S)

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.

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

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

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 BeF_{2},
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

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

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

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

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

Number | Atomic Mass | |

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.

written 26 August 2014, updated 9 May 2018