Don Herbison-Evans
donherbisonevans@outlook.com
26 August 2014, revised 3 June 2017
Introduction
No-one has actually weighed electrons, and maybe their gravitational mass is not the same as their inertial mass as Einstein assumed in his Equivalence Principle. The Eotvos Experiment indirectly implies that electrons are subject to gravity, but it would be nice actually to weigh them.
Electrons constitute the highest proportion of the mass of molecules that are composed of atoms of the lowest atomic number. 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. Thus the use of H, He, Li, B, C, O, N, can be complicated by this fractionation, but Be and F have only one stable isope so considered below as possible choices for this experiment.
Electrons only contribute about 549 micrograms per Grm Dalton, but balances are now commercially available that can measure several grams to an accuracy of 1 microgram, such as the Sartorius Cubis MSE6.6S-0CE-DM balance.
In the following considerations, we shall use
N = Avogadro's number
and
Atomic numbers (number of electrons per atom):
H,D 1
Be 4
F 9
and
Atomic (inertial) masses derived from mass spectrometry:
H 1.00782505 Daltons
D 2.01410178
Be 9.01218315
F 18.99840316
e 0.00054858
Be to BeF_{2} with Weighty Electrons
The simplest suitable reaction seems to be that of the conversion of metallic Beryllium to Beryllium Fluoride.
1 gm atom of Be containing N atoms:
= 9.01218315 grams
1 gm molecule of BeF_{2}
= 9.01218315 + 2 x 18.99840316 = 47.00898947 grams
Normalising : 1 gram of Beryllium would produce
47.00898947 / 9.01218315 = 5.21616002 grams BeF_{2}
Be to BeF_{2} with Weightless Electrons
1 gm atom of Be containing N atoms minus the mass of 4.N electrons:
= 9.01218315 - 4 x 0.00054858 = 9.01218315 - 0.00219432 = 9.00998883 grams
1 gm molecule of BeF_{2} minus the 4.N electrons in the Be,
and 2 x 9.N electrons in the 2F:
= 47.00898947 - 22 x 0.00054858 = 47.00898947 - 0.01206876 = 46.99692071 grams
Normalising : 1 gram of Beryllium would produce
46.99692071 / 9.00998883 = 5.21609090 grams BeF_{2}
The difference in the weight of BeF_{2} produced by by 1 gram of Beryllium is
5.21616002 - 5.21609090 = 69.12 micrograms BeF_{2}
BeH_{2} to BeF_{2} with weighty Electrons
N molecules of BeH_{2} will weigh
= 9.01218315 + 2 x 1.00782505 = 11.02783325 grams
So reducing to unit weight of BeH_{2},
1 gram of BeH_{2} will produce
47.00898947 / 11.02783325 = 4.26275846 grams BeF_{2}
BeH_{2} to BeF_{2} with weightless Electrons
1 gm molecule of BeH_{2} has 6 electrons/molecule,
with electron inertial mass
= 6 x 0.00054858 = 0.00329148 grams
N molecules of BeH_{2} (nuclei) will weigh
= 11.02783325 - 0.00329148 = 11.02454177 gms
So reducing to unit weights:-
1 gram of BeH_{2} will produce
46.99692071 / 11.02454177 = 4.26293643 grams BeF_{2}
BeD_{2} to BeF_{2} with weighty Electrons
N molecules of BeD_{2} will weigh
= 9.01218315 + 2 x 2.01410178 = 13.04038671 gramss
1 gram of BeD_{2} will produce
47.00898947 / 13.04038671 = 3.60487695 grams BeF_{2}
BeD_{2} to BeF_{2} with weightless Electrons
1 gm molecule of BeD_{2} has 6 electrons/molecule,
with electron inertial mass
= 6 x 0.00054858 = 0.00329148 grams
N molecules of BeD_{2} (nuclei)
= 13.04038671 - 0.00329148 = 13.03709523 grams
So reducing to unit weights:-
1 gram of BeD_{2} will produce
46.99692071 / 13.03709523 = 3.60486135 grams BeF_{2}
Discussion
The results are shown in table 1:
produced | 1 gram Be | 1 gram BeH_{2} | 1 gram BeD_{2} |
electrons | |||
electrons | |||
micrograms |
Using inertial masses, 1gm Be with HF would produce 5.21616002 gms BeF2 if electrons are subject to gravity, and 5.21616002 if not, a difference of 69.12 micrograms easily measured by good balances. Has anyone done this?
One would of course work with arbitrary but accurately weighed amounts of Be, BeH_{2}, and BeD_{2}, and scale the results accordingly. These calculations suggest that taking the above BeH_{2} reactions, the difference between heavy and weightless electrons in producing BeF_{2} would be about 178 micrograms per gram of BeH_{2}.
The H in the BeH_{2} would not have to be isotopically pure to get a measurable result, but the purer it is isotopically, the greater the difference will be in the answers. Be and F are monoisotopic so there are no fractionation problems to be considered in these reactions, but spray, impurity levels, and corrosion are problems that would need solving.
One might alternatively consider other reactions involving the other monoisotopic elements:
Na, Al, P, Sc, Mn, Co, As, Y, Nb, Rh, I, Cs, Pr, Tb, Ho, Tm, Au,
although the neutron imbalance of the heavier elements would make the result harder to measure.
For the general case, consider the transformation of molecule A with molecular mass 'A' Daltons into a molecule B with molecular mass 'B' Daltons. Let 'a' be the number of electrons in A, and 'b' be the number electrons in B. Let 'e' be the mass of an electron in Daltons. Then if electrons have gravitational weight: then 1 gram of A will produce B/A grams of B. If electrons have no gravitational mass: (A-ae) grams of A will produce (B-be)grams of B. So 1 gram of A will produce (B-be)/(A-ae) grams of B. The difference 'x' will be
So for BeH_{2 -> BeF2 }