Johann Marinsek 2015
Refutation of binding energy formula EB = ∆ mc2
1: Uranium fission violates energy conservation
2: Impossibility:
Alteration from C+-12.00 to H+-1.00 scale results in altered binding energies
Ontological excursus:
Atoms are not comprised of a nucleus and
extra nuclear electrons in orbitals!
In the course of fission the artful uranium electron orbital structure must
implode.
When the first fragmentation is finalized the electrons reside at the surfaces of
Xe-140 and Sr-94.Then two reactions proceed with β- radiation
1: Xe-140 → Cs-140 → Ba-140 → La-140 → Ce-140;
2: Sr-94 → Y-94 → Zr-94
These intermediate fragments are not considered further...
Imagine the supernatural resurrection of Ce and Zr orbital shells:
Ce and Zr allegedly possess after fission of U-235 their artful particular electron
orbital structures... Obviously, this is physically impossible.
Any electron must know where to go: 58 electrons must pilot to Ce and 40
electrons must pilot to Zr... An invisible hand must pilot them.
But there is no invisible hand in physics...
Uranium fission demonstrates the failure of the Bohr atom:
There are no extra nuclear electrons!
U –235 + n → intermediate fragments... → Ce-140 + Zr-94 + 2n + Q
(Figures from wikipedia show the old-fashioned electron shells. But electron
orbitals also cannot solve the problem.)
Uranium fission is not an empirical confirmation
of binding energy formula EB = Δ mc2
According to relativistic formula EB = Δ mc2 uranium fission shows a
wonderfull proliferation of energy!
The nuclear reaction is:
235
U + n → 140Ce + 94Zr + 2n + Q (energy released) [MeV]
The recipe to calculate the binding energy of an element
A
Z
XN is:
1: Calculate the sum of inert masses of electrons and protons: Z × 1,007825
2: Calculate inert mass of all neutrons: N × 1,008665
3: Add both numbers and subtract inert mass of the element. Obtain ∆m (mass
defect)
4: Multiply mass defect ∆m with 931.49 and you get binding energy.
Example 235U:
Q [MeV] = (m[235U] + m[n] – m[140Ce] – m[94Zr] - 2m[n]) × 931.49 = 208.2
Element
A
ZXN
∑me + mp =
Z ×1,007825
∑ mN =
N × 1.008665
∑ me+mp+mN
mi
me+mp+mN- mi
= ∆m
235
92
235.043933
92.7199
144.23695
236.95685
1.913
U143
∆m ×
931.49
MeV/c2
1782
EB = Δ mc2 delivers invalid binding and nuclear reaction energies. Why?
Under the line of the nuclear reaction we insert the binding energies
[in MeV ] according to formula E = ∆ mc2. Note that we took into
consideration only binding energies, not rest energies. (The Q-value is
approximately confirmed experimentally.)
235
U
1782
+ n
+?
→
≠
140
Ce + 94Zr
1173 + 815
+ 2n + Q
+ ? + 208
Result:
After fission the calculated binding energies of the fragments Ce and Zr add up
to
173 + 815 = 1988 MeV. The source of fission energies (kinetic energies and
binding energies of the fragments) is the binding energy of the parent uranium
atom = 1782 MeV
Surprisingly the binding energies of the fragments plus the kinetic enrgy (208
MeV) of the fragments surmount the binding energy of the parent uranium
atom!
For energies there is a huge imbalance: 1782 ≠ 2196 [MeV].
A wonderfull proliferation of 414 MeV energy discredits relativistic binding
energy calculation!
Graph of Finkelnburg (1956): Binding energy as a function of mass number A.
Assumption: U-240 decays into 2 fragments with A = 120.
Binding energies of fragments = 2000 MeV, binding energy of uranium = 1800
MeV. Finkelnburg did not realize this impossibiity and therefore the obvious
refutation of EB = Δ mc2 .
EB=1000 +1000
Uranium fission:
Binding energies of fragments greater
than binding energy of uranium
EB=1800
Current physics does not attribute a binding energy to n.
Even a correction due to (low) neutron binding energy cannot solve the problem.
The neutron is not an elementary particle but a composed one.
The formula EB = Δ mc2 for binding energy is based on the mass/energy
conversion formula E = mc2 that relates energy to rest mass.
Here it is shown that EB = Δ mc2 is mistaken because applied to fission it
violates the energy conservation bookkeeping rule!
All particles of U (92 electrons and protons, 143 neutrons) plus the neutron that
causes the reaction remain conserved.
Cigar-shaped 235U decays into two pieces. During the break, if uranium electrons
are arranged in orbital’s, must crash into the disintegrating nucleus. Electrons of
the fragments, do they undergo a resurrection into the predetermined orbital’s of
quantum mechanics?
Only a part of the binding energy of 235U is transformed into radiation energy,
namely the binding energy that glues Ce and Zr together.
The alleged mass-energy conversion formula E = mc2 cannot provide a recipe to
calculate
1: the binding energy for the fission fragments;
2: kinetic energies of fragments;
3: radiation energy.
(Intermediate fragments of the fission process are not considered here.)
The source of the binding energies of Ce and Zr as well for the released energy
is the binding energy of uranium!
The uranium binding energy
minus 208 MeV energy
release (= kinetic energy of
fragments) =
= roughly the binding energy
of fragments Ce + Zr.
The necessity of this energy
balance shows the analogue in
macrophysics:
A body consists of
4 masses each m.
They are connected by 4
constrained strings.
The potential energies of
springs are 100 and 900 MeV,
respectively, see their
locations.
Unfix now the horizontal
strings with 100 MeV energy
each!
Left and right part of the body
drift apart.
Each fragment possesses now
a kinetic energy of 100 MeV
and a binding energy of
900MeV.
Energy conservation before and after fission:
binding energy: 2×900 +2×100 —>
2×100 kinetic energy + 2×900 binding energy.
Relativistic uranium binding energy calculation violates energy conservation rule!
Wechsel der Ionen-Bezugsmasse: statt C+= 12.00 auf H+ = 1.00
Dadurch ergäbe sich eine Vergrößerung der Bindungsenergie. Bindungsenergie
somit nicht eindeutig, Formel EB = Δ mc2 somit falsch.
Element
Ion!
H-1 = p
D-2
He-4
C-12
Ni-58
U-235
MS Referenzmasse
C+ = 12.000
Referenzmasse
p + = 1.00
1.0072764668
2.0135532
4.0026
12.000
57.935342
235.043933
1.000
1.999
3.974
11.913
57.51682
233.345976
Bindungsenergie von U-235; A = 235, Z = 92, N = 143
EB= (Z mp+ Z me+(A-Z) mn – m(A, Z)) × 931,5 (MeV)
EB = 92.05047 +144,23908 - 233.345976 =
= (236.28955 – 233.345976) × 931,5 =
= 2.943574× 931.5 = 2741.939 MeV !
Bindungsenergie Ni-58= 896 MeV = 15 MeV per Nukleon
Vergleiche Theorie-Experiment
Measurements of released energies: (rough) kinetic energies of fragments: source [kik]
Fragment amu
Kinetic energy MeV
Nd-142
79,5
Nb-93
106
Ce-140
80.5
Zr-90, Zr-94
104.5
Kr-89
102.5
Ba-144
77.5
Comparison:
measured energies versus energies according to E = mc2 calculation
142
140
140
144
Reaction: measured
Nd+90Zr =
Ce+93Nb =
Ce +94Zr =
Ba +89Kr =
kinetic energy MeV
= 184
= 186.5
= 185
= 180
Kinetic energy neutrons
6
5
5
5
measured kinetic energy + = 220
= 221
220
= 215
30 MeV radiation
E = mc2 calculation:
= 196
= 187
= 208
= 177
kinetic energy + radiation
Energy (MeV) distribution in fission reactions www.science.uwaterloo.ca/~cchieh/cact/nuctek/fissionenergy
Kinetic energy of fission fragments
Kinetic energy of fission neutrons
-6
Prompt (< 10 s) gamma ray energy
Gamma ray energy from fission products
Beta decay energy of fission products
Energy as antineutrinos (ve)
167 MeV
8
8
7
7
7
Energy From Uranium Fission according to hyperphysics (Mulligan)
Form of Energy Released
Kinetic energy of two fission fragments
Immediate gamma rays
Delayed gamma rays
Fission neutrons
Energy of decay products of fission
fragm. rays
Gamma
Beta particles
Neutrons
Average total energy released
Energy Released (MeV)
168
7
3-12
5
...
7
8
12
215 MeV
According to Bogdansky [bog] kinetic energy of fragments is 173 and radiation
is 34 MeV.
For comparisons we take for kinetic energy of fragments and neutrons
184 + 6 = 190 MeV
30 MeV is taken as the median energy of radiation.
(Radiation energy is probably for any reaction in the same range.)
So 30 + 190 = 220 MeV is a reasonable approximate energy release of U-235
fission reactions. Computed values for reaction energies differ in a larger range
of 177 - 208 MeV (mean 192)!
One cannot say that fission measurements are an experimental confirmation of
EB = Δ mc2 ! And different sources offer different magnitudes...
Apropos:
U-235 + n
U-238 + n
Why?
→
fission of U
→ no fission of U
Conjecture:
Of course, low energy neutrons cannot split neither 235U nor 238U atoms.
Neutrons simply bounce off from 238U.
Why can a low energy neutron activate the fission process of 235U ?
Every atom is an oscillator with many Eigen frequencies. 235U and 238U possess
different Eigen frequencies. In the case of 235U the oscillating neutron
incorporates into 235U and „pitches“ at once a resonance frequency that excites
the emerging 236U atom:
235
U
+ n →
236
U* → ....
fragments + radiation
The excited 236U* agitates the aether, that is radiation. Intermediate fragments
decay and are β–emitters. Source of radiation is potential energy stored in 235U.
In principle, it is possible to split also 238U. But a resonance frequency to start
the process must be known.
Did Nobel price winners Cockcroft and Walton
confirm experimentally E = Δ mc2 ? No!
Cockcroft and Walton: The first alleged experimental validation of E = Δ mc2
according to: http://chem.chem.rochester.edu/~chm132tr/lectures/lecture_16.pdf
Cockroft and Walton in 1932 took a beam of protons accelerated to
high energy by a particle accelerator... and bombarded lithium nuclei:
The nuclear reaction considered was:
p + 7 Li → 2 α + Q (released energy).
From the masses of the reactants and products, due to E = ∆ mc2 it is allegedly
possible to calculate the energy liberated in the splitting process:
Calculation of Cockcroft with recent values for masses:
Masses: 7 3Li = 7.016005, p = 1.00727647, 4 2He = 4.0026033
Mass defect Δm = 7.016005 + 1.00727647 - 2×(4.0026033)
= 0.01807547 amu.
This corresponds to 0.01807547 amu × 931,49 MeV/amu = 16.84 MeV.
The claim is that the formula EB = Δ mc2 is confirmed because the calculated Qvalue is in good agreement with the observed 17,2 MeV of Cockcroft and
Walton.
But this no a proof for the relativistic Q-value because Cockcroft ignored the
kinetic energy of the proton in order to calculate the Q-value!
Secondly, energy conservation of binding energies and released energy not
shown!
If we assume that the observed kinetic energy for alpha particles (17 MeV) is
approximately accurate, then we can calculate the energy balance for binding
energies in order to estimate the kinetic energy of the proton. Cockcroft in his
Nobel lecture mentioned protons with 280 kilovolt energy.
Binding and kinetic energies considered:
The main source of the released energy is the binding energy of Li. Regarding
estimated binding energies of both the reactants and the products as well as the
kinetic energies of p and the 2α’s we get the following result for the energy
imbalance:
p
p
Kinetic energy
Binding energy
MeV ?
10 MeV ?
7
→
2α
+Q
Binding energy
→
Binding energy
Released energy
39 MeV ?
→
56 MeV ?
17 MeV
Li
But this calculation is fictitious because it was shown that relativistic binding
energies are meaningless.
The main experimental problem is the measurement of kinetic energies.
References
[eon] Einstein, A., Times, Space and Gravitation. In Out of my Later Years, N.Y. 1950
[bog] Bogdansky, D., Nuclear energy..., 2nd ed. N.Y. 2004
[fin] Finkelnburg, W. Einführung in die Atomphysik, Berlin 1956
[lei] Leibniz, Neues System der Natur: Fünf Schriften zur Logik und Metaphysik, Reclam,
Stuttgart 1975
[rob] Roboz, John, Introduction to Mass Spectrometry, Wiley, New York, 1968
[pau] Paus, H., Physik, München 1995
[kik] Kiker, W.E., Correlated Energy and Time-of-Flight Measurements of Fission
Fragments. Thesis. OAK RIDGE NATIONAL LABORATORY
U. S. ATOMIC ENERGY COMMISION
http://www.ornl.gov/info/reports/1964/3445604514693.pdf
[lea] Leachman R., B.,
Velocities of Fragments from Fission of U-233, U-235, and Pu-239
Phys. Rev. 87, 444–447 (1952)
[spi] Spiering, Ch., Auf der Suche nach der Urkraft. Frankfurt/M 1986
[coc] Cockcroft, John. Nobel lecture 1951
http://nobelprize.org/nobel_prizes/physics/laureates/1951/cockcroft-lecture.pdf