Binding energy and nuclear fission

In summary: That's an example of spallation reaction. One would not expect to see a fission reaction with that result.In summary, a nuclei is held together by the strong force - which is strong, but has a short range - and energy = mass. When the masses of the products of a reaction are less than the reactants, the difference in mass is manifest in the kinetic energy of the products, and is exothermic, i.e., produces excess energy.
  • #1
Leo Liu
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If I am not mistaken, binding energy is the energy required to separate the constituents of the nucleus, and is always a positive number. However, if splitting elementary particles in the nucleus apart requires energy, then why do fission reactions release energy?
 
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  • #2
Simple logic: if your reasoning leads to something that isn't right, your starting point is wrong ...

Leo Liu said:
and is always a positive number

by the way, an energy is not a number !

##\ ##
 
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  • #3
BvU said:
Simple logic: if your reasoning leads to something that isn't right, your starting point is wrong ...
Source:
1651494919008.png


Edit: "The total mass of the bound particles is less than the sum of the masses of the separate particles by an amount equivalent (as expressed in Einstein’s mass–energy equation) to the binding energy."
 
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  • #4
So ?
What if you do the math ?
 
  • #5
BvU said:
So ?
What if you do the math ?
1651495483337.png


Joking aside, I don't actually know how to do the math. Can you please help?
 
  • #6
The thing is, you always need energy to split a nucleus into lone nucleons.
However, if you split a nucleus into, not exclusively lone nucleons but into one or more other nuclei (plus lone nucleons left over), you can compare the binding energy of original nucleus with the binding energy of final nucleus, or sum of the binding energies of final nuclei if there are several final nuclei.
 
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  • #7
Leo Liu said:
Source:
View attachment 300904

Edit: "The total mass of the bound particles is less than the sum of the masses of the separate particles by an amount equivalent (as expressed in Einstein’s mass–energy equation) to the binding energy."
The note about total mass of bound particles is the key. The sum of the masses of the fission products is less than the masses of the exited 236U nucleus.

hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fisfrag.html
http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin.html#c4

Pick two complementary fission products, add their masses, and subtract the mass of the 236U nucleus, or sum of masses of 235U + n.
 
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Astronuc said:
The note about total mass of bound particles is the key. The sum of the masses of the fission products is less than the masses of the exited 236U nucleus.

hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fisfrag.html
http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin.html#c4

Pick two complementary fission products, add their masses, and subtract the mass of the 236U nucleus, or sum of masses of 235U + n.
Wow this result is impressive. May I ask you to try to explain the reason behind the drop in the total mass, please?
 
  • #9
Leo Liu said:
Wow this result is impressive. May I ask you to try to explain the reason behind the drop in the total mass, please?
Total mass drops if, and because, the total binding energy increases.
You could pick a pair of complementary fission products where the mass does not decrease. Such as Pa-234 and D (plainly complementary fission products!). You would find that those fission products only have nonzero yield with fast neutrons.
 
  • #10
Leo Liu said:
May I ask you to try to explain the reason behind the drop in the total mass, please?
Nuclei are held together by the strong force -- which is strong, as the name says, but has a very short range. So in bigger nuclei it has relatively more difficulty overcoming the Coulomb repulsion. Hence the downward slope for larger ##Z## in your picture in post #3. And energy = mass.

##\ ##
 
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  • #11
Leo Liu said:
May I ask you to try to explain the reason behind the drop in the total mass, please?
Where the masses of the products of a reaction are less than the reactants, the difference in mass is manifest in the kinetic energy of the products, and is exothermic, i.e., produces excess energy. The nuclei of fission products are more tightly bound than the original U-235 (or Pu-239,-241). Fertile (or fissionable) atoms like U-238 or Pu-240, require fast neutrons to induce fission, otherwise, they tend to emit gamma radiation instead of fissioning.

Masses of U and Pa isotopes, respectively
https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=U&ascii=html&isotype=all
https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Pa&ascii=html&isotype=all

235.0439301(19) u
1.00866491595 u
236.052595 sum of U-235 + n

234.0433072 u
2.013553212745 u

Mass difference in U-235 + n - (Pa-234 + d) = -0.004265397 u (-3.973191522 MeV) indicating an endothermic reaction, or threshold of ~ 4 MeV.

236.0455682 u NIST
236.0455619 u https://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=U-236

Masses in u, or amu:
Mass d, https://physics.nist.gov/cgi-bin/cuu/Value?mdu|search_for=atomnuc
Mass n, https://physics.nist.gov/cgi-bin/cuu/Value?mnu|search_for=atomnuc

Mass e, https://physics.nist.gov/cgi-bin/cuu/Value?are|search_for=meu

https://physics.nist.gov/cgi-bin/cuu/Value?muc2mev

https://physics.nist.gov/cgi-bin/cuu/Results?search_for=atomnuc
https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl

snorkack said:
You could pick a pair of complementary fission products where the mass does not decrease. Such as Pa-234 and D (plainly complementary fission products!). You would find that those fission products only have nonzero yield with fast neutrons.
That's an example of spallation reaction. One would not expect to see a fission reaction with that result.

On the other hand, we do find examples of ternary fission in which deuterons, tritons or alpha particles are released in addition to the two major fission products and neutrons. Tritons are a concern, especially where stainless steel claddings (both austenitic and ferritic/martensitic) are used since steel is relatively transparent to hydrogen isotopes, and in the case of tritons, they neutralize to form tritium in the coolant, which was a concern in earlier LWRs that used stainless steel cladding, and would be a concern for LWRs using FeCrAl or F/M claddings, e.g., HT9, or similar alloys.
 
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  • #12
BvU said:
Nuclei are held together by the strong force -- which is strong, as the name says, but has a very short range. So in bigger nuclei it has relatively more difficulty overcoming the Coulomb repulsion. Hence the downward slope for larger ##Z## in your picture in post #3. And energy = mass.

##\ ##
So more pions relative to the number of protons in the heavier nucleus?
 
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  • #13
Misinformation infraction issued
Astronuc said:
That's an example of spallation reaction. One would not expect to see a fission reaction with that result.
Which is kind of a matter of definition. Endothermic reactions, like
U-235+n=Pa-234+d
U-235+n=U-234+2n
are classified as "spallation" not "fission" because they lose rather than release energy.
 
  • #14
snorkack said:
are classified as "spallation" not "fission" because they lose rather than release energy.
I don't believe that losing energy (or endothermic reaction) is a qualifier for a reaction being classified as 'spallation', but perhaps it could be. Spallation reactions do release some energy, but it's more the case they transform energy. Usually, in a spallation reaction, one or more of the products is similar in size to the incident (projectile) particle that hits a nucleus, e.g., (n,p), (p,n), (n,d), (n,α), or even (p, #n), where the # depends in the energy of the incident p and the target.

In fission, after a neutron (or incident particle) is absorbed, the excited nucleus splits (fissions) into two nuclei of similar size. Having a symmetric fission, e.g., product of two palladium (Pd) atoms is possible, but has lower probability than two nuclei of different species, although the probability of a symmetric fission increases with energy of the neutrons. Pd is a concern, because like Ni, it attacks SiC, which has been proposed for some advanced nuclear fuel concepts.

Some discussions on spallation vs fission.
Contrasted to fission, spallation differs in the following ways: a) in the nuclear debris remaining after the reactions; b) in the amount of energy deposited per neutron produced; c) in the amount of gamma-ray energy produced; and d) in the energy distribution of the emitted neutrons.
Source: https://inis.iaea.org/collection/NCLCollectionStore/_Public/23/015/23015552.pdf

https://www.ill.eu/fileadmin/user_u...017/02-2017-01-31_Berke-GonzalezRodriguez.pdf

Physics and Technology of Spallation Neutron Sources
https://www.osti.gov/etdeweb/servlets/purl/20020531
 
  • #15
Astronuc said:
In fission, after a neutron (or incident particle) is absorbed, the excited nucleus splits (fissions) into two nuclei of similar size. Having a symmetric fission, e.g., product of two palladium (Pd) atoms is possible, but has lower probability than two nuclei of different species, although the probability of a symmetric fission increases with energy of the neutrons. Pd is a concern, because like Ni, it attacks SiC, which has been proposed for some advanced nuclear fuel concepts.
In which case, if fission is really symmetric, you would get two nuclei of Pd-118. Which promptly decay to Sn.
Something conspicuous is the big yield gap at precisely around the peak of binding energy curve.
The one cluster decay of U-235 which is well characterized with a known branching ratio is to (neutron-rich, unstable) Ne-24, with branching ratio at 8*10-12. The total branching ratio of "fission" is 2*10-9, which means that fission product yield of Ne-24 is about 0,4% of all fissions. Quite comparable with fissions around the nuclear mass of 110. In contrast, there seem to be very few fission products near the maximum binding energy - nuclear mass 50...60...
 
  • #16
snorkack said:
In which case, if fission is really symmetric, you would get two nuclei of Pd-118. Which promptly decay to Sn.
However, the fission of uranium usually releases at least two neutrons, or sometimes 1, or 3, or 4, but on average slightly more than 2. So one might expect 2 atoms of 117Pd, which would promptly decay to 117Ag > 117Cd > 117In > 117Sn. Meanwhile, these nuclides may absorb a neutron. Furthermore, one might find 116Pd, 118Pd rather than two 117Pd. However, Pd is much less probable than other fission products.

More stable forms of Pd would come from decay of Mo, Tc, Ru and Rh radionuclides.
snorkack said:
The one cluster decay of U-235 which is well characterized with a known branching ratio is to (neutron-rich, unstable) Ne-24, with branching ratio at 8*10-12. The total branching ratio of "fission" is 2*10-9, which means that fission product yield of Ne-24 is about 0,4% of all fissions.
This is incorrect. Where did one get such information?
 
  • #17
Astronuc said:
This is incorrect. Where did one get such information?
Cluster decay:
https://en.wikipedia.org/wiki/Cluster_decay
note that:
it is given as "branching ratio" (relative to the fission to α and Th-231)
out of the 4 cluster decays presented for U-235, only Ne-24 has a definite branching ratio - 1 other has only upper bound and 2 more not even that.
Fission:
https://en.wikipedia.org/wiki/Spontaneous_fission
the "fission" probability is given in %, again relative to the fission to α, as 2*10-7 % - which converts to branching ratio of 2*10-9, right?
And the Ne-24 branching ratio of 8*10-12 means 0,4% fission yield of the total fission branching ratio of 2*10-9. Correct?
 
  • #18
Cluster decay and spontaneous fission are decay modes that compete at low probability with alpha decay, and should not be confused with neutron induced fission, which is what happens in a nuclear reactor.

Checking the NNDC at Brookhaven National Laboratory (BNL), one finds for 235U:
ɑ = 100.00%
SF = 7.0E-9% (or 7.0 E-11), which is less than 2.0·E−7% or 2.0E-9, assuming they are ratios with respect to the same basis.
28Mg = 8.E-10% (or 8.0 E-12)
20Ne = 8.E-10% (or 8.0 E-12)
25Ne ≈ 8.E-10% (or 8.0 E-12)

Surprisingly, 24Ne does not appear, but taking the number from the Wikipedia article, 8.06E-12, these decay modes are exceedingly rare compared to alpha decay.

Furthermore, in a reactor core producing 3.4 GW of thermal power requires a fission rate of approximately 1.06 E20 fissions per second. Decays of 235U are approximately 4.0 E11 decays/sec, of which 1E-10 are spontaneous fission (SF) or cluster decay (CD), or the rate of SF/CD is about 40 per second, which contributes a fraction of 3.8E-19 of fissions in typical nuclear reactor of that rating containing 193 assemblies, 264 fuel rods per assembly, 2 kgU per fuel rod and 5 wt% enrichment.

I will do a recheck later, but that's about the right order of magnitude. The rate of SF or CD is insignificant in a nuclear reactor where the fissioning is principally neutron induced with the intermediate step being the formation of an excited 236U nuclide, although fast fissions of 238U are also part of the process, not to mention the fast and thermal fissions of various transuranic elements that are produced through transmutation processes.

I am curious about the discrepancy in some numbers between what NNDC published and those found in the Wikipedia article. I will have to investigate.

Edit update note: corrected number of fissions per sec from 1.06 E22 to 1.06 E20, and fraction 4 E-21 to 3.8 E-19.
 
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  • #19
SF is, however, vital for a nuclear reactor which is in a subcritical state. Especially high subcritical.

The mean number of neutrons emitted by a fission event is a bit over 2. But what is the median? Are fission events producing no neutrons at all, or say 5, common?

Because when a reactor is high subcritical, it should be cold between individual random neutron avalanches, right? So there is a large probability that a single avalanche dies out right at the start - a SF event just happens to produce 0 neutrons, or does produce 2 but both fly out or are absorbed?

The propagation of avalanches, and probability to die out once started, derives from properties of induced fission. But the initiation of avalanches, the first step, depends on the specific properties of spontaneous fission - the number and energy distribution of SF neutrons, as opposed to thermal neutron fission and fast neutron fission neutrons. And the baseline exact probability of SF in the first place.
 
  • #20
snorkack said:
SF is, however, vital for a nuclear reactor which is in a subcritical state. Especially high subcritical.

The mean number of neutrons emitted by a fission event is a bit over 2. But what is the median? Are fission events producing no neutrons at all, or say 5, common?
SF of 235U, 239Pu, 240Pu, 241Pu, . . . is NOT vital for a nuclear reactor, even in the cold zero power, because there are insufficient neutrons produced, as calculated above, and SF contributes an insignificant, minuscule amount of neutrons to the neutron production.

The preferred startup source of neutrons these days is 252Cf (half-life 2.465 yrs; ɑ = 96.91%, SF = 3.09% ) of which a small amount (grams) is used. Before 252Cf, there were a variety of (α,n) sources in which an alpha-decaying radionuclide, e.g., 241Am or 226Rn was intimately mixed with Be (9Be) to produce neutrons. A secondary source is provided by 123Sb + Be, in which during reactor operation, 123Sb captures a neutron to become 124Sb, which decays to 124Te, which subsequently releases a gamma ray of energy ~1.64 MeV, which induces a photoneutron reaction in the Be.

The objective of a startup neutron source is to provide sufficient neutrons to be 'detectable' with a count rate of something like 2 cps (counts per second) at the detector.

With the move to high burnup cores in which there is an ample amount of transuranic radionuclei (240Pu, Am, Cm, . . .), some utilities implement a 'sourceless startup', since neutron sources are a hassle and represent additional radioactive waste and disposal burden. Sourceless startups rely on a sufficient inventory of nuclides that undergo spontaneous fission, but still, SF does not contribute much to the neutron generation or operation of the reactor once it has achieved low power.

snorkack said:
Because when a reactor is high subcritical, it should be cold between individual random neutron avalanches, right? So there is a large probability that a single avalanche dies out right at the start - a SF event just happens to produce 0 neutrons, or does produce 2 but both fly out or are absorbed?
We usually only refer to subcritical (keff < 1), critical (keff = 1) or supercritical (keff > 1)). The idea of using a neutron source in a reactor is to approach criticality with some idea of how close one is to criticality, without unknowingly exceeding criticality. The number of neutrons (on average) emitted by SF is given in the table in the Wikipedia article one cited. For 252Cf the average is 3.73, which is roughly twice that of 235U. Note also the half-lives of those nuclides.
 
  • #21
Astronuc said:
Sourceless startups rely on a sufficient inventory of nuclides that undergo spontaneous fission, but still, SF does not contribute much to the neutron generation or operation of the reactor once it has achieved low power.
But in a core with all fresh fuel that has never undergone chain reaction the only source of neutrons (if no additional n source is present) is SF is it not?
I assume there also must be some small fraction of fission fragments within U 238/235 mix as SF over the years has caused there to be some but what would be the percentage of neutrons given off from fresh fuel in terms of how much of it comes from SF and how much from the decay of some fission fragments already within the fuel do to earlier SF within the mass?
 
  • #22
artis said:
But in a core with all fresh fuel that has never undergone chain reaction the only source of neutrons (if no additional n source is present) is SF is it not?
I assume there also must be some small fraction of fission fragments within U 238/235 mix as SF over the years has caused there to be some but what would be the percentage of neutrons given off from fresh fuel in terms of how much of it comes from SF and how much from the decay of some fission fragments already within the fuel do to earlier SF within the mass?
This depends on the criticality state of the assembly.
Note that most of the fission fragments only ever undergo beta decay. The fission fragments which do undergo neutron decay are all of them short lived - the longest lived of them is Br-87, with half-life just 56 s. So the fission fragments that produce neutrons are produced over the minutes, not years.
The yield of n emitters in thermal neutron fission of U-235 is given as 0,64% out of 200 %. If so, in an assembly with prompt k of 0,9900, 1 spontaneous fission would on average induce 100 prompt fissions, but 280 fissions total... meaning that 180 fissions out of 280 (64%) would be due to delayed neutrons. Whereas if k is 0, then the neutron yield from delayed neutrons would be about 0,34%
 
  • #23
artis said:
But in a core with all fresh fuel that has never undergone chain reaction the only source of neutrons (if no additional n source is present) is SF is it not?
Yes, which is why neutron sources are placed in fresh fuel in order to monitor approach to criticality during startup.
artis said:
I assume there also must be some small fraction of fission fragments within U 238/235 mix as SF over the years has caused there to be some but what would be the percentage of neutrons given off from fresh fuel in terms of how much of it comes from SF and how much from the decay of some fission fragments already within the fuel do to earlier SF within the mass?
Fission products from SF would be very low, since during processing of ore to concentrated oxide and oxide to UF6, undesirable elements are removed. There are restrictions on elements in nuclear fuel ceramics, which are measured for each batch of fuel.

As I indicated in a previous post, the number of neutrons produced from SF are exceedingly small. I estimated about 1 in 1019, and even if it was an order of magnitude greater, 1 in 1018, that's still insignificant. The preferred neutron source is 252Cf, with a secondary source of Sb-Be backed up by presence of transuranic nuclides that are produced during operation. Only after 3 years of operation, either 3 annual cycles or two 18-month cycles, or roughly a burnup of about 30-35 GWd/tU is there substantial TU isotopes that one could rely less on a primary startup neutron source.

The delay neutron fraction for 235U is 0.0065, or 0.65% of neutrons from SF. So, if SF rate from 235U is 40 fissions/sec (in the entire core), the delayed neutrons would be 0.26 neutrons/s. One could work out a similar number for 238U, which has a longer half-life, but larger branching ratio for SF than 235U.

One should keep in mind that fresh fuel will contain burnable absorbers to 'hold down' the reactivity of fresh fuel. Fuel designers use burnable poisons mixed in the fuel ceramic (in select fuel rods in an assembly), and also in special components that are placed in guide tubes of PWR assemblies, in addition to the soluble poison (boric acid) in the cooling water, as well as the control rods in the core. In contrast, BWR fuel incorporates burnable poisons in selected fuel rods in the assembly, otherwise, reactivity is controlled by the control rods inserted in the core.
 
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Thread closed temporarily for Moderation...

After a Mentor discussion, the thread is re-opened.
 
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  • #25
snorkack said:
The yield of n emitters in thermal neutron fission of U-235 is given as 0,64% out of 200 %. If so, in an assembly with prompt k of 0,9900, 1 spontaneous fission would on average induce 100 prompt fissions, but 280 fissions total... meaning that 180 fissions out of 280 (64%) would be due to delayed neutrons. Whereas if k is 0, then the neutron yield from delayed neutrons would be about 0,34%
This is not correct after "The yield of n emitters in thermal neutron fission of U-235". One does not compare 0.64% to 200%, or 2.

The fraction of delayed neutrons is usually designated by β, which is approximately 0.0064 or 0.0065 for 235U for thermal neutrons. The β is dependent on the fissile nuclide, e.g., for 233U β = 0.00261 and for 239Pu β = 0.0021. However, some fissions occur in 238U due to fast neutrons, and in reality β > 0.007, the greater the better for reactor control.

The fraction, β, represents a fraction of the total neutrons produced from fissions occurring at the same point in time. The lag of seconds up to 56 ~seconds allows one to make small changes in reactivity (removing a control element or control rod, typical in a BWR) or diluting/reducing boron in the coolant gradually (in a PWR).

The statement "in an assembly with prompt k of 0,9900, 1 spontaneous fission would on average induce 100 prompt fissions, but 280 fissions total... meaning that 180 fissions out of 280 (64%) would be due to delayed neutrons" is just wrong. I have never heard prompt k, or prompt keff, and certainly not for an individual assembly in a reactor core (we do refer to prompt critical, which is a reactor condition to be avoided). The value of k (or keff) applies to the entire system. One will not find 64% of fissions due to delayed neutrons, and it is not clear where one gets 100 fissions, but 280 total.

One neutron causes on fission, which releases 2 or 3 (on average about 2.2 to 2.3) neutrons. Some will be absorbed by structural materials, e.g., cladding, grids, guide tubes, in-core instrumentation or by the coolant (certainly the boron in the coolant in a PWR, but also the H in the H2O), or by a control rod in BWR, or by a burnable absorber in the fuel or in a special element. The fuel components, 235U and 238U may absorb neutrons without fissioning, and simply emit a gamma ray instead, and then fission products, particularly Xe-135, absorb thermal neutrons. And finally, a small fractions of neutrons simply exit the core (i.e., leak out) and do not return.

Furthermore, 235U is consumed, or depleted during reactor operation, and some if it is replaced by 239Pu, 240Pu, 241Pu, 242Pu and heavier transuranic elements. Both 238U and 240Pu are important with respect to the Doppler broadening of absorption resonances, which mitigate thermal reactor transients.

I'm not sure from where the last statement, "Whereas if k is 0, then the neutron yield from delayed neutrons would be about 0,34%. I've never heard of k = 0, except for a defueled, or mostly empty, or a fully depleted core.

The OP was asking about binding energy and how it related to energy produced from fission. That question has been answered.
 
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  • #26
Astronuc said:
The fraction of delayed neutrons is usually designated by β, which is approximately 0.0064 or 0.0065 for 235U for thermal neutrons. The β is dependent on the fissile nuclide, e.g., for 233U β = 0.00261 and for 239Pu β = 0.0021. However, some fissions occur in 238U due to fast neutrons, and in reality β > 0.007, the greater the better for reactor control.
To keep conversion simply, let's say we have 1000 neutrons at some time T and population P within a reactor that is at the very start aka low power end of a power up procedure. For a fresh fuel assuming the beta 0.0064 would that mean that of those 1000 fission neutrons 6 will be the delayed ones?

Here is the tricky part let me see if I recall correctly, the delayed fraction is subtracted from the total not added to it right? So for my 1000 neutrons the prompt fraction would be 994 and delayed 6?
Or it doesn't matter? Could I do it by saying my prompt fraction is 1000 and delayed still 6 and then the total is 1006? Because either way it adds up to 100%
If I had another population down the line at a higher power that consisted of say 10k neutrons then my delayed fraction would now consist of 64 neutrons

The delayed neutron count increases as the total neutron flux increases but the proportionality stays the same so the overall K of the core can be maintained at barely over 1 if delayed fraction is included?
IIRC at steady state irrespective of total power produced the prompt fraction will always be slightly less than 1 say 0.994 and the delayed will always add up the total to 1?

But if my last statement is correct then it got me thinking, how does one produce a transient where the core enters a mode where prompt neutron fraction reaches above 1 and the core starts to accelerate on prompt neutrons alone like in Chernobyl? I would think one can achieve this by simply having not enough neutron absorption in the core during a neutron flux increase (like the one in Chernobyl that happened when Xe concentration decreased sufficiently without enough moderator present) or removing too much neutron absorption during operation at steady flux. When thinking about this one begins to see why negative feedback coefficient is important for the physics of the core, especially if one removes too much absorption too fast for whatever reason.
 
  • #27
Astronuc said:
This is not correct after "The yield of n emitters in thermal neutron fission of U-235". One does not compare 0.64% to 200%, or 2.

The fraction of delayed neutrons is usually designated by β, which is approximately 0.0064 or 0.0065 for 235U for thermal neutrons. The β is dependent on the fissile nuclide, e.g., for 233U β = 0.00261 and for 239Pu β = 0.0021. However, some fissions occur in 238U due to fast neutrons, and in reality β > 0.007, the greater the better for reactor control.

The fraction, β, represents a fraction of the total neutrons produced from fissions occurring at the same point in time.
You are right. I confused two possible ways to measure delayed neutron production: delayed neutrons per fission event (which produces 2 fragments and a variable number of prompt neutrons) or delayed neutrons per neutron. The quoted value of approximately 0.0064 or 0.0065 for 235U for thermal neutrons is applicable per neutron.

Note that the delayed neutrons have lower energy spectrum than prompt neutrons - which means that they have less effect on the nuclei that undergo fission due to fast neutrons.
Astronuc said:
The statement "in an assembly with prompt k of 0,9900, 1 spontaneous fission would on average induce 100 prompt fissions, but 280 fissions total... meaning that 180 fissions out of 280 (64%) would be due to delayed neutrons" is just wrong. I have never heard prompt k, or prompt keff, and certainly not for an individual assembly in a reactor core (we do refer to prompt critical, which is a reactor condition to be avoided). The value of k (or keff) applies to the entire system. One will not find 64% of fissions due to delayed neutrons, and it is not clear where one gets 100 fissions, but 280 total.
If your keff is 0,9964 counting delayed neutrons or 0,9900 on account of prompt neutrons alone then on average one fission event should produce an avalanche producing 100 fissions through prompt neutrons. But one shake is what, 10 ns? The avalanche is on average over in a microsecond.
If the Wikipedia article on SF (quoting Shultis et al, Fundamentals of Nuclear Science and Engineering, 2008) is correct on conversions and only wrong about the source lifetime of U-235 then the "neutron production" of U-235 is given as 0,0003 per g*s... which means 0,3 per kg*s, or 0,16 fission events per kg/s. A bare sphere critical mass of HEU, at 50 kg, would then produce just 8 SF events per second - one per 125 milliseconds. When it is barely subcritical, then a SF event would through subcritical multiplication produce a large number of fissions but die out in microseconds, leaving the core devoid of free neutrons and unresponsive to any changes in geometry/reactivity while cold... while prone to react in microseconds to the stochastic decay events.

Now, the delayed neutron decay events are separated from the fission that induced them by the lifetime of the precursor... ranging to tens of seconds. Shortest "group" is quoted as 230 ms half-life. Which means that in high supercritical range, where the core is not subject to a neutron flux but to overlapping avalanches, the core would have avalanches initiated by delayed neutron events. Once an avalanche is initiated, it would look the same whether started by SF or delayed neutron event, but delayed neutrons would show by avalanche frequency over and above the SF background.
Astronuc said:
I'm not sure from where the last statement, "Whereas if k is 0, then the neutron yield from delayed neutrons would be about 0,34%. I've never heard of k = 0, except for a defueled, or mostly empty, or a fully depleted core.
0,34% was mistake from assuming 0,64% to be per fission and converting to per neutron, whereas it was per neutron to begin with. My point was to compare the state of k=0 (no subcritical multiplication at all) to the situation where k is near but below 1, so the neutron flux is dominated by subcritical multiplication.
The OP was asking about binding energy and how it related to energy produced from fission. That question has been answered.
The wording was "why do fission reactions release energy".
One possible answer is "because reactions which do not release energy are called something else than fission, such as spallation". Another possible (but not quite correct) answer is "because reactions which do not release energy do not happen".

Trying to explain fission by reference to binding energy curve, my tangent was to point out low yield of fission products around the peak of binding energy curve. Why don´t we see more of reactions like:
U-235=Hf-187+Ca-48
U-235=Gd-171+Ni-64
U-235+n=4V-59?
 
  • #28
snorkack said:
Which means that in high supercritical range, where the core is not subject to a neutron flux but to overlapping avalanches, the core would have avalanches initiated by delayed neutron events. Once an avalanche is initiated, it would look the same whether started by SF or delayed neutron event, but delayed neutrons would show by avalanche frequency over and above the SF background.
I agree that delayed neutrons can start fission just like SF neutrons but how can a core "not be subject" to a neutron flux when in "supercritical" state?
From what I understand I would think that for a reactor core with any neutron flux that is within the resulting design power band (thermal) of the core , it would be impossible to tell different neutrons apart empirically as all you register is just a neutron flux that is either constant or decreasing or increasing.

You could maybe tell the delayed from the SF apart only in some subcritical fuel mass like that of the HEU bare sphere a far as I see it.
 
  • #29
artis said:
I agree that delayed neutrons can start fission just like SF neutrons but how can a core "not be subject" to a neutron flux when in "supercritical" state?
Look at the derivation above. For the critical mass of 50 kg HEU, the SF rate is just 8 events per second by high estimate - and there is a far lower estimate offered.
If your prompt k is 0,5000 then your total k should be what, 0,5032? Which means that 1 SF event causes on average 2 (1/0,5000) fissions in short term, and in a long term 2,013 fissions (1/0,4968). But you will not observe these 16 fissions per seconds in 16 randomly distributed microseconds out of the million microseconds in the second. No, in a shake (10 ns, which is 0,01 microseconds) a neutron flies out of the core or else causes a fission. An induced fission happens in the same microsecond as the SF that induced it. There are 16 fissions in the second, but only 8 microseconds in the second when any fissions happen, and on average 125 millisecond intervals when no fissions happen and no neutrons are in the core.
Raise prompt k to 0,9500, and total k will be 0,9561. Then one SF event causes on average 20 fissions promptly and 23 fissions total, counting almost 3 fissions added due to delayed neutrons. But these almost 3 are in avalanches of average 20 fissions in the same microsecond just the same, so maybe 9 avalanches per second instead of 8. Still on average over 100 milliseconds with no fission events and no neutrons around.
Now, what happens if during these 100 milliseconds with no neutrons in core, your prompt k should increase from 0,95 to 1,05? Like movement of nearby reflectors such as a collapsing pile of beryllium bricks?
Next time there is a random SF event, you might observe, not 20 fissions, and not 40 or 400, but 1020 fissions.
It is true that this is a potentially uncomfortable situation and precautions often include intentional neutron sources, to ensure that something happens already when k is 0,995 or 1,005, not many milliseconds later.
artis said:
From what I understand I would think that for a reactor core with any neutron flux that is within the resulting design power band (thermal) of the core , it would be impossible to tell different neutrons apart empirically as all you register is just a neutron flux that is either constant or decreasing or increasing.
In high subcritical state, yes. What you see is the outcome of avalanches, which is determined by the conditions for the final stages of avalanches, not the originating event (intentionally added neutron source, SF, delayed neutron from fission product, photoneutron due to fission product gammas, α-n reaction, cosmic ray...). And what you could see is the frequency of avalanches.

Even low subcritical states, I think are not well fitted for direct observation of the neutron emission events. Something which is said to be a headache to measure well, but vital in principle, is the exact spectrum of delayed neutrons.
 
  • #30
artis said:
To keep conversion simply, let's say we have 1000 neutrons at some time T and population P within a reactor that is at the very start aka low power end of a power up procedure. For a fresh fuel assuming the beta 0.0064 would that mean that of those 1000 fission neutrons 6 will be the delayed ones?
Yes, roughly a delayed neutron fraction 0.0064 means that of 1000 neutron at a given instant, approximately 6 would be delayed. I'm not sure what one means by "at the very start aka low power end of a power up procedure." A fresh core starts at cold zero power (CZP) then gradually comes to hot zero power by running pumps to get flow and establish temperature at hot zero power (HZP). From there, approach to criticality begins, and since there are no delayed neutrons and very few neutrons from SF, neutron sources are used to produce enough neutrons to count. From criticality, control rods are moved (withdrawn) gradually (and in the case of a PWR, boron concentration in the coolant reduced), just enough to get a small increase in power (fissions). This occurs in steps to hot full power (HZP). There are periods of holds, or constant power, where the reactor instrumentation is checked to verify accuracy and consistency of readings in temperature and power (neutron flux) distribution.

artis said:
Here is the tricky part let me see if I recall correctly, the delayed fraction is subtracted from the total not added to it right? So for my 1000 neutrons the prompt fraction would be 994 and delayed 6?
Or it doesn't matter? Could I do it by saying my prompt fraction is 1000 and delayed still 6 and then the total is 1006? Because either way it adds up to 100%
If I had another population down the line at a higher power that consisted of say 10k neutrons then my delayed fraction would now consist of 64 neutrons
The convention is to normalize the neutron population to 1. Remember, we consider about 1E20 neutrons per second, of which 6.4 E17 would be delayed, if all fissions originated in 235U. In reality, since fissions occur in 238U, the delayed neutron fraction is greater. I've seen different numbers, so I will contact a colleague and try to get a typical number for his power reactors.
Meanwhile, the updated final safety analysis report for some of the reactors indicates an allowable range for βeff, 0.0044 - 0.0075.
artis said:
The delayed neutron count increases as the total neutron flux increases but the proportionality stays the same so the overall K of the core can be maintained at barely over 1 if delayed fraction is included?
IIRC at steady state irrespective of total power produced the prompt fraction will always be slightly less than 1 say 0.994 and the delayed will always add up the total to 1?
The delayed neutron population may vary slightly. As the neutron flux increases, the fission rate increases (as does the thermal energy), or as the neutron flux decrease, the fission rate decreases, which is how one operates a reactor. Usually power reactors operate with a constant power, within an allowable band. The delayed neutron fraction is more or less constant with constant power, but may change based on neutron energy spectral changes and change in core composition.

If βeff is say 0.007, then for 1000 neutrons, 993 would be prompt and 7 would be delayed for k = 1.

artis said:
But if my last statement is correct then it got me thinking, how does one produce a transient where the core enters a mode where prompt neutron fraction reaches above 1 and the core starts to accelerate on prompt neutrons alone like in Chernobyl? I would think one can achieve this by simply having not enough neutron absorption in the core during a neutron flux increase (like the one in Chernobyl that happened when Xe concentration decreased sufficiently without enough moderator present) or removing too much neutron absorption during operation at steady flux. When thinking about this one begins to see why negative feedback coefficient is important for the physics of the core, especially if one removes too much absorption too fast for whatever reason.
One is describing a reactivity insertion accident (RIA) scenario in which some amount of reactivity is introduced into the system. Reactivity (that is, positive reactivity) can be introduced by various means, e.g., inadvertent dilution of boron in the coolant (in a PWR), or by inadvertent removal of a control rod in either PWR or BWR. In PWRs, one is concerned about control rod ejection under different conditions, and in BWR, one is concerned about control rod drop. When reactor is shutdown, all control rods are usually in the core, or some could be removed, but one would never remove enough control rods to the point where the core would be critical.

Chernobyl was a special case where besides being a water-cooled, graphite moderated reactor, it was taken way outside of its design allowances into a dangerous configuration, and we see the results. At Chernobyl usual safety systems were disabled. When the transient began, as the Xe-135 decayed, the insertion of the control rods exacerbated the situation because the control rods had graphite tips, which acted as a moderator, but also displaced cooling water, which in a graphite reactor absorbs neutrons.
 
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Related to Binding energy and nuclear fission

1. What is binding energy?

Binding energy is the amount of energy that holds the nucleus of an atom together. It is the difference between the mass of an atom and the sum of the masses of its individual protons and neutrons. This energy is released when an atom undergoes nuclear fusion or fission.

2. How is binding energy related to nuclear fission?

Nuclear fission is the process of splitting an atom's nucleus into smaller fragments. This process releases a large amount of energy in the form of heat and radiation. The energy released is a result of the decrease in binding energy between the smaller nuclei compared to the larger nucleus.

3. What is the role of binding energy in nuclear power plants?

Nuclear power plants use nuclear fission to generate electricity. The heat produced from the fission process is used to create steam, which then drives turbines to generate electricity. The binding energy released during fission is what provides the heat needed to generate this electricity.

4. How is binding energy calculated?

The binding energy of an atom can be calculated using the famous equation E=mc², where E is the energy, m is the mass, and c is the speed of light. By measuring the mass of the nucleus before and after a nuclear reaction, the change in mass can be used to calculate the binding energy.

5. Can binding energy be harnessed for other purposes besides nuclear power?

Yes, binding energy can also be harnessed for medical and industrial purposes. In medicine, it is used in nuclear medicine for diagnostic imaging and cancer treatment. In industry, it is used for radiography, sterilization, and food preservation.

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