Can Iodine Be a Viable Alternative for Fissionable Isotopes?

[bananan]

might be more expensive than uranium or plutonium, but storage costs and environmental impact and anti-nuke activism might be muted if there was an isotope of any element, say an isotope of iodine, that when fissioned, gives off energy AND short-lived radioactive isotopes.

so, say, hypothetically speaking, iodine captures a neutron from say plutonium, and becomes a radioactive isotope of iodine, which fissions into non-radiactive iron isotope and radioactive tritium isotope with a short half life.

the upfront cost of using iodine might be more expensive, but the overall cost might be less expensive.

 

[Astronuc]

Nothing below U-233 is readily fissionable by low to intermediate energy neutron absorption.

Pa-231 is fissile with thermal and epithermal neutrons, but has a very low cross-section < 1b, and more on the order of 1-100 mb. Th-232 is effectively fissionable with fast neutrons only, >1 MeV and higher.

http://oldserver.ba.infn.it/~ntof/proposte/TH_PROPv7.pdf

Pu-241 has a half-life of 14.290 y and is fissile. The isotope Americium-241 (which results from the 14-year half-life decay of Pu-241 which accumulates in reactor grade plutonium with increasing burnup) emits highly penetrating gamma rays, increasing the radioactive exposure of any personnel handling the material.

See Morbius's discussion of 'fissionable' vs 'fissile'.

 

[Morbius]

might be more expensive than uranium or plutonium, but storage costs and environmental impact and anti-nuke activism might be muted if there was an isotope of any element, say an isotope of iodine, that when fissioned, gives off energy AND short-lived radioactive isotopes.

so, say, hypothetically speaking, iodine captures a neutron from say plutonium, and becomes a radioactive isotope of iodine, which fissions into non-radiactive iron isotope and radioactive tritium isotope with a short half life.

the upfront cost of using iodine might be more expensive, but the overall cost might be less expensive.

bananan,

First I'd like to address an issue of terminology.

The term you want to use above is not "fissionable" but "fissile".

"Fissile" means that the nuclide is fissioned by a neutron of ANY energy.

"Fissionable" means that the nuclide can be fissioned by a neutron, but only if it has
kinetic energy above a nuclide-dependent threshold.

For example, U-238 is actually "fissionable" because it will fission with fast neutrons
with energy in excess of about 1 MeV. However, U-238 can't be used as a reactor
fuel alone.

So allow me to rephrase your question into "are there any fissile isotopes of any
element..."

Consistent with Astronuc's reply above; for fissile isotopes, you have your choice of
U-233, U-235, or Pu-239.

Unfortunately, there's no nuclide that fits your criteria above that produces energy,
produces only short-lived daughters, is acceptable to the anti-nukes... If there was,
then we'd be using it.

Dr. Gregory Greenman
Physicist

 

[bananan]

bananan,

First I'd like to address an issue of terminology.

The term you want to use above is not "fissionable" but "fissile".

"Fissile" means that the nuclide is fissioned by a neutron of ANY energy.

"Fissionable" means that the nuclide can be fissioned by a neutron, but only if it has
kinetic energy above a nuclide-dependent threshold.

For example, U-238 is actually "fissionable" because it will fission with fast neutrons
with energy in excess of about 1 MeV. However, U-238 can't be used as a reactor
fuel alone.

So allow me to rephrase your question into "are there any fissile isotopes of any
element..."

Consistent with Astronuc's reply above; for fissile isotopes, you have your choice of
U-233, U-235, or Pu-239.

Unfortunately, there's no nuclide that fits your criteria above that produces energy,
produces only short-lived daughters, is acceptable to the anti-nukes... If there was,
then we'd be using it.

Dr. Gregory Greenman
Physicist

I understand from wiki that India is using Thorium as a supply fuel. Could there be a combination of Thorium, uranium and plutinion that when placed with spent radiactive fuel, might cause it to decay quicker?

 

[Morbius]

I understand from wiki that India is using Thorium as a supply fuel. Could there be a combination of Thorium, uranium and plutinion that when placed with spent radiactive fuel, might cause it to decay quicker?

bananan,

In the "Thorium cycle", you use Th-232 as a fertile material to breed fissile U-233.

Then you burn the U-233 for power and make more U-233.

Putting combinations of different materials together doesn't affect the radioactivity.

The radioactive decay characteristics depend on what's going on in the nucleus.

The nucleus has no way of knowing what's also in the mix. In terms of length scale;
all those other materials are a LONG, LONG, LONG way away from the nucleus.

Dr. Gregory Greenman
Physicist

 

[Astronuc]

This might be helpful.

Some Physics of Uranium
http://www.uic.com.au/uicphys.htm

Thorium fuel was used in at least two reactors in the US - Shippingport and Indian Point 1. IIRC, the fuel did use U-235 dispersed in thoria.

 

[bananan]

bananan,

In the "Thorium cycle", you use Th-232 as a fertile material to breed fissile U-233.

Then you burn the U-233 for power and make more U-233.

Putting combinations of different materials together doesn't affect the radioactivity.

The radioactive decay characteristics depend on what's going on in the nucleus.

The nucleus has no way of knowing what's also in the mix. In terms of length scale;
all those other materials are a LONG, LONG, LONG way away from the nucleus.

Dr. Gregory Greenman
Physicist

THank you Dr. Greenman,

But those radiactive elements, which are the waste products and the objections of greenies, are they Fissionable (I hope I used the right term this time), in that if they are in the pressence of high-neutron flux, as a result of either uranium or plutonium, can either capture a neutron and transform into another element or split directly, into waste products with shorter decay?

Alternatively, would it be possible to combine a source of high neutron flux uranium or plutonium, with a lower atomic number material, say carbon-14, so that should carbon 14 capture a neutron, it gives off energy and decay or transform into something nonradioactive.

Thanks

 

[Astronuc]

Many fission products already decay rapidly. It's the longer living isotopes which present a problem, e.g. Cs-137 and Sr-90, but these decay with about about 29-30 year half-life.

http://www.nndc.bnl.gov/chart/ - one can select a section and zoom in on the appropriate radionuclide.

Any nucleus can be 'transmuted' into a new isotope by neutron capture, and that is one reason nuclear fuel has to be periodically removed from a reactor - the fission producst compete with the fissile material for neutrons - and isotopes of Xe and Kr build up causing pressurization of the fuel and swelling of the ceramic or metal matrix.

One idea has been to recover the U, Pu and other transuranics and burn those in a so-called actinide burner. This approach gets rid of some of the long-lived radionuclides and utilizes the thermal energy.

These might be of use - http://www.uic.com.au/nip.htm

 

[Morbius]

THank you Dr. Greenman,

But those radiactive elements, which are the waste products and the objections of greenies, are they Fissionable (I hope I used the right term this time), in that if they are in the pressence of high-neutron flux, as a result of either uranium or plutonium, can either capture a neutron and transform into another element or split directly, into waste products with shorter decay?

bananan,

There are two classes of materials in nuclear waste. One class is the "fission products";
these are the remnants of nuclear fuel that has fissioned. They are all fairly short-lived;
as Astronuc points out, the longest lived fission products are Sr-90 and Cs-137 which
have half-lives of 29 years and 30 years, respectively.

The other class of materials in nuclear waste, are the "transuranics" or "actinides".
These are the long lived wastes, and there are both "fissile" and "fissionable" nuclides
in this group. This group would include Pu-239, which is "fissile" and has a half-life
of about 24,000 years. You correctly surmise that the rest are "fissionable".

The way to get rid of these long-lived components of nuclear waste is to reprocess the
nuclear waste and recycle the long-lived components back to the reactors to be burned
as fuel, as you suggest above.

This was always the intention of the nuclear power program in the USA; to reprocess
spent fuel and recycle the long-lived waste components back to the reactor. The USA
was setup to do this in the early '70s, but the "greenies" went to court to block it;
claiming that there must be an environmental impact report before the USA could
authorize reprocessing.

So the AEC, and its successor ERDA, did an environmental impact statement called the
GESMO - Generic Environmental Statement for Mixed Oxide. [ When reprocessed
actinides are returned to the reactor as fuel, they are in oxide form, and mixed with
fresh uranium dioxide to form a fuel known as "mixed oxide" or "MOX". ]

Shortly thereafter, the "greenies" got Congress to OUTLAW the reprocessing and
recycling of nuclear waste in the USA!

One has to understand, the "greenies" evidently don't want a solution to the nuclear
waste problem. They want to shutdown nuclear power in the USA. That's why they
have opposed recycling nuclear waste. They have opposed Yucca Mountain.
They oppose power companies even shipping the waste out of the power plant.

The strategy of the "greenies" is to back-up the nuclear fuel cycle so that nuclear
power plants won't have any place to put spent fuel. If the power company doesn't
have any place to put spent fuel; then they can't unload their last core of spent fuel
and reload the reactor with fresh fuel. The nuclear power plant will have to shutdown;
and the "greenies" will have accomplished their purpose.

A solution to the nuclear waste "problem" is the LAST thing the "greenies" are interested in.

Nations like Great Britain, France, and Japan routinely reprocess and recycle spent
nuclear fuel in exactly the manner you suggest without the problem of obstruction by
the "greenies".

If ALL the electricity used by a family of four for 20 years were generated by
nuclear power; the accumulated nuclear waste due to that electricity generation would
fit in a shoebox if it was not reprocessed. If it was reprocessed, then the equivalent
amount of nuclear waste for the family of 4 for 20 years would fit in a pill bottle or
shot-glass.

Most of the volume and mass of nuclear waste [ >90%] is U-238; no more dangerous
or radioactive than the day it was dug out of the ground.

Dr. Gregory Greenman
Physicist

 

[bananan]

Dear Dr. Gregory Greenman,

I'm surprised that these facts are not usually presented on the nuclear debate. While I'm infavor of environmental issues, I think nuclear might be a good way to address global warming.

Is it possible to reprocess the radiactive long-lived waste, both fissile and fissionable, on-site? Greenies and after 911 are afraid of terrorists/accidents if it is done off site. Also, I understand U-238 can be bred into plutonium.

IMHO, improving an existing technology that exists now (fission) is better than the countless billions poured on a pipe dream (fusion). I've argued in other forums that if we invest in billions to recycle nuclear waste rather than nuclear fusion, then we can reduce CO2 emissions now, rather than wait another 100 years for fusion to become online (if it does at all).

I've read extensively on nuclear technology. I understand that pebble bed technology, while safe, is a once-through cycle that produces considerable waste. It was done experimentally at South Africa.

I am under the impression that long-lived nuclear waste disposal, in peaceful Western democracies, is the primary objection to nuke, with safety (i.e Chernobyl) being second.

bananan,

There are two classes of materials in nuclear waste. One class is the "fission products";
these are the remnants of nuclear fuel that has fissioned. They are all fairly short-lived;
as Astronuc points out, the longest lived fission products are Sr-90 and Cs-137 which
have half-lives of 29 years and 30 years, respectively.

The other class of materials in nuclear waste, are the "transuranics" or "actinides".
These are the long lived wastes, and there are both "fissile" and "fissionable" nuclides
in this group. This group would include Pu-239, which is "fissile" and has a half-life
of about 24,000 years. You correctly surmise that the rest are "fissionable".

The way to get rid of these long-lived components of nuclear waste is to reprocess the
nuclear waste and recycle the long-lived components back to the reactors to be burned
as fuel, as you suggest above.

This was always the intention of the nuclear power program in the USA; to reprocess
spent fuel and recycle the long-lived waste components back to the reactor. The USA
was setup to do this in the early '70s, but the "greenies" went to court to block it;
claiming that there must be an environmental impact report before the USA could
authorize reprocessing.

So the AEC, and its successor ERDA, did an environmental impact statement called the
GESMO - Generic Environmental Statement for Mixed Oxide. [ When reprocessed
actinides are returned to the reactor as fuel, they are in oxide form, and mixed with
fresh uranium dioxide to form a fuel known as "mixed oxide" or "MOX". ]

Shortly thereafter, the "greenies" got Congress to OUTLAW the reprocessing and
recycling of nuclear waste in the USA!

One has to understand, the "greenies" evidently don't want a solution to the nuclear
waste problem. They want to shutdown nuclear power in the USA. That's why they
have opposed recycling nuclear waste. They have opposed Yucca Mountain.
They oppose power companies even shipping the waste out of the power plant.

The strategy of the "greenies" is to back-up the nuclear fuel cycle so that nuclear
power plants won't have any place to put spent fuel. If the power company doesn't
have any place to put spent fuel; then they can't unload their last core of spent fuel
and reload the reactor with fresh fuel. The nuclear power plant will have to shutdown;
and the "greenies" will have accomplished their purpose.

A solution to the nuclear waste "problem" is the LAST thing the "greenies" are interested in.

Nations like Great Britain, France, and Japan routinely reprocess and recycle spent
nuclear fuel in exactly the manner you suggest without the problem of obstruction by
the "greenies".

If ALL the electricity used by a family of four for 20 years were generated by
nuclear power; the accumulated nuclear waste due to that electricity generation would
fit in a shoebox if it was not reprocessed. If it was reprocessed, then the equivalent
amount of nuclear waste for the family of 4 for 20 years would fit in a pill bottle or
shot-glass.

Most of the volume and mass of nuclear waste [ >90%] is U-238; no more dangerous
or radioactive than the day it was dug out of the ground.

Dr. Gregory Greenman
Physicist

 

[Morbius]

Is it possible to reprocess the radiactive long-lived waste, both fissile and fissionable, on-site? Greenies and after 911 are afraid of terrorists/accidents if it is done off site. Also, I understand U-238 can be bred into plutonium.

bananan,

YES - on site reprocessing is indeed feasible.

That is one of the principal features of the Integral Fast Reactor [ IFR ] that I worked
on in the early part of my career when I was at Argonne National Laboratory.

Here is an interview my former boss, Dr. Charles Till; did with PBS's Frontline in a program
entitled "Nuclear Reaction" hosted by Pulitzer Prize winning author Richard Rhodes
[ "The Making of the Atomic Bomb" ]:

http://www.pbs.org/wgbh/pages/frontline/shows/reaction/interviews/till.html

As Dr. Till mentions, the fuel for the IFR is in metalic form, as opposed to the oxide
ceramic found in most reactors. By keeping the fuel in metalic form, it is particularly
easy to do the reprocessing. Instead of needing a chemical processing plant, as one
needs with oxide fuel; IFR fuel is reprocessed using metallurgical techniques, namely
"halide slagging" followed by "electrorefining".

As Dr. Till mentions, the plutonium / actinide containing output of this process can NOT
be used as nuclear weapons fuel, but CAN be recycled back to the IFR to be burned.
Therefore the IFR doesn't impose a proliferation risk.

Because the metallurical process is much simpler than chemical reprocessing, the
reprocessing plant can be placed on-site, as you suggest. The plutonium / actinides
never leave the high-radiation region of the plant; so there is no opportunity for theft.

The following web page, courtesy of the Nuclear Engineering Department at the
University of California - Berkeley mentions the on site nature of the IFR's
reprocessing system under the heading of "Diversion":

http://www.nuc.berkeley.edu/designs/ifr/anlw.html

In addition, the IFR was "inherently safe". It did not rely on engineered systems to
shutdown and cool the reactor. All that is needed to keep the IFR safe, is that the
laws of Physics work - which they ALWAYS do.

As stated in the article; the IFR is "almost too good to be true" as Richard Rhodes
phrased it. It's really too bad that President Clinton canceled the IFR in 1994.

Dr. Gregory Greenman
Physicist

 

[theCandyman]

Ronaldo Szilard, Nuclear Science and Engineering director of INL, came to give a talk my my university today. He was talking about the NGNP, and later I had asked if it was similar to the IFR design. Would it also be able to process waste on site?

 

[bananan]

bananan,

YES - on site reprocessing is indeed feasible.

That is one of the principal features of the Integral Fast Reactor [ IFR ] that I worked
on in the early part of my career when I was at Argonne National Laboratory.

Here is an interview my former boss, Dr. Charles Till; did with PBS's Frontline in a program
entitled "Nuclear Reaction" hosted by Pulitzer Prize winning author Richard Rhodes
[ "The Making of the Atomic Bomb" ]:

http://www.pbs.org/wgbh/pages/frontline/shows/reaction/interviews/till.html

As Dr. Till mentions, the fuel for the IFR is in metalic form, as opposed to the oxide
ceramic found in most reactors. By keeping the fuel in metalic form, it is particularly
easy to do the reprocessing. Instead of needing a chemical processing plant, as one
needs with oxide fuel; IFR fuel is reprocessed using metallurgical techniques, namely
"halide slagging" followed by "electrorefining".

As Dr. Till mentions, the plutonium / actinide containing output of this process can NOT
be used as nuclear weapons fuel, but CAN be recycled back to the IFR to be burned.
Therefore the IFR doesn't impose a proliferation risk.

Because the metallurical process is much simpler than chemical reprocessing, the
reprocessing plant can be placed on-site, as you suggest. The plutonium / actinides
never leave the high-radiation region of the plant; so there is no opportunity for theft.

The following web page, courtesy of the Nuclear Engineering Department at the
University of California - Berkeley mentions the on site nature of the IFR's
reprocessing system under the heading of "Diversion":

http://www.nuc.berkeley.edu/designs/ifr/anlw.html

In addition, the IFR was "inherently safe". It did not rely on engineered systems to
shutdown and cool the reactor. All that is needed to keep the IFR safe, is that the
laws of Physics work - which they ALWAYS do.

As stated in the article; the IFR is "almost too good to be true" as Richard Rhodes
phrased it. It's really too bad that President Clinton canceled the IFR in 1994.

Dr. Gregory Greenman
Physicist

Hi,
Well I've been reading the wiki article on generation IV
http://en.wikipedia.org/wiki/Generation_IV_reactor

I'm a little surprised that none will be online at least until 2030 given that fission nuclear plants do exist. If fusion could be performed beyond the break-even point tomorrow, I infer it won't go commericial until 2050. Personally, I'd rather see money diverted from fusion research into fission reactor designs. (That, and the stupid war in Iraq and war on Terrorism). I, uh, vote democrats, and I'm rather disappointed it was a democrat that killed IFR.

I'm surprised that of the designs proposed once-through cycles are seriously considered, as greenies really dislike radioactive waste. Which of the designs of Gen IV do you and nuclear physics community think is most promising?

* 1.1.1 Very-High-Temperature Reactor (VHTR)
* 1.1.2 Supercritical-Water-Cooled Reactor (SCWR)
* 1.1.3 Molten Salt Reactor (MSR)

1.2 Fast reactors

* 1.2.1 Gas-Cooled Fast Reactor (GFR)
* 1.2.2 Sodium-Cooled Fast Reactor (SFR)
* 1.2.3 Lead-Cooled Fast Reactor (LFR)

 

[Astronuc]

Which of the designs of Gen IV do you and nuclear physics community think is most promising?

* 1.1.1 Very-High-Temperature Reactor (VHTR)
* 1.1.2 Supercritical-Water-Cooled Reactor (SCWR)
* 1.1.3 Molten Salt Reactor (MSR)

1.2 Fast reactors

* 1.2.1 Gas-Cooled Fast Reactor (GFR)
* 1.2.2 Sodium-Cooled Fast Reactor (SFR)
* 1.2.3 Lead-Cooled Fast Reactor (LFR)

It's a matter of selecting the right structural materials for a 60 year lifetime - which has yet to be demonstrated, and then convincing a utility to accept the risk of a multibillion dollar/euro investment.

Even with new Gen-III plants utilities are faced with obstacles in building them, and much of that has to do with the disposition of spent fuel.

Of the list above, I'd say MSR and Lead (or Pb-Bi) are least likely, and SCWR is problematic with respect to material performance. As temperatures and pressures increase, so do the challenges to materials performance.

 

[Azael]

Astro or Morbius. Now when Bush has openly stated his war feelings for nuclear power. Is there any plans to restart projects like the IFR over there in the states?

 

[Morbius]

Astro or Morbius. Now when Bush has openly stated his war feelings for nuclear power. Is there any plans to restart projects like the IFR over there in the states?

Azael,

The Bush Administration has put forth an initiative called the GNEP -
Global Nuclear Energy Partnership:

http://www.gnep.energy.gov/

http://fpc.state.gov/fpc/61808.htm

It's a fairly new initiative, just really getting started.

How this initiative will fair with the new composition of the Congress is anybody's guess.

Dr. Gregory Greenman
Physicist

 

[Morbius]

I, uh, vote democrats, and I'm rather disappointed it was a democrat that killed IFR.

bananan,

Bill Clinton and Al Gore CAMPAIGNED on the fact that federal monies were given to
national labs to work on nuclear programs. They promised to put an end to that
practice and shutdown research on nuclear power. That's what they did - and said
so at the time.

Clinton stated that nuclear power research "wasn't needed" in his first
State of the Union address in 1993:

http://www.presidency.ucsb.edu/ws/index.php?pid=47232

"We are eliminating programs that are no longer needed, such as
nuclear power research and development. We're slashing subsidies
and canceling wasteful projects."

--President William J. Clinton, February 17, 1993

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

bananan,

Bill Clinton and Al Gore CAMPAIGNED on the fact that federal monies were given to
national labs to work on nuclear programs. They promised to put an end to that
practice and shutdown research on nuclear power. That's what they did - and said
so at the time.

Clinton stated that nuclear power research "wasn't needed" in his first
State of the Union address in 1993:

http://www.presidency.ucsb.edu/ws/index.php?pid=47232

"We are eliminating programs that are no longer needed, such as
nuclear power research and development. We're slashing subsidies
and canceling wasteful projects."

--President William J. Clinton, February 17, 1993

I am quite impressed with the IFR - (not only because the project was directed by a Canadian) and I think you make an excellent case for its reconsideration.

But to be fair to Clinton, cancellation of the IFR was done as part of a massive economic restructuring to balance the enormous deficit that had been built up (and which pales in comparison to the present Bush administration deficit). If he had not cut it one could well have asked why are they cutting these other programs (which Clinton describes as having been difficult decisions in the full quote of the speech to Congress which you excerpt) when they are keeping this expensive IFR nuclear energy project?

So a good argument could be made that the IFR was more a casualty of previous administrations that spent (or failed to adequately tax) irresponsibly. Clinton did what any responsible leader interested in keeping the country strong should have done: cut spending and raise taxes to balance the books.

AM

 

[Morbius]

But to be fair to Clinton, cancellation of the IFR was done as part of a massive economic restructuring to balance the enormous deficit that had been built up (and which pales in comparison to the present Bush administration deficit). If he had not cut it one could well have asked why are they cutting these other programs (which Clinton describes as having been difficult decisions in the full quote of the speech to Congress which you excerpt) when they are keeping this expensive IFR nuclear energy project?

So a good argument could be made that the IFR was more a casualty of previous administrations that spent (or failed to adequately tax) irresponsibly. Clinton did what any responsible leader interested in keeping the country strong should have done: cut spending and raise taxes to balance the books.

Andrew,

BALONEY!

If you look back at the budgets of the Department of Energy - it went UP under
Clinton. There was no great cutback! There was plenty of money spent on
programs that are now dead because they didn't pan out - all the so-called
"green energy". Now THAT was WASTE!

The IFR was cut for ONE and ONLY ONE reason - it was NUCLEAR.

You need to go back and read the speeches by Al Gore at the time!

Al Gore trumpeted the fact that the Clinton Administration canceled anything and
everything that was NUCLEAR.

The policies of the Clinton Administration were some of the most anti-nuclear of
any previous Administration. That's most likely due to the influence of Al Gore.

It costs more to dismantle a facility than it takes to just shut it down and
"mothball". If the Clinton Administration was interested in saving money;
they could have mothballed EBR-II; which served as the IFR prototype.

NO - they ordered it DISMANTLED - and PAID for doing so! They wanted
it dismantled so that it would be difficult to reverse their decision.

I'm sorry if it hurts your partisan sensibilities; but you need to face facts!

The Clinton Administration was anti-nuclear and PROUD of it!

One can't come back later and claim it was just budget pressures; because the
Clinton Administration was quite up-front at the time that they wanted to shutdown
ALL nuclear programs and research. [ The Supercollider died too. ]

Dr. Gregory Greenman
Physicist

 

[pauldusa]

Hello I'm a beginer at this, but would like to share my point with all of you smart ones out there.
here goes, I understand that Energy given off is based on the Electrons of a atom, and the quantity of electons in a action, is what its all about.
I understand that the medal mercury, Hg, 80, has the most electrons per element, Non-Radioative,but the last safe choice, just before the unsafe elements, if we could just use mercury as a energy source. I would enjoy the pro's help with my concept.
Thanks

 

[Morbius]

Hello I'm a beginer at this, but would like to share my point with all of you smart ones out there.
here goes, I understand that Energy given off is based on the Electrons of a atom, and the quantity of electons in a action, is what its all about.
I understand that the medal mercury, Hg, 80, has the most electrons per element, Non-Radioative,but the last safe choice, just before the unsafe elements, if we could just use mercury as a energy source. I would enjoy the pro's help with my concept.
Thanks

Paul,

I'm afraid it's more complicated than that.

When you talk about "energy given off" - what energy are you talking about.

In the case of nuclear energy; the energy that one can get by fissioning or fusing
the nuclei of elements; the amount of energy available has absolutely nothing to
do with the number of electrons.

The electrons are important in chemical reactions; not nuclear reactions. Even then,
how many electrons there are is not really important either. When you have a bunch
of chemicals, the laws of Physics and Chemistry tell you what combinations those
elements can make. Each of the combinations has a certain energy associated with
them - the "binding energy".

If you can take chemicals in one energy state, and rearrange them into another
arrangement that has lower energy - then net energy is released.

For example, let's take the combusion of Hydrogen to form Water:

2 H2 + O2 ---> 2 H2O

That is two molecules of Hydrogen plus one molecule of Oxygen gives you two
molecules of water.

If you look at the reactants, the Hydrogen molecules and the Oxygen molecule;
there is a certain amount of energy in the Hydrogen-Hydrogen bonds of the two
Hydrogen molecules plus the energy in the Oxygen-Oxygen bond of the Oxygen
molecule.

When we rearrange them into water; you now have two Hydrogen-Oxygen bonds in
each of the two water molecules - for a total of 4 Hydrogen-Oxygen bond.

It turns out that those 4 Hydrogen-Oxygen bonds represent LESS energy than the
total energy in the Hydrogen-Hydrogen and Oxygen-Oxygen bonds you had before
the reaction.

That difference in energy is released as Heat Energy - and that's why you can get
energy out of burning hydrogen.

It's not as simple as just counting electrons.

When you say you want to use Mercury as an energy source; what do you have in
mind? You don't get energy just from an element; you get energy from a chemical
or nuclear reaction.

Do you have a specific chemical reaction with Mercury in mind?

There's no nuclear reaction with Mercury that's going to give you energy and be
self-sustaining. Mercury isn't a nuclear fuel source.

Dr. Gregory Greenman
Physicist

 

[pauldusa]

Dr. Gregory Greenman thanks for you help, here is my idea, since Hg has the most electrons, like the match idea, if you could gather all the energy from just one single match,you would have all the energy to power the homes in the world forever. I study many areas of research, I understand your idea on a chemical reaction, Maybe I'm trying to apply my most electons element to getting energy out of, can you plug in a concept of mercury used as a secondary storage battery, being better than Lithium or with Lithium the latest battery of today, instead using mercury, since it loaded with a lot of electrons.
Thanks for your time

 

[Morbius]

Dr. Gregory Greenman thanks for you help, here is my idea, since Hg has the most electrons, like the match idea, if you could gather all the energy from just one single match,you would have all the energy to power the homes in the world forever. I study many areas of research, I understand your idea on a chemical reaction, Maybe I'm trying to apply my most electons element to getting energy out of, can you plug in a concept of mercury used as a secondary storage battery, being better than Lithium or with Lithium the latest battery of today, instead using mercury, since it loaded with a lot of electrons.
Thanks for your time

paudusa,

First in a "lithium" batery; which is more properly called a "lithium-ion" battery; you are
not getting energy out of lithium.

A particular chemical reaction is being used to store energy and release it later. You
can see the formula for that reaction at:

http://en.wikipedia.org/wiki/Lithium_ion_battery

The reaction that stores energy involves not just Lithium, but also Cobalt (Co),
Carbon (C) and Oxygen (O).

It's called a "lithium battery" or a "lithium-ion battery"; but that's just a convention;
the real chemistry behind this battery is more involved than just the element lithium
and how many electrons it has.

Read the article from Wikipedia. There's a lot of things that go into the design of a
battery; like how much charge it can store, whether it suffers from "memory effect",
the temperature sensitivity...

As the article states, it took a lot of good chemists working on this for 20 years to
come up with the lithium battery. I don't think just using another element with more
electron automatically means that you get a better battery.

Dr. Gregory Greenman
Physicist

 

[Morbius]

So a good argument could be made that the IFR was more a casualty of previous administrations that spent (or failed to adequately tax) irresponsibly. Clinton did what any responsible leader interested in keeping the country strong should have done: cut spending and raise taxes to balance the books.

Andrew,

It's blatently partisan to say that the IFR was a casualty of previous Administrations.

First, the IFR was canceled during a period where the Democrats controlled BOTH
houses of Congress. One can't say that the fiscal policies of spending and taxing
were the fault of one party or the other. The budgets and spending prior to the
cancellation of the IFR were passed by a Congress controlled by Democrats,
and a White House controlled by Republicans until 1993.

BOTH parties have to share the responsibilities for the fiscal health of the US Gov't
at that time.

However, programs like the IFR are really a "drop in the bucket" compared to the
overall budget of the USA.

It was clear from the very beginning of the Clinton Administration that it
was opposed to everything and anything nuclear.

The options for the IFR were to mothball it; stop work and place the facility in a
condition where it could sit unused; but could be restarted at a future date. That would
be the least expensive option.

The second option would be to continue the research program; which is more expensive.

The third option is the MOST expensive; DISMANTLE the facility. Although this is the
most expensive option; it makes it much more difficult for a future Administration to
restart the project; because they have no facility, and have to build a new one. This is
the option the Clinton Administration chose for the IFR.

The Clinton Administration was also seeking to shutdown the DOE's nuclear laboratories.
The first lab in the gunsight was Lawrence Livermore. Clinton's first Secretary of Energy,
Hazel O'Leary; appointed a task force which was composed mostly of "greenies" and led
by Motorola CEO Robert Galvin; hence the task force was called the "Galvin Commission".
They wanted to either shutdown LLNL, or turn it into a "green lab" which was a hub for
technology transfer to industry, and LLNL nuclear physicists could design blades for
wind turbines.

Fortunately, LLNL was saved by the herculean efforts of its then Director Dr. Bruce Tarter:

http://www.llnl.gov/llnl/history/directors.html

Under the section devoted to Dr. Tarter's tenure as Director, called "New Challeges",
one reads:

"In 1994, the Galvin Commission was called by former Energy Secretary Hazel O'Leary
to explore whether the Laboratory should be closed down. Others thought the Laboratory
should relinquish all weapons work, and some thought it should become a hub for
technology transfer. Tarter felt the Lab's primary mission should remain weapons work,
particularly through the use of new tools such as the National Ignition Facility and
terascale computers. It was a controversial position, but his tenacious efforts led the
Galvin Commission, along with President Clinton, to agree that the nation needed
laboratories such as Lawrence Livermore."

I can cite you chapter and verse of the anti-nuclear efforts of the Clinton Administration.

The gutting of the nuclear research capabilites of the USA for both power production
and national security is a legacy of the Clinton Administration, the damage of which is
still being repaired.

It's what the Clinton Administration wanted; and they got it. They were proud of this
stance at the time. Many of Clinton's followers believe it was the right thing to do; or
had wanted him to go even farther.

However, if one supports nuclear power and nuclear research for both power production
and national security; one can't say now that the Clinton Administration was aligned with
those priorities.

My friends who are ardent Democrats, but that support nuclear power and nuclear
research are not proud of the Clinton Administration in that matter. Most see that
as the influence of Al Gore, and for that reason did not support his Presidential
aspirations.

Dr. Gregory Greenman
Physicist

 

[selfAdjoint]

Dr. Greenman, you are properly indignant at the Clinton Administration's behavior, and so am I. They had at that time what they considered a powerful client; the green anti-nuclear movement. Just like some other social fads that have faded with time, it's hard to appreciate today just how petty and fanatical they were. They told each other lies about the existing reactors. Nova produced a program about TMI with a sinister voiceover, implying, but never quite stating, that people in the region wer dying like flies of cancer and other radiation-linked illnesses. The anti-nukes wanted to shut down a reactor as unsafe every time some normal problem was found (one time that I remember it was a broken perimeter fence! Well, "children could get in and be poisoned"). And the Clintons threw them that bone. It was truly shameful. Republican administrations haven't done that, but I haven't seen them follow up their casual talk about encouraging nuclear options as a global warming action. Nobody has clean hands.

 

[Morbius]

. Nova produced a program about TMI with a sinister voiceover, implying, but never quite stating, that people in the region wer dying like flies of cancer and other radiation-linked illnesses.

selfAdjoint,

EXACTLY. I'm familiar with the Nova program. The implications that TMI caused
health problems and deaths were thoroughly discredited by studies such as the
Rogovin Report following the accident at TMI.

The residents around Three Mile Island sued Metropolitan Edison. The judge granted
"summary judgment"; i.e. threw their case out of court because they didn't have any
evidence of harm. There was no factual matter for a jury to decide.

The ruling of Judge Sylvia Rambo can be read at the aforementioned
Frontline website for the program that featured Dr. Till's interview:

http://www.pbs.org/wgbh/pages/frontline/shows/reaction/readings/tmi.html

From Judge Rambo's conclusion:

"The parties to the instant action have had nearly two decades to muster evidence in
support of their respective cases. As is clear from the preceding discussion, the
discrepancies between Defendants, proffer of evidence and that put forth by Plaintiffs in
both volume and complexity are vast. The paucity of proof alleged in support of Plaintiffs,
case is manifest. The court has searched the record for any and all evidence which
construed in a light most favorable to Plaintiffs creates a genuine issue of material fact
warranting submission of their claims to a jury. This effort has been in vain."

Dr. Gregory Greenman
Physicist

 

[Astronuc]

Dr. Gregory Greenman thanks for you help, here is my idea, since Hg has the most electrons, like the match idea, if you could gather all the energy from just one single match,you would have all the energy to power the homes in the world forever.

In a neutral atom, the number of electrons balances the number of protons in the nucleus. So Hg (Z=80) does not have the most electrons. Tl (Z=81) has two stable isotopes (Tl-203 and 205), Pb (Z=82) has 4 stable isotops (Pb-204, 206, 207 and 208), and Bi (Z=83) has one table isotope (Bi-209) which is the heaviest stable isotope.

Heavier nuclides like Th-232 (Z=90) and U-238 (Z=92) obviously have more electrons, and although radioactive, they have very long half lives which implies very low activity.

http://www.webelements.com/ - periodic table

http://www.nndc.bnl.gov/chart/ - chart of nuclides

Having electrons however does not imply readily useful or usable energy. Most electrons are bound to the atom and only the outermost several participate in the formation of chemical bonds.

 

[Andrew Mason]

Andrew,

It's blatently partisan to say that the IFR was a casualty of previous Administrations.

I don't vote Democrat or Republican. I'm Canadian.

I am just looking at the reasons given for ending the IFR. The reasons stated were economic. I don't know if that was true. But if the U.S. had a balanced budget it would have been harder to make the argument for cutting a program.

First, the IFR was canceled during a period where the Democrats controlled BOTH houses of Congress. One can't say that the fiscal policies of spending and taxing were the fault of one party or the other. The budgets and spending prior to the cancellation of the IFR were passed by a Congress controlled by Democrats, and a White House controlled by Republicans until 1993.
BOTH parties have to share the responsibilities for the fiscal health of the US Gov't at that time.

I think the Democrats were trying to balance the books and President George "read my lips" Bush vetoed new taxes. What you need is a law that says government cannot run deficits. It works very well up here.

However, programs like the IFR are really a "drop in the bucket" compared to the overall budget of the USA.

Everything is a drop in the bucket when you are talking about government. Cuts have to start somewhere and the easiest thing to cut is something that will not affect the voting public in any noticeable way.

It was clear from the very beginning of the Clinton Administration that it was opposed to everything and anything nuclear.

The options for the IFR were to mothball it; stop work and place the facility in a condition where it could sit unused; but could be restarted at a future date. That would be the least expensive option.

But then there would be all this pressure to restart it again.

The second option would be to continue the research program; which is more expensive.

The third option is the MOST expensive; DISMANTLE the facility. Although this is the most expensive option; it makes it much more difficult for a future Administration to restart the project; because they have no facility, and have to build a new one. This is the option the Clinton Administration chose for the IFR.

I think you have to include development, not just research. Unless you know how long the research AND DEVELOPMENT program would have taken to complete, it is a little hard to say which would be more expensive.

I can cite you chapter and verse of the anti-nuclear efforts of the Clinton Administration.

http://wesupportlee.blogspot.com/2006/09/al-gores-speech.html" It doesn't sound to me like he is all that anti-nuke. He sounds fairly sanguine about it. There is some irony in his comments, however, as I am sure you will note.

Presumably the reason the IFR was dismantled was because the Clinton administration did not think it would be restarted and something had to be done with it. That is unfortunate as it appears to have been a very good prospect for solving the problems that make nuclear power unattractive for many: nuclear security and disposal of the spent fuel.

It has never made sense to me to be fundamentally anti-something. You can be in favour of protecting human safety and preserving the environment. And if nuclear technology endangers human health and/or the environment one might oppose a nuclear development.

Current nuclear technology has those problems and before nuclear power proliferates uncontrollably around the world, I would like to see these problems solved. In that sense, I don't disagree with Al Gore, (but I note the irony that he was part of the administration that ended many efforts to find those solutions).

AM

 

[Morbius]

I am just looking at the reasons given for ending the IFR. The reasons stated were economic. I don't know if that was true. But if the U.S. had a balanced budget it would have been harder to make the argument for cutting a program.

Andrew,

The US hasn't had a balanced budget in DECADES! How did all this
research, much of it paid by a Government running a deficiet; get done?

I think the Democrats were trying to balance the books and President George "read my lips" Bush vetoed new taxes. What you need is a law that says government cannot run deficits. It works very well up here.

You aren't up on your history. George H.W. Bush went back on his no new
tax pledge in an agreement with the Democratic Congress. Many Republicans
were very upset about that. It's oft said that contributed to his loss in '92.

Additionally, it's often said that the Clinton Administration "balanced the
budget". They did so ONLY if you take the Social Security Program out
of the budget. If you count the intake of money by Social Security, but
not the outlay; THAT budget ran a surplus. However, if you put the
Social Security outlays into the budget; then the USA still ran a deficiet
ALL through the '90s, and still does.

Everything is a drop in the bucket when you are talking about government. Cuts have to start somewhere and the easiest thing to cut is something that will not affect the voting public in any noticeable way.

Some cuts are like "eating your seed corn" It solves a short term problem
but engenders long term failure.

I think you have to include development, not just research. Unless you know how long the research AND DEVELOPMENT program would have taken to complete, it is a little hard to say which would be more expensive.

WRONG - Argonne was doing DEVELOPMENT too. They had a complete
program with the prototype reactor. In fact, IFR was more in the
Development stage when it got cancelled. The research phase was
pretty much complete in the mid-80s. That's when they did the tests
that Dr. Till talked about.

http://wesupportlee.blogspot.com/2006/09/al-gores-speech.html" It doesn't sound to me like he is all that anti-nuke. He sounds fairly sanguine about it.

That article demonstrates Gore's IGNORANCE of nuclear power. His comment
about reactors coming in only one size - "extra large"; for example. If you
know about reactors, you will know they can be made in many sizes.

The Big Rock Point reactor in Michigan was 60 MWe. Dresden I in Illinois was
300 MWe. Palisades in Michigan is about 750 MWe. Diablo Canyon units are
about 1100 MWe.

The NRC's website gives you the power levels of ALL the plants ever built.
Gore should have done his homework before displaying his ignorance in a
speech.

The reactors were sized the way they were because that's how the power
companies ordered them. One likes to use a reactor as a "base-load" unit;
and about 1000 MWe is typical of what utilities, at the time, where ordering
as base load units; be it nuclear or coal-fired.

The reactor venders will build you a nuclear power plant of any size you
want.

Presumably the reason the IFR was dismantled was because the Clinton administration did not think it would be restarted and something had to be done with it. That is unfortunate as it appears to have been a very good prospect for solving the problems that make nuclear power unattractive for many: nuclear security and disposal of the spent fuel.

It has never made sense to me to be fundamentally anti-something. You can be in favour of protecting human safety and preserving the environment. And if nuclear technology endangers human health and/or the environment one might oppose a nuclear development.

When Clinton and Gore were in the White House; they were against everything
and anything nuclear. I work for a Lab that the Clinton Administration tried to
shutdown. As I pointed out above, our Director and others saved the day
for us. The previous Director was fired when he suggested that the Lab had
a mission in trying to combat and prevent terrorism.

Other labs weren't so lucky. A colleague of mine attended a meeting on the
Global Nuclear Energy Partnership where it was assessed what each Lab
could bring to the table. The current expertise at those labs are a mere
"ghost" of what they once were.

The national laboratory system in the USA took decades, since the '40s to
build up to where they were in the late '80s and early '90s. They no longer
have the nuclear expertise they once had. That's the Clinton / Gore legacy.

Current nuclear technology has those problems and before nuclear power proliferates uncontrollably around the world, I would like to see these problems solved. In that sense, I don't disagree with Al Gore, (but I note the irony that he was part of the administration that ended many efforts to find those solutions).

The idea that nuclear power technology promotes proliferation is a MYTH!

Yes - experienced weapons designer can use "reactor grade" plutonium for
a weapon. But that requires knowledge and techniques that a nascent
proliferator doesn't have. No nuclear weapon state ever got to be a nuclear
weapon state because of their nuclear power program.

That's another of Al Gore's misrepresentations in the article you link above.
Gore claims that in his 8 years in the White House all the nuclear proliferants
were related to nuclear reactor programs.

What he implies, disengenuously; is that these proliferants were using
nuclear power reactors for producing weapons material. Not a single one
of the proliferators in the '90s was using a power reactor. They had all
designed and built "production reactors" for the sole purpose of making
nuclear weapons material.

North Korea didn't have a nuclear power program that gave them the materials
for a nuclear weapon. No - North Korea set out years ago to make nuclear
weapons and built a nuclear reactor for that purpose.

There's ZERO history of a nuclear power program leading to a nuclear weapons
program. So why does Al Gore imply in his speech linked above that there is?

Because Al Gore LIES! Al Gore doesn't have the moral scrupples to tell the
truth about nuclear power. He could have said that nuclear power is a viable
power source for generating non-greenhouse gas polluting electrical energy,
as does Greenpeace-founder Patrick Moore in his testimony to Congress:

http://www.greenspirit.com/logbook.cfm?msid=70

However, no he has to insinuate that it has problems. The first 3 problems he
lists that he supposes could be solved; are already solved problems.

He says that there might be scalable designs to meet the economics problem;
but not any time soon. I have news for him:

http://www.gnep.energy.gov/pdfs/06-GA50506-07.pdf

http://hulk.cesnef.polimi.it/

From the College of Engineering at the University of Michigan:

http://www.engin.umich.edu/class/ners211/pro01/

So why doesn't Al Gore tell you the truth?

Dr. Gregory Greenman
Physicist

 

[Morbius]

In my area, we have a radio program with a host who is a PhD scientist.
People call in and ask questions about science. Every so often, he gets
a caller who can't understand why all hybrid cars don't have little wind
turbines on the roof. They could recover "free" energy from the apparent
wind of the car as it goes down the road. He wonders why auto engineers
can't see such an "obvious" energy saver.

The host explains how the wind turbine would be a self-defeating idea.
However, the caller persists that he has a great idea, and wonders why
the PhD. scientist can't see how wonderful his idea is. The host tells the
guy his idea is a "shovel with a rope handle", but the caller still persists
in trying to convince the scientist it is obviously a good idea.

Or when a politician by the name of Lewis Strauss became Chairman of the
Atomic Energy Commission in 1953; he was convinced that the AEC was
aiding other nations get nuclear weapons because the AEC sold them
radioisotopes.

Strauss testified to Congress on his concerns. He was followed by the
"Father of the Atomic Bomb", J. Robert Oppenheimer.

It never ceases to amaze me how those that know NOTHING about nuclear
weapons and nuclear technology, can pontificate about what technologies
are needed to obtain nuclear weapons.

Robert J Oppenheimer said it best in testimony before Congress:

"You can't get me to say that you don't use radioisotopes in making
an atom bomb. You can not get me to say that you don't use a shovel
in making an atom bomb - in fact you do use a shovel in making an
atomic bomb. The question is one of relative importance. I would
put radioisotopes as far less important than shovels, and only
slightly more important than vitamins!"

I guess some people believe that scientists are stupid.

I saw a seminar that Helen Caldicott gave at Georgetown
Univerisyt on C-SPAN. She said she could easily have
caused a melt down at Indian Point. She was on Bobby
Kennedy's boat in the Hudson River by the plant. She said
that if she had brought along some fertilizer like Timothy
McVeigh; they could have driven the boat into where the
plant gets its cooling water - it needs 1/2 million gallons
a day.

She said they could have destroyed the intake, and without
a coolant flow of 1/2 million gallons per day; the reactors
would melt down!

Did she ever consider that if they did that; the operators,
if not the automatic systems, would SHUTDOWN the
reactor so it didn't need 1/2 million gallons a day?

She evidently things engineers and operators are so
stupid that they would leave the reactor at full-power
when she had taken out the coolant for the condensor.

GIVE ME A BREAK!

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

might be more expensive than uranium or plutonium, but storage costs and environmental impact and anti-nuke activism might be muted if there was an isotope of any element, say an isotope of iodine, that when fissioned, gives off energy AND short-lived radioactive isotopes.

so, say, hypothetically speaking, iodine captures a neutron from say plutonium, and becomes a radioactive isotope of iodine, which fissions into non-radiactive iron isotope and radioactive tritium isotope with a short half life.

the upfront cost of using iodine might be more expensive, but the overall cost might be less expensive.

We appear to have strayed from the topic of your original post.

Since the energy that is released in fission is not energy from the nuclear force but the coulomb energy stored in the nucleus, fission will only release energy when the nuclear binding force approaches the coulomb (proton-proton) repulsion forces. Such nuclei are necessarily large (as the nuclear force diminishes rapidly with distance, the outer nucleons are held less forcefully in the nucleus) and so are less stable than smaller nuclei. In such a case, a little energy added to the nucleus will overcome the nuclear binding force and release the huge coulomb energy stored in the nucleus (in that sense, the energy released is electromagnetic and not nuclear in origin).

For light elements (iron or lighter), it takes more energy to separate nucleons than to put them together (fusion) which is why stars work. For elements heavier than iron, fusion actually requires more energy than it releases, which is why nuclear reactors work.

Now I don't see why you couldn't, with a powerful enough neutron source, cause an element like gold to fission and release energy. But neutrons from other fission reactions do not provide that level of energy. You would need some kind of powerful neutron accelerator (which is difficult because it is hard to get neutrons moving because they are not electrically charged). I would be interested to see Astronuc's or Morbius' views on this.

AM

 

[Morbius]

We appear to have strayed from the topic of your original post.

Since the energy that is released in fission is not energy from the nuclear force but the coulomb energy stored in the nucleus, fission will only release energy when the nuclear binding force approaches the coulomb (proton-proton) repulsion forces.

Andrew,

The nuclear binding force ALWAYS exceeds the coulomb ( proton-proton ) repulsion forces regardless of
how big or how small the nucleus is, and independent of whether the nuclide is fissile, fissionable, or neither.

Your first misunderstanding is that the energy released in fission certainly IS due to the strong nuclear force,
NOT stored Coulomb energy. The Coulomb repulsion does add some "unbinding energy"; but every nucleon
in the nucleus deepens the nuclear potential well. For example, if I took a nucleus like U-233, and I add two
neutrons to make U-235, I will deepen the nuclear potential well and the difference in binding energy between
U-233 and U-235 is about 12 MeV. However, I have not stored any additional Coulomb energy, since U-235 has
the same 92 protons as does U-233. The mass deficit, and hence the energy released in fission is different for
U-233 and U-235 even though they have the same stored Coulomb energy.

Such nuclei are necessarily large (as the nuclear force diminishes rapidly with distance, the outer nucleons are held less forcefully in the nucleus) and so are less stable than smaller nuclei. In such a case, a little energy added to the nucleus will overcome the nuclear binding force and release the huge coulomb energy stored in the nucleus (in that sense, the energy released is electromagnetic and not nuclear in origin).

This also is incorrect. You really can't speak of the nucleons in the nucleus as a packed ball made of lesser balls;
so that there are some on the outside. It is more correct to imagine the nucleons like planets in orbit - some with
orbits more elliptical than others. They are not a series of nested concentric "circles" like the planets around the Sun.

Nuclear force is roughly two orders of magnitude greater than the Coulomb force and the ratio is given by the
inverse of the "fine structure constant."

Even if a neutron has essential zero kinetic energy, when it falls into the nuclear potential well of the nucleus,
you've added the energy of the neutron relative to the bottom of the nucleus' nuclear potential well, a well that's
deeper due to the additional nucleon.

If's this NUCLEAR energy that "upsets the apple cart" and induces a fission. The Coulomb energy is
MINISCULE by comparison. That's why you can fission a fissile nuclide like U-235; with "thermal" neutrons -
neutrons that have energies in thermal equilibrium with the material it's in.

At room temperature, a thermal neutron has 0.025 eV of kinetic energy. That 0.025 eV of kinetic energy
isn't going to overcome any nuclear binding energy which are MeV - many MILLIONS of times greater. It's the
fact that the neutron is falling into this deep nuclear potential well that disrupts the nucleus and causes a fission.

[Think about a stone sitting on the ledge of a deep well. You add a little push to the stone so it falls into the
well and goes CRASH. It's not the energy you gave the stone in pushing it that is the origin of the sound energy
you heard; it's the gravitational potential energy of the stone that turned into sound energy.]

Contrary to your assertion above, the energy released is definitely NUCLEAR, not electromagnetic in origin.
Electomagnetic force is a VERY MINOR player inside the nucleus.

For light elements (iron or lighter), it takes more energy to separate nucleons than to put them together (fusion) which is why stars work. For elements heavier than iron, fusion actually requires more energy than it releases, which is why nuclear reactors work.

The fact that an isotope of Iron, namely Fe-56 is the most stable nucleus is due to BOTH the "unbinding"
energy of the Coulomb repulsion which scales like Z(Z-1)/2 which is the number of pairs of mutually repulsing
protons and the additional binding energy per nucleon which offsets the minimum of the parabola representing
the "unbinding energy" to Z = 26, which is Iron. The particular isotope Fe-56 is also affected by the effects of
spin-coupling [ pairing ], as well as asymmetry effects due the Pauli exclusion of the two sets of fermions.

Now I don't see why you couldn't, with a powerful enough neutron source, cause an element like gold to fission and release energy. But neutrons from other fission reactions do not provide that level of energy. You would need some kind of powerful neutron accelerator (which is difficult because it is hard to get neutrons moving because they are not electrically charged). I would be interested to see Astronuc's or Morbius' views on this.

Gold has a very low fission probability. It's not as simple as just blasting away at a nucleus with neutrons and
if you have enough energy you get fission. The various "channels" as they are called as to what reactions are
permitted and their relative probabilities are dictated by the laws of quantum mechanics.

You need to conserve not only energy [ aka "mass" ], but also momentum, angular momentum, nuclear spin...
Some reaction channels are forbidden because they don't satisify one or more conservation equations.

If you hit Gold with high energy neutrons, you get spallation. You get reactions like (n,n'), (n,2n), (n,3n).. that is a
high energy neutron goes in - you get either another neutron out, or 2 neutrons, or 3 neutrons... but you don't get
much in the way of fission.

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

Your first misunderstanding is that the energy released in fission certainly IS due to the strong nuclear force, NOT stored Coulomb energy. The Coulomb repulsion does add some "unbinding energy"; but every nucleon in the nucleus deepens the nuclear potential well by about 40 MeV. For example, if I took a nucleus like U-233, and I add two neutrons to make U-235, I will deepen the nuclear potential well by about 80 MeV. However, I have not stored any additional Coulomb energy, since U-235
has the same 92 protons as does U-233. The mass residual, and hence the energy released in fission is different for U-233 and U-235 even though they have the same stored Coulomb energy.

Thanks very much for the very detailed reply. This is very useful - for me at any rate!

Since the average binding energy per nucleon peaks at iron, the binding energy of each additional nucleon added to heavier nuclei must progressively decrease. You seem to be saying that the binding energy of all nucleons is 40 MeV, the same for all, which does not seem to fit with the decreasing binding energy per nucleon.

I appreciate that the binding energy of a neutron captured by a U-235 nucleus for example, will increase the energy of the nucleus. That is definitely due to the nuclear force. And in a fissile nucleus like U-235, that energy will cause the energy of some nucleons to exceed their binding energy and the nucleus will split. But I understand that the nucleus does not split in all cases where U-235 absorbs a neutron. Doesn't that necessarily mean that the nuclear energy from that neutron capture is of the same order of magnitude as the binding energy of the two parts of the nucleus that fission?

Once the nucleus splits and the two parts move apart a tiny amount - a few Fermi units, the nuclear force virtually disappears. At that point there is tremendous coulomb repulsion between the two parts that will cause the parts to fly away with great energy, so the coulomb force definitely contributes to the production of heat from nuclear fission.

Contrary to your assertion above, the energy released is definitely
NUCLEAR, not electromagnetic in origin. Electomagnetic force
is VERY MINOR player inside the nucleus.

Ok for fission. But the energy of nuclear particles (beta and alpha particles) from spontaneous decay must come from the proton-proton coulomb repulsion. Would you agree then that radioactivity is not energy from the nuclear force?

Gold has a very low fission probability. It's not as simple as just blasting away at a nucleus with neutrons and if you have enough energy you get fission. The various "channels" as they are called as to what reactions are permitted and their relative probabilities are dictated by the laws of
quantum mechanics.

You need to conserve not only energy [ aka "mass" ], but also momentum, angular momentum, nuclear spin... Some reaction channels are forbidden because they don't satisify one or more conservation equations.

If you hit Gold with high energy neutrons, you get spallation. You get reactions like (n,n'), (n,2n), (n,3n).. that is a high energy neutron goes in - you get either another neutron out, or 2 neutrons, or 3 neutrons... but you don't get much in the way of fission.

Thanks for this. What if you hit lighter nuclei with fast protons. Have we been able to turn lighter elements into gold? (I'm not suggesting it would be cost effective).

AM

 

[Morbius]

Since the average binding energy per nucleon peaks at iron, the binding energy of each additional nucleon added to heavier nuclei must progressively decrease. You seem to be saying that the binding energy of all nucleons is 40 MeV, the same for all, which does not seem to fit with the decreasing binding energy per nucleon.

Andrew,

No - I am saying that each nucleon contributes about 40 MeV to the nuclear potential well;
that's NOT the binding energy.

First, when we are talking about the "binding energy per nucleon"; we are talking about
an average. Each nucleon has a binding energy dependent upon its energy, just as
each electron in an atom has its own "ionization potential" - the amount of energy you
have to add to that electron to kick it out of the atom.

However, we are not considering the energy of the individual nucleons based on their
position in the nuclear shell structure. We're just considering an average for convenience.
We could just as well talk about the total binding energy - which is just the average
multiplied by the number of nucleons.

In order to get the true binding energy for a nucleus; you really have to do a full
solution of the equations of quantum mechanics with a model for the nuclear potential.
That's complicated.

However, there is a rough formula that gives you the total binding energy for a nucleus
with "Z" protons and "N" neutrons. The first two terms of the formula are:

Binding Energy = 15.8 MeV * (N + Z) - 0.174 MeV * Z * (Z-1) / (N+Z)^(1/3) ...

There are more terms; but these two suffice for the present. Each nucleon, be it a
proton or neutron makes an equal contribution to the first term. The protons make a
contribution to the second term which is an "unbinding energy" because the sign is
negative. This second term increases in magnitude roughly as Z squared. So for
the heavier nuclei, the ones with higher Z; it is this negative "unbinding energy" that
decreases the binding energy per nucleon as the atomic number Z increases.

I appreciate that the binding energy of a neutron captured by a U-235 nucleus for example, will increase the energy of the nucleus. That is definitely due to the nuclear force. And in a fissile nucleus like U-235, that energy will cause the energy of some nucleons to exceed their binding energy and the nucleus will split. But I understand that the nucleus does not split in all cases where U-235 absorbs a neutron. Doesn't that necessarily mean that the nuclear energy from that neutron capture is of the same order of magnitude as the binding energy of the two parts of the nucleus that fission?

Not at all. I don't understand why your claim above would necessarily be so.

Once the nucleus splits and the two parts move apart a tiny amount - a few Fermi units, the nuclear force virtually disappears. At that point there is tremendous coulomb repulsion between the two parts that will cause the parts to fly away with great energy, so the coulomb force definitely contributes to the production of heat from nuclear fission.

I didn't say that the Coulomb force didn't contribute.

The fact that the nuclear force "virtually disappears" as the fission fragments fly apart
while the Coulomb force has a longer range doesn't mean the contribution to the energy
due to the nuclear force is small. It just means that the nuclear force has less time to
transfer its energy to the particles.

Consider the following analogy. We have a coil of wire with a DC current flowing through
it. There is a magnetic field surrounding the coil and that field contains a certain amount
of energy. We are going to ramp down the current flow to zero, and the magnetic field
is going to collapse. When the magnetic field collapses, it induces a current flow in the
wire.

We are going to do the experiment twice; one ramping down the field slowly, the other
ramping down the field fast. When we ramp down the field fast, does that mean that
we get less magnetic field energy out because of the short time?

No. When the current drops to zero, there will be no magnetic field, and hence no
magnetic field energy. When you ramp down the current flow fast, and the magnetic
field collapses faster, it induces a higher voltage, and higher current flow in the wire.

When you ramp down the current flow fast; it's true there's less time for the magnetic
field energy to come out - so it comes out at at faster rate - with higher voltages and
current flow.

So even though the nuclear force has a shorter distance to transfer the nuclear energy
to the fission fragments - it has to transfer that energy at a higher rate. It can do that
because the nuclear force is over two orders of magnitude more powerful than the
Coulomb force.

Ok for fission. But the energy of nuclear particles (beta and alpha particles) from spontaneous decay must come from the proton-proton coulomb repulsion. Would you agree then that radioactivity is not energy from the nuclear force?

Radioactivity certain IS energy from the nuclear force.

Consider the alpha decay of U-238. U-238 ---> Th-234 + He4

If you look up the nuclear masses of the above reactant and products; you will seen that
you have "LOST" about 0.00458 amu of mass which is the equivalent of about 4.27 MeV
of energy. The result of the alpha decay of U-238 is an alpha particle with 4.27 MeV of
energy. [ The Th-232 gets a little energy to conserve momentum, but because it is so
much more massive, the bulk of the energy goes to the alpha particle.]

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

Andrew,

No - I am saying that each nucleon contributes about 40 MeV to the nuclear potential well; that's NOT the binding energy.

First, when we are talking about the "binding energy per nucleon"; we are talking about an average. Each nucleon has a binding energy dependent upon its energy, just as each electron in an atom has its own "ionization potential" - the amount of energy you have to add to that electron to kick it out of the atom.

However, we are not considering the energy of the individual nucleons based on their position in the nuclear shell structure. We're just considering an average for convenience. We could just as well talk about the total binding energy - which is just the average multiplied by the number of nucleons.

In order to get the true binding energy for a nucleus; you really have to do a full solution of the equations of quantum mechanics with a model for the nuclear potential. That's complicated.

However, there is a rough formula that gives you the total binding energy for a nucleus with "Z" protons and "N" electrons. The first two terms of the formula are:

Binding Energy = 15.8 MeV * (N + Z) - 0.174 MeV * Z * (Z-1) / (N+Z)^(1/3) ...

There are more terms; but these two suffice for the present. Each nucleon, be it a proton or neutron makes an equal contribution to the first term. The protons make a contribution to the second term which is an "unbinding energy" because the sign is negative. This second term increases in magnitude roughly as Z squared. So for the heavier nuclei, the ones with higher Z; it is this negative "unbinding energy" that decreases the binding energy per nucleon as the atomic number Z increases.

I didn't say that the Coulomb force didn't contribute. I'm saying MOST of the energy is from the nuclear force.

The fact that the nuclear force "virtually disappears" as the fission fragments fly apart while the Coulomb force has a longer range doesn't mean the contribution to the energy due to the nuclear force is small. It just means that the nuclear force has less time to transfer its energy to the particles.

This is very useful to me. Thanks.

[I think you meant N "neutrons" not "electrons"]. So, putting in the values for U-235 (N = 143 and Z= 92) I get:

E_{binding} = 15.8*(235) - .174*(92*91)/235^{.33} = 3713 - 236 = 3477 \text{MeV}

So am I correct in concluding that the electromagnetic energy is about 7% (236/3713) of the total energy?

...So even though the nuclear force has a shorter distance to transfer the nuclear energy to the fission fragments - it still has more energy to transfer than the Coulomb force, and has to transfer that energy at a higher rate. It can do that because the nuclear force is over two orders of magnitude more powerful than the Coulomb force.

I understand your analogy. The fact that nuclear force is much greater offsets the fact that it operates over a shorter distance, so the energy well is still very deep.

What I am having difficulty understanding is how that relates to the energy that is released when the nucleus splits. The nuclear force is trying to keep the nucleus together so energy is needed to overcome it. The greater the binding energy of a nucleon, the more energy is needed to separate it from the nucleus so less energy is released in the form of kinetic energy of the fission parts.

Radioactivity certain IS energy from the nuclear force.

Consider the alpha decay of U-238. U-238 ---> Th-234 + He4

If you look up the nuclear masses of the above reactant and products; you will seen that you have "LOST" about 0.00458 amu of mass which is the equivalent of about 4.27 MeV of energy. The result of the alpha decay of U-238 is an alpha particle with 4.27 MeV of energy. [ The Th-232 gets a little energy to conserve momentum, but because it is so much more massive, the bulk of the energy goes to the alpha particle.]

That's MUCH more than can be attributed to Coulomb repulsion. Most of the energy of the alpha is due to the difference in mass deficits.

Since E=mc^2, the sum of the rest masses of the alpha particle and Th-234 must be less than the U-238 nucleus since energy is released. But that would be the case whether it was electromagnetic energy or nuclear energy that was given off. The mass of an atom increases when it absorbs a photon and an electron moves to a higher energy level - just not very much. It is perceptible in the case of nuclear events becuase there is so much more energy involved in the release of energy from the nucleus.

(By the way, I didn't mean to include beta particles, which is obviously not due to proton-proton repulsion. I was really thinking just of alpha decay).

AM

 

[Morbius]

This is very useful to me. Thanks.

[I think you meant N "neutrons" not "electrons"].

Yes - I meant neutrons - corrected above.

So, putting in the values for U-235 (N = 143 and Z= 92)

E_{binding} = 15.8*(235) - .174*(92*91)/235^{.33} = 3713 - 236 = 3477 \text{MeV}

So am I correct in concluding that the electromagnetic energy is about 7% (236/3713) of the total energy?

I gave you only two terms of a much more complex expression. There are other terms
not included. These terms such as the asymmetry term are also negative and tend to
"unbind" the nucleus.

But if you want to compare the "unbinding energy" due to electrostatic repulsion to the
attractive part of the nuclear force; you can use those terms.

The complete expression is:

E_B = 15.8 A - 18.3 A^{(2/3)} - 0.714 {Z (Z-1) \over A^{(1/3)}} - 23.2 { (N - Z)^2 \over A } + \delta

where \delta = 0 for an odd A nucleus like U-235

I understand your analogy. The fact that nuclear force is much greater offsets the fact that it operates over a shorter distance, so the energy well is still very deep.

What I am having difficulty understanding is how that relates to the energy that is released when the nucleus splits. The nuclear force is trying to keep the nucleus together so energy is needed to overcome it. The greater the binding energy of a nucleon, the more energy is needed to separate it from the nucleus so less energy is released in the form of kinetic energy of the fission parts.

The nuclide with the greater binding energy is going to be a product nuclide.
The reactant nuclide has less binding energy.

If we call our zero energy level; the total amount of energy of the constituent particles;
then the bound state nucleus has a negative energy on that scale, and the magnitude
is the binding energy.

So when a nuclide goes from a state with lesser binding energy to a state with greater
binding energy; it means the product nuclide has an energy level that is even lower
than that of the reactant nuclide. The difference in the energy level of the reactant and
product - which is also equal to the difference in the binding energies is the "Q" of the
reaction - the energy available to be distributed as kinetic energy.

If a nucleus has greater binding energy; it means that the energy of that nucleus is
LOWER on a absolute energy scale than a nucleus that has lower binding energy.

Don't think of binding energy as an amount of energy in the nucleus - think of it
as an energy deficit - in fact it's equal to a quantity called the "mass deficit"
converted to energy units as per Einstein's famous equation.

A more stable nucleus - which is one with higher binding energy - will be
a product of an exothermic reaction. A less stable nucleus; one with lesser
binding energy - will be the reactant.

Since E=mc^2, the sum of the rest masses of the alpha particle and Th-234 must be less than the U-238 nucleus since energy is released. But that would be the case whether it was electromagnetic energy or nuclear energy that was given off.

Yes - what one really needs to calculate is the Coulomb potential energy for the
separated charges. [That's becaus I saved a post before I was finished because
I didn't want to lose the work so far like I did when Mozilla "evaporated" on me.]

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

If we call our zero energy level; the total amount of energy of the constituent particles; then the bound state nucleus has a negative energy on that scale, and the magnitude is the binding energy.

So when a nuclide goes from a state with lesser binding energy to a state with greater binding energy; it means the product nuclide has an energy level that is even lower than that of the reactant nuclide. The difference in the energy level of the reactant and product - which is also equal to the difference in the binding energies is the "Q" of the reaction - the energy available to be distributed as kinetic energy.

If a nucleus has greater binding energy; it means that the energy of that nucleus is LOWER on a absolute energy scale than a nucleus that has lower binding energy.

Right. Similar to gravitational potential energy being negative (0 at infinity) because the force is attractive so matter gains kinetic energy as it is 'caught' in the force field.

Don't think of binding energy as an amount of energy in the nucleus - think of it as an energy deficit - in fact it's equal to a quantity called the "mass deficit" converted to energy units as per Einstein's famous equation.

Yes, I realize that the binding energy represents the energy required by a nucleon to get over the lip of the binding energy well. So if it is given that amount of kinetic energy it is home free, especially if it is a proton, in which case it gets an electromagnetic propulsive kick as well.

That is my whole problem here. If I understand what you are saying (and I do not doubt that you are right, it may be just me) that the neutron's capture by the nucleus gives the neutron such enormous kinetic energy that it is enough not only to get about 90 nucleons over the top of their binding energy well, but has enough left over to give the two fission parts kinetic energy that greatly exceeds the energy from proton-proton coulomb repulsion.

[That's because I saved a post before I was finished becauseI didn't want to lose the work so far like I did when Mozilla "evaporated" on me.]

Yeah. That can be a problem. I copy a long reply to Notepad and save it, then paste it back in later.

AM

 

[Morbius]

Right. Similar to gravitational potential energy being negative (0 at infinity) because the force is attractive so matter gains kinetic energy as it is 'caught' in the force field.

Andrew,

Exactly.

Yes, I realize that the binding energy represents the energy required by a nucleon to get over the lip of the binding energy well. So if it is given that amount of kinetic energy it is home free, especially if it is a proton, in which case it gets an electromagnetic propulsive kick as well.

Yes - if the proton can escape - more likely an alpha particle - it gets additional energy
due to the electrostatic repulsion of the alpha and the remaining nucleus.

That is my whole problem here. If I understand what you are saying (and I do not doubt that you are right, it may be just me) that the neutron's capture by the nucleus gives the neutron such enormous kinetic energy that it is enough not only to get about 90 nucleons over the top of their binding energy well, but has enough left over to give the two fission parts kinetic energy that greatly exceeds the energy from proton-proton coulomb repulsion.

There's more to it than just the energy. For example, let's take a target nucleus that
has Z protons and N neutrons. A = Z + N. Let's assume Z is odd and N is even.
Then A will be odd, and the \delta in the expression above is zero. The nucleus will
have a binding energy given by the above.

Now the nucleus captures a neutron. We now have to compute a new binding energy
for the compound nucleus. In this case, the value of A --> A + 1; so the new A is even.
The number of neutrons N --> N+1; which is now odd. When A is even; but both Z and
N are odd; the value of \delta is negative. [The \delta term is called the "pairing term"; and is
due to a quantum mechanical effect.]

So in effect, the compound nucleus "looses" some binding energy. [ This doesn't mean
that any real energy is being destroyed; it just means that the ground state energy for the
compound nucleus is higher than it would have been otherwise.]

So you now have a nucleus that not only has a bunch of energy that the new neutron
brought in - but also its ground state energy has changed. These two effects coupled
together may mean that the new compound nucleus is unstable. The nucleus has to
now find a stable state.

Most likely, a nucleus that absorbs a neutron will be unstable with respect to \beta- decay.
That extra neutron will turn into a proton, an electron, and an anti-neutrino; and the latter
two will be ejected and the nuclide transmutes to one with the next higher atomic number
Z+1 due to the new proton.

Note that the daughter nuclide with atomic number Z+1 will have more electrostatic repulsion
than the parent which had atomic number Z. There will be MORE Coulomb "unbinding energy".
However, the nuclear effects are more important than the Coulomb effect. Since the original
value of Z was odd; the new value Z+1 will be even. The neutron number N will go back to
its original value which is even. So the daughter nucleus will have an even number of protons,
and an even number of neutrons. This is more important for stability of the nucleus than is
the extra Coulomb repulsion. The nucleus is more stable even with the additional Coulomb
"unbinding energy" because the nuclear effects "trump" the Coulomb effects.

So the extra energy the neutron brings in doesn't have to be enough to get out of the
"old" potential well of the target nucleus - it has to be enough to get out of the well of
the "new" nucleus.

You really have to due a quantum mechanical description of the nucleus with all the
shell structure; just like one has to do with electrons in an atom.

Yeah. That can be a problem. I copy a long reply to Notepad and save it, then paste it back in later.

Yes - also discovered that if you make a mistake in a "tex" equation, and you edit it;
the system doesn't re-evaluate the tex expression after you edit it. You have to copy
the entire post to an editor [ I use Emacs instead of Notepad because I'm working in
Unix], delete the faulty post, and post a new reply using the saved text from the editor.

Dr. Gregory Greenman
Physicis

 

[Andrew Mason]

Thanks very much for all of this. It has been very helpful and a very enjoyable discussion..

I have to run but let me just put this thought out:

In order to release energy, the fission parts have to have more binding energy than the original nucleus (as in fusion: the binding energy of the product - fused nucleus - has to be greater than the original parts if the fusion releases energy). This is possible for U because the fission parts can both be heavier than iron so the total binding energy of the fission parts is higher (which I think is the short answer to the original question).

But isn't the reason the binding energies of the fission parts increase due to electrostatic repulsion? That is, there is a smaller negative electrostatic repulsion term in the binding energy equation for the fission parts compared to original U-235.

Yes - also discovered that if you make a mistake in a "tex" equation, and you edit it; the system doesn't re-evaluate the tex expression after you edit it. You have to copy the entire post to an editor [ I use Emacs instead of Notepad because I'm working in Unix], delete the faulty post, and post a new reply using the saved text from the editor.

Now I understand why your posts have these extra carriage returns! The system DOES re-evaluate the tex expression, it just doesn't reload your screen. If you use the advanced editor it should reload it for you. Otherwise, just reload the screen in your browser and the revaluated tex graphic will show up..

AM

 

[Morbius]

In order to release energy, the fission parts have to have more binding energy than the original nucleus

Andrew,

The fission products have to have more binding energy than the COMPOUND nucleus -
not the orignal nucleus. Just as in the case with the neutron capture; it's the binding
energy of the compound nucleus that is important.

What happens after the nucleus absorbs a neutron; and which decay channel it follows
is dependent on the compound nucleus; not the original nucleus.

One option for decay is for the nucleus to eject the new neutron. That's a reaction
called "compound elastic scatter" - scatter because the neutron may not come out
going the same direction as when it went in. [When it comes out not going the
same direction, the target nucleus has to recoil to conserve momentum, hence
the ejected neutron doesn't have all the energy it came in with - some goes to
the kinetic energy of the recoiling target. The target nucleus is left in the same
internal state that it had before the collision.]

The nucleus could also keep some of the energy of the neutron, and be left in an
excited state after ejecting the neutron. In this case, the neutron will have lost
even more energy than it does in an elastic collision. In this case, the reaction is
"compound inelastic scatter". All inelastic scatter is compound inelastic scatter.
However, not all elastic scatter is compound elastic scatter; because there is
also the possiblility of "potential scatter" [sometimes called "shape scatter"]; in
which the neutron scatters off the nuclear potential of the target nucleus without
forming a compound nucleus.

But isn't the reason the binding energies of the fission parts increase due to electrostatic repulsion? That is, there is a smaller negative electrostatic repulsion term in the binding energy equation for the fission parts compared to original U-235.

Yes - that's PART of it. However, I again refer you to the previous post where I discussed
\beta- decay. There the nucleus decayed in a manner which INCREASES the
electostatic repulsion in favor of nuclear effects.

I'm just saying that you can't say that the path to fission depends on just the
electrostatic effects. That's part - but not the whole story. If the nucleus does fission;
then some of the energy of the fission products is due to their repulsion.

However, the nuclear effects are more important than the Coulomb effect, and it is often
the case, as we see in \beta- decay that the nuclear effects will trump the Coulomb effect.
So you can't say that fission is energetically selected based on just the Coulomb repulsion.

Although U-235 is fissile, it will fission with low energy "thermal" neutrons; U-238 is
fissionable, it will fission only with neutrons with kinetic energy above a certain
threshold.

However, for low energy neutrons; U-238 will not fission. But U-238 has the same
92 protons that U-235 has. If it was only about the Coulomb energetics, the U-239
compound nucleus formed when U-238 absorbs a low energy neutron, could split into
a couple of fission fragments of lower Z; and thus lower the stored electrostatic
repulsion energy in a manner identical to the way the U-236 formed by the absorption
of a neutron by U-235; will decay by fission.

But U-239 created by low energy neutrons doesn't fission! Instead U-239 forms first
Np-239, which then forms Pu-239. So we go from Z=92 ultimately to Z=94. The
electrostatic repulsion "unbinding energy" is proporttional to Z*(Z-1). In going from
Z=92 to Z=94; the Coulomb repulsion "unbinding energy" INCREASES by 4.4%

The nucleus could minimize the Coulomb repulsion energy by fissioning; and
instead it INCREASES the Coulomb "unbinding energy" by 4.4%

If it was only about minimizing the Coulomb energy; then U-238 should fission with
low energy neutrons like U-235 does! But U-238 doesn't fission with low energy neutrons;
it takes an alternate route which INCREASES Coulomb "unbinding energy". So whether
a nucleus fissions or not is not due solely to the Coulomb energetics.

Now I understand why your posts have these extra carriage returns! The system DOES re-evaluate the tex expression, it just doesn't reload your screen. If you use the advanced editor it should reload it for you. Otherwise, just reload the screen in your browser and the revaluated tex graphic will show up..

I've heard the carriage return comment on other forums too. Evidently Mozilla Firefox
under Unix / Linux works differently than Explorer under Windows.

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

However, the nuclear effects are more important than the Coulomb effect, and it is often the case, as we see in \beta- decay that the nuclear effects will trump the Coulomb effect. So you can't say that fission is energetically selected based on just the Coulomb repulsion.

Although U-235 is fissile, it will fission with low energy "thermal" neutrons; U-238 is fissionable, it will fission only with neutrons with kinetic energy above a certain threshold.

However, for low energy neutrons; U-238 will not fission. But U-238 has the same 92 protons that U-235 has. If it was only about the Coulomb energetics, the U-239 compound nucleus formed when U-238 absorbs a low energy neutron, could split into a couple of fission fragments of lower Z; and thus lower the stored electrostatic repulsion energy in a manner identical to the way the U-236 formed by the absorption of a neutron by U-235; will decay by fission...

If it was only about minimizing the Coulomb energy; then U-238 should fission with low energy neutrons like U-235 does! But U-238 doesn't fission with low energy neutrons; it takes an alternate route which INCREASES Coulomb "unbinding energy". So whether a nucleus fissions or not is not due solely to the Coulomb energetics.

But... the addition of 3 more neutrons increases the size of the nucleus and average distance between protons, so it reduces the Coulomb potential within the nucleus (effectively increasing the nuclear binding energy). That is (partly) why it does not fission with low energy neutrons. It needs more energy to overcome the higher nuclear binding energy. But I appreciate that there are some quantum effects with that \delta term which is >0 for U-238.

I knew that I had read Feynman somewhere saying that fission energy was electrical energy. After a bit of digging, I found the passage:

" There is another question: "What holds the nucleus together?" In a nucleus there are several protons, all of which are positive. Why don't they push themselves apart? It turns out that nuclei there are, in addition to electrical forces, nonelectrical forces, called nuclear forces, which are greater than the electrical forces and which are able to hold the protons together in spite of the electrical repulsion. The nuclear forces, however, have a short range - their force falls off much more rapidly than 1/r2. And this has an important consequence. If a nucleus has too many protons in it, it gets too big, and will not stay together. An example is uranium, with 92 protons. The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all of the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just "tapped" lightly (as can be done by sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical reuplsion. The energy which is liberated is the energy of the atomic bomb. The energy is usually called "nuclear" energy, but it is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces.

Now I agree that he oversimplifies things a little here. I think he is ignoring the fact that the kinetic energy of the free neutrons that are produced in fission obviously cannot be the direct result of coulomb repulsion since there is none with a neutron. I am also aware that this was written 43 years ago and that Feynman was not a nuclear physicist - but he was a pretty smart guy. So I am thinking he can't be that far off here.

AM

 

[Morbius]

But... the addition of 3 more neutrons increases the size of the nucleus and average distance between protons, so it reduces the Coulomb potential within the nucleus (effectively increasing the nuclear binding energy).

Andrew,

The formula above accounts for that because of the cube root of "A"
in the denominator.

{Z(Z - 1) \over {A^{1/3}}

Lets look at the individual terms in the expression for binding energy:

E_B = 15.8 A - 18.3 A^{(2/3)} - 0.714 {Z (Z-1) \over A^{(1/3)}} - 23.2 { (N - Z)^2 \over A } + \delta

________________U-235______U-238_______Difference

"Volume"________3713.0______3760.4_________+47.4

"Surface"________-696.7______-703.4__________-6.8

"Coulomb"________-968.8______-964.3_________+4.7

"Asymmetry"______-256.8______-284.2________-27.5

"Pairing"_____________0.0_________0.2_________+0.2

Total_____________1790.7_____1808.7________+18.0

Except for the "pairing term" which is usually pretty small for large nuclei;
the Coulomb term is the LEAST important of any of the major term.

The nuclear "Volume" term is an order of magnitude larger than the
Coulomb term. It's only because some of the nuclear terms have
opposite signs, that the Coulomb term is as important as it is; which
it is still a minority player.

It's the extra NUCLEAR binding energy that makes the U-238 more stable!

Now I agree that he oversimplifies things a little here. I think he is ignoring the fact that the kinetic energy of the free neutrons that are produced in fission obviously cannot be the direct result of coulomb repulsion since there is none with a neutron. I am also aware that this was written 43 years ago and that Feynman was not a nuclear physicist - but he was a pretty smart guy. So I am thinking he can't be that far off here.

Feynman is correct - but he's simplifying things a bit. The addition of a
free neutron to a nucleus is more than "tapping" slightly - because the
free neutron is falling into a deep potential well.

I again refer to my analogy with the stone and the well. Yes - you tap the
stone slightly to get it to fall into the well - but the loud sound you hear
didn't come from the energy imparted by your tap.

It came from the potential energy of the stone.

Likewise, U-235 is really not ready to fall apart, which is why it has a
half-life of 705 MILLION years. [U-238 is 4.5 BILLION]

It's the energy of a free neutron falling into a nuclear potential that
blows the nucleus apart.

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

It's the energy of a free neutron falling into a nuclear potential that
blows the nucleus apart.

I can accept that since the binding energy of the additional neutron is released inside the nucleus that the nucleus will become excited. What I can't quite see is how this energy (which I calculate using the formula you have provided to be about 6.4 MeV) would be sufficient to break apart the nucleus, let alone propel the fission parts with such energy.

AM

 

[Morbius]

I can accept that since the binding energy of the additional neutron is released inside the nucleus that the nucleus will become excited. What I can't quite see is how this energy (which I calculate using the formula you have provided to be about 6.4 MeV) would be sufficient to break apart the nucleus, let alone propel the fission parts with such energy.

Andrew,

Then you have to do a Quantum Mechancal treatment.

It's more complicated than just comparing the energy to the binding energy. After all,
a neutron brings in a few MeV; and the total binding energy is on the order of 2 GeV.
[The binding energy formulas are only approximate anyway.] You might not think that
a few MeV is enough to disturb a nucleus that has a couple GeV in binding energy;
but it does - which is why certain nuclides will fission.

As I stated in one of my previous posts - if you want to know how the nucleus is
going to react - then you need to do an analysis of the compound nucleus via
quantum mechanics.

Quantum Mechanics puts limits on how the nucleus can decay, and more importantly;
what states are stable. There are more variables to be considered than just the energy.
You need to conserve angular momentum, nuclear spin...

A given reaction "channel" as it is called may be favored energetically - that is, the
resultant energy state of the products will be minimized. However, if you can't conserve
angular momentum, or spin... then that channel will be forbidden.

When a nucleus absorbs a neutron, you might be tempted to say that the small increase
in energy relative to the binding energy of the nucleus should result in the nucleus doing
nothing - we would just have a stable compound nucleus. However, that new configuration
of protons and neutrons may not be stable from the quantum mechanical point of view -
there's no "stationary state" for that configuration - so it has to decay; regardless of what
you might think would happen if you consider the energetics alone.

Dr. Gregory Greenman
Physicist

 

[Andrew Mason]

When a nucleus absorbs a neutron, you might be tempted to say that the small increase in energy relative to the binding energy of the nucleus should result in the nucleus doing nothing - we would just have a stable compound nucleus. However, that new configuration of protons and neutrons may not be stable from the quantum mechanical point of view - there's no "stationary state" for that configuration - so it has to decay; regardless of what you might think would happen if you consider the energetics alone.

If there is no stationary state for the compound nucleus, of ^{235}U + n how can ^{236}U exist at all? Is there path that quantum mechanics permits for the excited ^{236}U nucleus to get rid of energy without fissioning?

AM

 

[Morbius]

If there is no stationary state for the compound nucleus, of ^{235}U + n how can ^{236}U exist at all? Is there path that quantum mechanics permits for the excited ^{236}U nucleus to get rid of energy without fissioning?

Andrew,

A "stationary state" is one that can exist "indefinitely" - i.e. a "stable" state.

If there is no "stationary state" - then the wave-function is time-dependent; that is whatever
state the quantum system is in will decay.

http://en.wikipedia.org/wiki/Ground_state

There are multiple ways for U-236 nucleus to de-excite; which gives rise to the MULTIPLE
reactions that a U-235 nucleus can have when struck by a neutron:

(n,n) elastic scatter
(n,n') inelastic scatter
(n,2n)
(n,3n)
(n,4n)
(n,fission)
(n,gamma)

U-235 doesn't always fission when hit by a neutron; fission is one possible reaction.

Dr. Gregory Greenman
Physicist

 

[Astronuc]

Is there path that quantum mechanics permits for the excited ^{236}U nucleus to get rid of energy without fissioning?

Neutron absorption in most nuclides is followed by emission of a gamma-ray, and the reaction is usually referred to as an (n,\gamma) reaction. IIRC, about 16-17% of neutron absorptions by U-235 result in U-236* which decays promptly by gamma emission, otherwise fission occurs.

 

[Morbius]

Neutron absorption in most nuclides is followed by emission of a gamma-ray, and the reaction is usually referred to as an (n,\gamma) reaction. IIRC, about 16-17% of neutron absorptions by U-235 result in U-236* which decays promptly by gamma emission, otherwise fission occurs.

Andrew and Astronuc,

Here's a plot of the total cross-section, along with elastic scatter, fission, and (n,gamma).

As can be seen; for low energy neutrons, fission is the dominant decay mode. However,
for high energy neutrons, above the resonace region; fission loses out to elastic scatter
by a wide marging.

 

[Astronuc]

I should have qualified my comment which refers to thermal neutrons. In the epithermal and higher energies, as Morbius indicated, the fission cross-section, i.e. the probability of fission, decreases in favor of scattering.