Binding energy

[andrewhaynes]

Hi

Just have a hypothetical question. The binding energy of Deuterium is 2.22MeV, if that is reached 18.5MeV is released.

What would happen if fired a 2.5MeV proton at D wouldn't it start a chain reaction, as gamma rays that would be released would be above the binding energy

[Carl Willis]

What is your source of information that says 18.5 MeV is released? Whatever it is, it's not reliable. The breakup of the deuteron into a proton and a neutron is an endothermic reaction, requiring 2.22 MeV: http://www.nndc.bnl.gov/qcalc/

-Carl

[andrewhaynes]

Hi
http://minimafisica.biodec.com/Members/ ... uteron.pdf , its says

"If the neutron in the deuteron were to decay to form a proton, electron
and antineutrino, the combined mass energies of these particles would
be 2(938.27 MeV) + 0.511 MeV = 1877.05 MeV"

If you fire a proton at 2.22MeV, the deuteron will break apart, I thought it would release the energy as a gamma ray

[Carl Willis]

The reaction you are talking about is the breakup of a deuteron by hitting it with an energetic proton:

H-2(p,np)H-1, Q = -2.22 MeV

To conserve both energy and momentum requires that the incident proton have considerably more than 2.22 MeV of energy. I will not belabor that issue, though. The process results in two unbound protons and one neutron moving in various directions but with a net momentum equal to that of the incident proton (assuming the deuteron was a stationary target). The kinetic energies of the particles sum to a total that is 2.22 MeV less than the incident proton's energy. There is no gamma radiation. Certainly there is not a source of 18.5 MeV of energy released here in any form. On the contrary, energy must be supplied to make the deuteron come unglued into its component nucleons.

The discussion in your link about the rest mass energies of the component particles is agreeable enough, but doesn't shed any light on your 18.5 MeV number!

-Carl

[andrewhaynes]

Is the total input energy required 3337.41KeV with the products 1H+n+1H ?
The 18.5MeV was spelling error was meaning 1877.05 MeV,but doesn't matter ,your statement makes sense about conservation of energy.

Thanks

[Dan Tibbets]

You need to familiarize yourself with the nuclear binding energy curves in their various configurations. Basically the binding energy is that energy that has to be put into a nucleus to free one nucleon. From Nickel 62 and smaller nuclei, energy has to be put in - endothermic, to remove that nucleon. The reverse of adding a nucleon releases kinetic energy in these lighter nuclei. That is why fusion of light elements releases KE. Things are opposite for heavier elements where fission releases energy. There are all sorts of permutations depending on the starting energy state of the nuclei, but this is the basic pattern.

Binding energy should not be confused with rest mass energy equivalence, the KE of the starting nuclei (speed), the missing mass, or mass deficit. They are inter related in many ways but can be miss leading when combined.

Nuclear binding energy (or more precisely the nuclear binding energy per nucleon) reflects the potential energy of the nucleus and it's relationship to possible daughter products or reactants. It is made up of two competing forces (ignoring the contribution of the Weak force which plays a major role in the possible neutron content of the nucleus), attractive Strong force and repellent Electromagnetic force. The net effect is that the Ni62 nucleus has the least potential energy(negative value by convention), the tightest packed nucleus, the greatest stability possible; and that is why it is the turn around point for the direction of energy flow. All other nuclei on either side have more potential energy (more positive values) and of course to get kinetic energy out you have to decrease the potential energy. The potential energy is basically the Strong force attractive energy summed between all of the nuclei (this is the negative number as by convention attractive energy/ force is assigned a negative number- applies to Strong force and gravity), added to the repulsive electromagnetic force between all of the nucleons. The range of the competing forces determines the relationship. The strong force is greater at close ranges, but there is always a range at which the electromagnetic force becomes greater. In light element fusion you are essentially gaining KE from the excess strong force potential energy stored in the nucleus , while with heavy element fusion you are harvesting the excess stored electromagnetic force in the nucleus. It can be confusing, but the use of negative numbers for the Strong force contribution makes the book keeping easier. This is why I prefer the Nuclear Binding Energy Graphs that take this into consideration.

Note that Iron 56 is often mentioned as the end point for energy release with building larger nuclei, but this reflects the possible/ likely nucleosynthesis processes in a star.

Dan Tibbets

[Dan Tibbets]

Nuclear binding energy graphs in various presentations:

https://www.google.com/search?q=nuclear ... 66&bih=602

The version that I find most usefull/ least confusing, as it ties into the potential energy changes that determines whether a reaction is endothermic or exothermic. Note that the binding energy per nucleon of a proton (hydrogen), and a free neutron are ~ zero. It uses the convention that the attractive force is a negative value.

https://www.google.com/search?q=nuclear ... 66&bih=602

Dan Tibbets

[andrewhaynes]

If the fuser instead of 20kv was 1.11MeV for one nuclei(or would you need the 2.22MeV), would you still be at a loss from the endothermic reaction, with proton+deuterium fusion (3He+γ 5493kev) gain the KE from the voltage difference, it should increase. I'm missing something...

Thanks

[Dan Tibbets]

The P+D reaction is exothermic and apparently does involve more energy than your ~ 1 MeV proton.

What you are missing is the likelyhood of the reaction proceeding. I don't know the fusion cross section curve for the P+D reaction, but presumably it is higher at 1 MeV than it is at 20 KeV (though it may be past the peak of the cross section curve). This improves the likelyhood of the of the reaction occuring, but remember, the likelyhood may be only once out of a thousand (or more collisions). Each of these non fusing collisions results in the the loss of energy through heating of the walls, etc. And there are radiation losses (Bremsstruhlung X-rays, black body radiation,etc.). This robs the system of a lot of energy.

The MeV energy of the reactants is drained faster than the added energy from the relatively rare exthermic fusion reaction. It is a losing game. The energy loss ends up mostly as waste heat. While some of the energy from this waste heat can be recovered though a steam cycle, etc. the net losses may still be more than 60% of the total heat output. You cannot keep up. To break even you may need to have energy recovery of 99% or more.

These numbers can be modified, at least in theory. This is where the Triple Product comes in. The relationship between density , temperature and losses need to meet a minimal product.

Fusion is easy. D-D, D-T, D-He3, P-B11 are all easier than P-D (I think), but fusion of these reactants in appropriate huge quantities is much harder. The reaction of D-T is easiest because it peaks at ~ 60 KeV and is at least 100 times more likely to occur than any other reaction and it results in the most fusion energy output per reaction. This gives you the most bang for the buck, but still there is considerable difficulty in overcoming the losses. You must have huge numbers of fusion events to have any hope of catching up to the losses, especially if you can harvest only ~30% of the energy output for useful electricity production through steam generation. Direct conversion, if possible for the P-B11 reaction or the D-He3 reaction helps the situation marginally as you might recover up to ~ 80% of the energy output (at least from the fusion energy output portion of the total).

Look at it this way. If one proton is accelerated to ~ 1 MeV and it participates in fusion with a D, then you have a net energy production. But you may need to accelerate 1000 protons (or even a million protons) to this energy before a fusion occurs. You get out a few MeV of fusion energy, but you put in ~ 1 GeV of energy. So you lose ~ 998 MeV of energy through non fusing collisions with walls, etc. and radiation losses. Even with energy recovery of this waste heat you are still down ~ 700 MeV. Getting reactants up to temperatures/ KE where exothermic fusion reactions become more likely is only a part of the picture. What is paramount is controlling the losses (improved particle/ energy containment) to remarkably good levels.

Dan Tibbets

[Richard Hull]

As I note frequently, fusion is all about loss in our world. As you start to gain energy due to fusion at X rate you are losing at (Xk) rate, always. Fusion, for power, demands that the k constant set by failure to contain, become zero. There are no free lunches in fusion outside of stars. Free luches are readily found in earth-bound fission...If we are talking nuclear energy.

For the fusor k= ~ 10e9

Free fusion lunches abound on earth as "ex-post fusion" chemical energy stores of coal, oil, wood and weather related events in hydro electric dams and solar panels (thermal and electric). These are all simple examples. "Old fusion" powers the bulk of the earth currently via potential energy chemical stores and fortunate weather related events. Fission is also stored fusion energy from many billions of years ago in some creational or super nova event if you wish to put a fine point on it.

It is just if we are talking totally, earth-bound nuclear processes where fission is the only free nuclear lunch on earth.

Richard Hull

[Carl Willis]

The reaction you mentioned earlier involved the disintegration of a deuteron into unbound nucleons. The H2(p,g)He-3 reaction is a different reaction, one that results in a bound nucleus (He-3). This reaction is exothermic and can in theory happen with protons of any energy. The reaction's cross section is far smaller than DD fusion, however.

-Carl

[andrewhaynes]

If the losses from non collisions is related to mean free path, if you started at 1MeV/40kv = 25000, couldn't you raise the pressure in the fuser by 25000, using this site http://hyperphysics.phy-astr.gsu.edu/hb ... enfre.html. Taking it to extreme, having the pressure match the diameter of the vessel, with the voltage matched.....

I was thinking about releasing neutrons through disintegration and maybe have 3H made reaction with 2H

Thanks

[andrewhaynes]

I was trying to make a neutron source, I think 1.112MeV is doable but 2.224MeV might be two high. Would the disintegration happen with the single nucleons or would it be 2.22.

[Carl Willis]

I don't think MeV protons are, in a word, "doable" in realistic hobby scenarios. At the least, proposals about using MeV beam energies aren't credible unless one has established a unique arrangement to borrow time on an accelerator, or already possesses the rare and enviable facility and technical background to accommodate such a project.

As I said earlier, you could not disintegrate deuterons with 2.22-MeV protons. A photon, say from radioactive decay of radium, can disintegrate the deuteron at about 2.22 MeV since the kinematic constraints caused by conservation of energy and momentum can be simultaneously satisfied at lower energies than with the proton. You need about 3 MeV to do this reaction with incident protons, and this reaction will compete with the aneutronic radiative capture (p,g) reaction that has also been mentioned.

Hobbyists thinking seriously about the cheapest, fastest, most effective way to get neutrons pretty much always come home to DD fusion. It is THE time-honored recipe. To consider something else, particularly an approach that requires an MeV beam, pretty much begs for some uniquely-justifying circumstances.

-Carl

[andrewhaynes]

Sure its a long way from 100kv to 1000kv. I still don't understand if the energy level is reached, if not from a proton then gamma ray, why it would be different. Would it be 2.22MeV or 1.11MeV which is the binding energy of one nucleons.

Thanks

[Carl Willis]

Read this:

http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html

[andrewhaynes]

Thanks Read that page found this link http://hyperphysics.phy-astr.gsu.edu/hb ... 5w.html#c1 , would it be 2.224 + rest mass of proton(934kev i think)?
On that link it says 1.29 will change a proton to a neutron, just curious if you had a 1cm spark gap with hydrogen in the middle and 1.5MV supply, would that accelerate the proton and change it to a neutron.

[Richard Hull]

No, it would not.

Richard Hull

[andrewhaynes]

Would the binding energy disintegration happen once its accelerated to 2.22MeV or would it happen when it collides with another atom?

If say a marx charged up two electrodes 1cm apart, in close to vacuum, with D2 in the middle would that accelerate electrons one way, and protons the other way?

With two electrode 1cm apart, you should only loss 30kv for the e-feild, how would you stop a cw or marx from not reaching there full voltage, wouldn't they top out at 30kv.

Thanks

[Richard Hull]

You'll not beat "the system" clutching at weak straws. Many, many before you have already come down this path with nothing to show at the end.

So far, ideas...Many.......Results, zero.

Only results feed th' bulldog.

Richard Hull

[andrewhaynes]

I've got some results http://4hv.org/e107_plugins/forum/forum ... php?147892 , after I started posting here.

[Carl Willis]

This thread is losing focus.

The original subject is a theory matter and belongs in this forum, but questions about Marx generators and pumps and so on should be put in the appropriate experimental forums. Also, if you want to discuss a project on this site, I would appreciate it if you would describe that project in its own thread, rather than just by reference to discussions over at 4HV.

To answer your last question about energetics in the breakup of the deuteron, think of the nucleus as a mechanical system of two magnetized steel balls stuck together that you want to take apart. You can throw the system of two balls, but they won't break apart until they hit something (possibly a third ball, representing another nucleon). Kinetic energy is relative--you define a reference frame, and then the velocity of the nucleus in that frame determines its kinetic energy in that frame. A deuteron could have 1, 10, 100, etc. MeV of kinetic energy depending on the observer, but whether or not it is capable of being taken apart in a collision depends only on relative velocity with respect to the other body in the collision.

-Carl

[Richard Hull]

To follow up on Carl. If you are looking for a pump, put that request in the trading post as a wanted item, since that is where everyone goes to buy and sell or make requests for information on hardware and other instruments.

Richard Hull