Fission Details: Atom, Velocity, Mass, Energy

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In summary, fission is a process in which the nucleus of an atom is split into two smaller nuclei, releasing a large amount of energy in the form of heat and radiation. This process is triggered by a high velocity neutron colliding with a heavy atom, causing it to become unstable and split apart. The mass of the resulting nuclei is slightly less than the original atom, with the missing mass converted into energy according to Einstein's famous equation, E=mc². Fission is the basis for nuclear power and weapons, and understanding its details is crucial for safe and efficient use of this powerful process.
  • #1
jaketodd
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I feel bad I've been asking so many questions, some of which don't really make sense. So I bought a 5 year membership to try to make it up to you guys; trying to give back.

Here's what I really need to know:

An example atom that is fissionable, along with the velocity of the nucleons before the fission, the mass of the nucleons before the fission, the kinetic energy of the nucleons before the fission, and the energy released by the fission minus the kinetic energy due to the opposing charges of the daughter nuclei (in other words, just the energy released due to mass converting to energy, and any binding energy released if that even applies (I am still learning here)).

I wish I could have just made my questions this simple to begin with, but I didn't understand enough. Now I do; thanks to all you guys and/or gals. Now that I've put a lot of thought into my above questions, I see that it's asking for a lot of data, I hope someone comes through for me.



Jake
 
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  • #2
jaketodd said:
I feel bad I've been asking so many questions, some of which don't really make sense. So I bought a 5 year membership to try to make it up to you guys; trying to give back.

Here's what I really need to know:

An example atom that is fissionable, along with the velocity of the nucleons before the fission, the mass of the nucleons before the fission, the kinetic energy of the nucleons before the fission, and the energy released by the fission minus the kinetic energy due to the opposing charges of the daughter nuclei (in other words, just the energy released due to mass converting to energy, and any binding energy released if that even applies (I am still learning here)).

I wish I could have just made my questions this simple to begin with, but I didn't understand enough. Now I do; thanks to all you guys and/or gals. Now that I've put a lot of thought into my above questions, I see that it's asking for a lot of data, I hope someone comes through for me.



Jake

Well, why don't you get the process started. Much of the info you asked for is available on Wikipedia ... some of it was even posted as replies to your previous thread on this. If you can choose a particular example, and work it through until the point where you get stuck, it will make it much easier to help you. I don't mind answering specific questions, but I don't have the time to construct an example from scratch ... it might not even be what you are looking for in the end.
 
  • #3
SpectraCat said:
Well, why don't you get the process started. Much of the info you asked for is available on Wikipedia ... some of it was even posted as replies to your previous thread on this. If you can choose a particular example, and work it through until the point where you get stuck, it will make it much easier to help you. I don't mind answering specific questions, but I don't have the time to construct an example from scratch ... it might not even be what you are looking for in the end.

Ok, here is as far as I can get, followed by what I can't figure out. There is a lot I don't know, and I really hope you don't find it too much to ask.

Let the example atom be Uranium 235.

From Wikipedia (http://en.wikipedia.org/wiki/Nuclear_fission#Output") there is 7 MeV in gamma rays + 4.8 MeV in neutrons released per fission event. This is ignoring the repulsion energy of the daughter nuclei. So the total energy released that I am concerned with is 11.8 MeV per fission of the Uranium 235 atom.

The first thing I don't know is if there is any additional energy released, such as binding energy.

From Wikipedia (http://en.wikipedia.org/wiki/Uranium-235" ) there are 143 neutrons, and 92 protons in the nucleus of Uranium 235.

Now here is where I ask a lot, but I honestly don't know how to figure these out:

The velocity of a single proton in the nucleus of Uranium 235 before fission.

The velocity of a single neutron in the nucleus of Uranium 235 before fission.

The energy released when two protons are split in fission of Uranium 235 (binding energy?).

The energy released when two neutrons are split in fission of Uranium 235 (binding energy?).

The energy released when one proton and one neutron are split in fission of Uranium 235 (binding energy?).

I was thinking maybe I could figure out these last three using ratios of the masses of protons and neutrons along with the ratio of the total energy released when Uranium 235 fissions to the number of protons and neutrons in the atom, but I can't seem to figure out how to represent that mathematically.

Thanks a million,

Jake
 
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  • #4
I think I figured out how to answer the last three questions if I can be provided with the answers to the other ones.

Thanks,

Jake
 
  • #5
Protons and neutrons don't split in fission. In fission, a nucleus separate into two nuclei, but the protons and neutrons remain intact. There are normally two or three (fission) neutrons that escape in addition to the two fission product nuclei. There are also prompt gamma rays emitted.

In the case of U-235, it's actually an excited U-236 (from U-235 + n) nucleus that fissions.
 
  • #6
The velocity of a single proton in the nucleus of Uranium 235 before fission.

The velocity of a single neutron in the nucleus of Uranium 235 before fission.

What do you mean? The neutrons and protons are all bound together and have the same velocity. Before fission you could say that the velocity it zero.
 
  • #7
Drakkith said:
What do you mean? The neutrons and protons are all bound together and have the same velocity. Before fission you could say that the velocity it zero.

The KE for the nucleons in an atom were posted here: https://www.physicsforums.com/showpost.php?p=3347668&postcount=3"

So can't we figure out the KE and/or velocity of the nucleons in Uranium 235?
 
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  • #8
jaketodd said:
The KE for the nucleons in an atom were posted here: https://www.physicsforums.com/showpost.php?p=3347668&postcount=3"

So can't we figure out the KE and/or velocity of the nucleons in Uranium 235?

The nucleons are quantum particles bound in a potential ... therefore you cannot associate their kinetic energies with classical velocities. Furthermore, estimating the kinetic energy by subtracting the binding energy from the well-depth is also not likely to be valid unless the potential has the form of a finite square-well to a good approximation (that may be correct .. I don't actually know the answer).
 
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  • #9
SpectraCat said:
The nucleons are quantum particles bound in a potential ... therefore you cannot associate their kinetic energies with classical velocities. Furthermore, estimating the kinetic energy by subtracting the binding energy from the well-depth is also not likely to be valid unless the potential has the form of a finite square-well to a good approximation (that may be correct .. I don't actually know the answer).

Thanks...

So nucleons have zero v? How can that be if they have KE? Is there an equation that tells some sort of v of nucleons given their KE?

Anyone know the answer?

And, does anyone know how much KE there is for the nucleons of Uranium 235 and how I can figure that out for an arbitrary atom?

Also, I know how to get v from phase velocity. So, if anyone knows how to calculate phase velocity of nucleons in an atom, I could get what I need I think.

Thanks,

Jake
 
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  • #10
Hrmm. Could the nucleons be treated similar to electrons in that they have something akin to "orbitals"? I know you can find the average speed of an electron in an orbital, so I could see it in the nucleus I guess. Still kind of weird to think about though.
 
  • #13
Drakkith said:
It doesn't really help me as I don't understand all the math and concepts used.

Neither do I. I just sent an email to the guy who posted that, seeing if he can help with this thread.

Anybody else want to jump in?

Thanks,

Jake
 
  • #14
I found something that might be useful: http://scienceworld.wolfram.com/physics/FermiVelocity.html"

and http://scienceworld.wolfram.com/physics/FermiEnergy.html"

But the equation for the velocity on the first page (the second equation on that page) has a negative number in a square root (when you sub-in the fermi energy defined on the second page), which my calculator doesn't know what to do with.

Help?

Maybe it's just for electrons...?
 
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  • #15
Jake, I just reviewed a few references on the nucleus, and here's my instant description of how fission occurs.

First, do you have much concept of quantum mechanics? There are two ideas you need: the idea of wavefunctions which provide a probabilistic description of what the particles are doing, and the idea of Pauli exclusion keeping particles of the same type apart. You see both of these in the electron shells of an atom. All the electrons are attracted towards the nucleus, but (1) they get quantum mechanically described as probability waves circulating around the nucleus (2) these waveforms are kept apart from each other by the exclusion principle. The result is that the electrons of an atom are described by a set of three-dimensional standing waves, clustered around the nucleus in "shells". A shell consists of a set of standing waves all with similar energy. Because of exclusion, only so many waves can occupy a given shell, then the next shell has to start filling up.

The same thing happens in the nucleus with protons and neutrons, except that here, there's no "nucleus-within-the-nucleus" to act as the central point; they are simply attracting each other (through the strong nuclear force; the positively charged protons repel each other electromagnetically). Nonetheless, there are again shells of waveforms, just as with the electrons further out.

Now in fission what happens is that an extra neutron falls into a nucleus and changes the balance of attraction and repulsion. All those waveforms shift and adjust, the whole nucleus oscillates violently until it takes on a dumbbell shape, and then this splits into two smaller nuclei.

So regarding your questions about the velocity of the nucleons, that would be calculated from the waveforms, it would only be an average velocity, even for the individual neutron or proton, and it would depend roughly on the shell layer they were in.

But I think the bottom line is that a nucleus is a complicated balance between nuclear attraction and electromagnetic repulsion, fission occurs when something tips the balance into instability (like an extra neutron falling into the nucleus - a stray proton would be kept away by electromagnetic repulsion unless it was moving very fast and aimed right at the nucleus, but a neutron, being neutral, doesn't have that issue), and the fission of a nucleus is complicated like the dripping of a tap is complicated. Gravity pulls one part of the drop away from the rest of the water, and this is the result of a tug-of-war between the gravitational force and the intermolecular attraction. The fission of a nucleus is similarly an effect decided by collective dynamics within the nucleus.
 
  • #16
jaketodd said:
An example atom that is fissionable, along with the velocity of the nucleons before the fission, the mass of the nucleons before the fission, the kinetic energy of the nucleons before the fission, and the energy released by the fission minus the kinetic energy due to the opposing charges of the daughter nuclei (in other words, just the energy released due to mass converting to energy, and any binding energy released if that even applies (I am still learning here)).

The nucleons are quantum mechanical, so a given nucleon doesn't have a well-defined velocity. This is an example of the Heisenberg uncertainty principle. Also, different nucleons are in different states, so they have different energies.

It sounds like you're trying to draw a distinction between "energy released due to mass converting to energy" and other forms of energy. E=mc2 applies to all forms of energy. It applies equally to kinetic energy, nuclear potential energy, and electrical potential energy.

-Ben
 
  • #17
Ah ok, I see now Mitchell. Thanks!
 
  • #18
Drakkith said:
What do you mean? The neutrons and protons are all bound together and have the same velocity. Before fission you could say that the velocity it zero.

No, they're all moving around inside the nucleus.

SpectraCat said:
Furthermore, estimating the kinetic energy by subtracting the binding energy from the well-depth is also not likely to be valid unless the potential has the form of a finite square-well to a good approximation (that may be correct .. I don't actually know the answer).

A square well is actually quite a decent approximation to the nuclear potential.

jaketodd said:
The KE for the nucleons in an atom were posted here: https://www.physicsforums.com/showpost.php?p=3347668&postcount=3"

So can't we figure out the KE and/or velocity of the nucleons in Uranium 235?

Yes, that should give you a reasonable estimate. Have you tried cranking out the calculation, or are there things you don't understand about how to carry it out?

jaketodd said:
Does the following post help at all? https://www.physicsforums.com/showpost.php?p=3347704&postcount=4"
This is basically the same thing as what Bill_K suggested, just phrased in fancier language.

Unfortunately, I think what you're going to find if you continue along these lines is that you will get nothing better than an order-of-magnitude estimate of the total mass-energy of the parent and daughter nuclei. Subtracting an order-of-magnitude estimate from an order-of-magnitude estimate to find a small difference is not going to give you a good result. In general, finding the total mass-energy of a nucleus based on a microscopic model of the interacting nucleons is not an easy problem. It can be done in principle if you know a lot of quantum mechanics and have an accurate model of the nuclear interaction, but in reality that's impractical because the calculations are too complicated and we don't have an accurate model of the nuclear interaction. The problem of how to do it in a practical way was not solved until 1968, when Strutinsky invented a trick that is now the standard tool for the purpose: Strutinsky, Nucl. Phys. A122 (1968) 1. A brief recap of the technique is given here: http://arxiv.org/abs/1004.0079

If all you want is a first-principles, ballpark estimate of the energy released in fission, try just calculating the electrical potential energy released. I believe it's quite close to the actual result ("close" meaning to within maybe a factor of 2).

-Ben
 
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  • #19
bcrowell said:
Yes, that should give you a reasonable estimate. Have you tried cranking out the calculation, or are there things you don't understand about how to carry it out?

I don't know how to work with wells. If you could give me the equation to figure out the KE for an arbitrary atom's nucleons, that would be great. Please define all terms. And the nucleons must have a velocity because of the forces bouncing them around, so how to get v would be really the other key I'm looking for. If Uranium 235 (236 with the extra neutron) could be used as an example, I will love you forever. :smile:

Thanks!

Jake
 
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  • #20
jaketodd said:
I don't know how to work with wells. If you could give me the equation to figure out the KE for an arbitrary atom's nucleons, that would be great. Please define all terms. And the nucleons must have a velocity because of the forces bouncing them around, so how to get v would be really the other key I'm looking for. If Uranium 235 (236 with the extra neutron) could be used as an example, I will love you forever. :smile:

I don't have an answer to the question about velocity, because IMO that question doesn't have an answer in the form in which you posed it. Also, I tried to explain why I don't think the method you're trying to use for calculating the energy release will work, and I suggested an alternate method, which is to calculate the electrical energy released. Why don't you go ahead and calculate that first, and then we'll at least be making some progress.

-Ben
 
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  • #21
bcrowell said:
I don't have an answer to the question about velocity, because IMO that question doesn't have an answer in the form in which you posed it. Also, I tried to explain why I don't think the method you're trying to use for calculating the energy release will work, and I suggested an alternate method, which is to calculate the electrical energy released. Why don't you go ahead and calculate that first, and then we'll at least be making some progress.

-Ben

Well I already know how much energy is released in fission of Uranium 235 (236 with the extra neutron). That was easy to find on Wikipedia. What I'm looking for is KE and v of the nucleons before fission. You said "they're all moving around inside the nucleus" earlier, so can you please help me find those two values? If "they're all moving around inside the nucleus," then there must be KE and v associated with that, at least to my humble reasoning.

I am sorry if I am testing your patience.



Jake
 
  • #22
jaketodd said:
Well I already know how much energy is released in fission of Uranium 235 (236 with the extra neutron). That was easy to find on Wikipedia. What I'm looking for is KE and v of the nucleons before fission. You said "they're all moving around inside the nucleus" earlier, so can you please help me find those two values? If "they're all moving around inside the nucleus," then there must be KE and v associated with that, at least to my humble reasoning.

I'm confused. What is your over-all goal? If your only goal was to find the KE and v of the nucleons before fission, then why were you asking about all the other stuff in #3...?

The reason I think there is no answer to your question about KE and v is the reason I gave in #16.
 
  • #23
bcrowell said:
I'm confused. What is your over-all goal? If your only goal was to find the KE and v of the nucleons before fission, then why were you asking about all the other stuff in #3...?

The reason I think there is no answer to your question about KE and v is the reason I gave in #16.

I think I figured it out. I'll post again if I have things I can't figure out.

Thanks!

Jake
 

Related to Fission Details: Atom, Velocity, Mass, Energy

What is fission?

Fission is a nuclear reaction in which the nucleus of an atom splits into smaller fragments, releasing a large amount of energy.

How does fission occur in an atom?

Fission occurs when a neutron collides with the nucleus of a large atom, causing it to become unstable and split into two smaller nuclei.

What factors affect fission velocity?

Fission velocity can be affected by the mass of the atom, the energy of the neutron, and the properties of the nucleus itself.

What is the relationship between mass and energy in fission?

In fission, a small amount of mass is converted into a large amount of energy according to Einstein's famous equation, E=mc².

What are the potential applications of fission?

Fission has been used in nuclear power plants to generate electricity, and it is also used in nuclear weapons. Research is also being done to explore the potential of fission as a source of clean, renewable energy.

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