Finding the distance between two nuclei right after fission

[guyvsdcsniper]

Homework Statement

Using energy considerations, calculate the distance between centers of the palladium nuclei just after fission,when they are momentarily at rest.

Relevant Equations

E=K+Eo

I am trying to solve part b of this problem. I know the particles are moving at first and are at rest right after fission per the question.

Using the relativistic Kinetic Energy equation and the energy between two charges I believe I can solve for r, the radius between the two charges.

I am a bit lost on what the V should be on the kinetic energy side. Would it be the answer I got from A? I don't think that makes much sense since I derived the V w/o the relativistic Kinetic equation because the speed was very low.

 

[haruspex]

I hesitated to respond because there are clearly things I do not understand about such processes, but maybe you can fill in those gaps.

What happened to all the electrons that would have surrounded the U235? Why don’t those swiftly cancel the electrostatic repulsion?

Why would the nuclei be momentarily at rest? Presumably the lost mass will have turned into KE, there being no other decay products, and the electrostatic repulsion will accelerate their separation. I can't see that gravitational attraction could be relevant.

If you solved a, please post your reasoning and working.

 

[dRic2]

What happened to all the electrons that would have surrounded the U235? Why don’t those swiftly cancel the electrostatic repulsion?

The splitting of the nucleus happens inside the nucleus. The electrons observe the whole process from the "outside" so if we assume a symmetric charge distribution they can not contribute to the electrostatic interactions between the newborn fragments (Gauss law). Even if the electronic charge was not symmetric the resulting effect could not be to shield the interaction between nuclei, but rather to influence the decay angle I guess (because nuclei would feel a stronger attraction along some direction).

Why would the nuclei be momentarily at rest?

Because (I think) the exercise asks to imagine that the nucleus of U235 is replaced at $t_0$ by two nuclei of Pd, created at some distance apart from each other, but at rest (at $t_0$). Once they are created they start to get accelerated.

Ps. Since the nuclei are supposed to be created at rest, I would simply equate the variation of rest mass with the Coulomb potential energy to find the distance.

 

[haruspex]

The splitting of the nucleus happens inside the nucleus.

I thought of that, but it says "a long way apart".

equate the variation of rest mass with the Coulomb potential energy to find the distance.

Why equate those? It is electrostatic repulsion, not attraction. They should accelerate apart until the electron fog comes into play.

 

[guyvsdcsniper]

I hesitated to respond because there are clearly things I do not understand about such processes, but maybe you can fill in those gaps.

What happened to all the electrons that would have surrounded the U235? Why don’t those swiftly cancel the electrostatic repulsion?

Why would the nuclei be momentarily at rest? Presumably the lost mass will have turned into KE, there being no other decay products, and the electrostatic repulsion will accelerate their separation. I can't see that gravitational attraction could be relevant.

If you solved a, please post your reasoning and working.

As far as the context behind the problem I also do not know too much. I am 2 weeks into this into to modern physics class and we have only talked about special relativity.

These are the results I came up with. A friend helped me with b.

I guess for b, The total energy is the U₂₃₆ and that is equal to the coulomb potential energy plus the rest energy for both Pd atoms. Then its just solving for r.

Our book gives us the equation E= K+E_o where E_o is the rest energy. There is no kinetic energy since the question states both Pd are at rest.

I can see why the Coulomb potential energy is included to the equation, I am just a bit thrown off by it. I guess its also considered a rest energy in the context of this problem?

 

[haruspex]

The total energy is the U236 and that is equal to the coulomb potential energy plus the rest energy for both Pd atoms.

As I wrote in post #4, that doesn’t actually make sense.
The fission converted some rest mass into KE. What happens to that KE? It is not converted into coulomb potential energy as the nuclei fly apart, since the coulomb potential energy is reducing, not increasing. Rather, coulomb potential energy will get converted into yet more KE. This will continue at least until the nuclei reach the U236's atomic radius. What happens then to the electron fog is, well, foggy.

 

[guyvsdcsniper]

As I wrote in post #4, that doesn’t actually make sense.
The fission converted some rest mass into KE. What happens to that KE? It is not converted into coulomb potential energy as the nuclei fly apart, since the coulomb potential energy is reducing, not increasing. Rather, coulomb potential energy will get converted into yet more KE. This will continue at least until the nuclei reach the U236's atomic radius. What happens then to the electron fog is, well, foggy.

Sorry I didint see your post. I guess you do make a valid point though. You have me thinking about this problem a little deeper. I will work on it, talk to my classmate, and post what I think.

 

[PeroK]

As I wrote in post #4, that doesn’t actually make sense.
The fission converted some rest mass into KE. What happens to that KE? It is not converted into coulomb potential energy as the nuclei fly apart, since the coulomb potential energy is reducing, not increasing. Rather, coulomb potential energy will get converted into yet more KE. This will continue at least until the nuclei reach the U236's atomic radius. What happens then to the electron fog is, well, foggy.

I admit I don't know how to do the detailed calculations for a problem like this either.

It also begs the question about the role of the strong nuclear force that keep the nucleus together in the first place. See, for example:

https://physics.stackexchange.com/q...rong-force-get-dominated-by-the-electrostatic

 

[haruspex]

I admit I don't know how to do the detailed calculations for a problem like this either.

It also begs the question about the role of the strong nuclear force that keep the nucleus together in the first place. See, for example:

https://physics.stackexchange.com/q...rong-force-get-dominated-by-the-electrostatic

Presumably the fission occurs because the electromagnetic repulsion overcomes the residual strong force. But that force declines very rapidly thereafter, being negligible after a few proton widths, so might not sap that much of the electric potential energy available.
If we crudely model the Pd proto-nuclei as hemispheres (and the parent U236 as a sphere of radius equal to 12 proton radii), the residual strong force has pretty much gone by the time they are a few proton radii apart. Their mass centres are 4.5 proton radii from the U236's centre, so at most half of the electrostatic potential is used to dissociate them.

 

[PeroK]

Presumably the fission occurs because the electromagnetic repulsion overcomes the residual strong force. But that force declines very rapidly thereafter, being negligible after a few proton widths, so might not sap that much of the electric potential energy available.

The reasons for the particular nuclear half-lifes and stability are buried in the detailed QCD calculations somewhere. Ultimately, we need the quark model to explain this. Thankfully, we don't need the reason for decay to do the basic calculation.

If we crudely model the Pd proto-nuclei as hemispheres (and the parent U236 as a sphere of radius equal to 12 proton radii), the residual strong force has pretty much gone by the time they are a few proton radii apart. Their mass centres are 4.5 proton radii from the U236's centre, so at most half of the electrostatic potential is used to dissociate them.

The basic model, ignoring the EM interaction and the electrons is that a mass $M$ splits into two equal masses $m$, where $M > 2m$. The masses $m$ have equal and opposite momenta, which can be determined from the energy-momentum (four-vector) conservation. As can the KE, of course.

Technically, the system is only two separate masses when the EM interaction is negligible (this is what large distance should mean). That allows us to calculate the eventual kinetic energy once the fission has completed and the new nuclei are far apart.

What we could do crudely is then apply a classical EM potential on top of this. That would give us the particles' KE at any distance from each other. That should work quite well (and only break down at relatively short distances where the strong force is involved).

As the EM potential energy tends to infinity, we could (I see this now) imagine an initial state where the nuclei are at rest a short distance apart (with classical masses $m$), no KE, but an EM PE. And, then the EM force pushes them apart until they reach their final KE far apart. This must be well beyond the radius of the original atom and its electron shells.

This is all nonsense, of course, but it does allow a calculation to be done.

PS Part D: do an accurate scale drawing of a nuclear fission in progress! That's an almost fanatical adherence to classical determinism that the student will have to unlearn at some point. The question ought to be honest and admit this is a crude classical model of nuclear fission.

 

[dRic2]

I thought of that, but it says "a long way apart".
Why equate those? It is electrostatic repulsion, not attraction. They should accelerate apart until the electron fog comes into play.

I think the physical model assumed in the exercise is not clear. In Krane's "Introductory Nuclear Physics" this example is actually discussed. Begin at the end of page 479 and figure 13.1 (see attachment)

As the author says it is a very crude model and the numbers are "not to be taken too seriously".

 

[PeroK]

As far as the context behind the problem I also do not know too much. I am 2 weeks into this into to modern physics class and we have only talked about special relativity.

These are the results I came up with. A friend helped me with b.

I guess for b, The total energy is the U₂₃₆ and that is equal to the coulomb potential energy plus the rest energy for both Pd atoms. Then its just solving for r.

Our book gives us the equation E= K+E_o where E_o is the rest energy. There is no kinetic energy since the question states both Pd are at rest.

I can see why the Coulomb potential energy is included to the equation, I am just a bit thrown off by it. I guess its also considered a rest energy in the context of this problem?

View attachment 296263

I agree with these numbers, so far as I can read them.

 

[Steve4Physics]

Hi @quittingthecult. Here are a few passing thoughts which overlap and hopefully add to what has already been said.

You are being asked to treat the system as two palladium nuclei created stationary with all the released energy initially stored as electrostatic potential energy. After a long time this energy is turned entirely into their kinetic energy due to the repulsive electrostatic force between them.

Your working looks OK but I haven't checked arithmetic. However, for neatness, I would define $\Delta E = (m_{236} – 2m_{118})c^2$ (the energy released) and use $\Delta E$ in the various equations.

There is some confusion about what level of precision is required. You are told to work to at least 6 significant figures but then told to use 1u = 1.66x10⁻²⁷kg whereas 1u=1.66053886 × 10⁻²⁷kg. And you are given an approximate formula for nuclear radius. Also note that if using ‘at least 6 significant figures’, c= 2.99792458 x10⁸m/s.

I think you only really need ‘at least 6 significant figures’ for $m_{236} – 2m_{118}$ because you are evaluating the difference between two similar values.

During fission, relevant inter-particle distances are ~10⁻¹⁵m. Since the atomic electrons are effectively at distances of ~10⁻¹⁰m, their effects are negligible. Similarly, ionisation energies of atomic electrons are negligible compared to energy released by fission. So atomic electrons can be ignored in these sorts of calculations.

Gravitational forces/energies are, as already noted, way too small to have any effect.

 

[Hall]

The part (a) is asking to assume all kinetic energy came from fission. And then part (b) is implying to equate it to the work done by Coulombic repulsion.

$$\int_{r}^{\infty} \frac{1}{4\pi \epsilon_{0}} (46e)^2 \frac{dx}{x^2} = \Delta E$$

I don't think I have proper list of adjectives with me to designate it one. But anyways, mass defect is responsible for kinetic energies of Palladium nuclei and electrical repulsion was responsible for the split.