Energy released by strong force

[granpa]

a chemical reaction might release 10 ev of energy.

a nuclear reaction releases millions of times that.

we are told that that is because the nuclear force is much stronger.

At first that seems to make perfect sense but upon closer inspection the numbers clearly dont work out.

 

The strong force is only 100 times as strong as electromagnetism and it only acts over a very short distance (about the size of the nucleus or 1/10,000 of an angstrom)

 

now energy = force * distance

 

so 100 * (1/10,000) = 0.01

 

even though the force is stronger it should release much less energy.

[swansont]

a chemical reaction might release 10 ev of energy.

a nuclear reaction releases millions of times that.

we are told that that is because the nuclear force is much stronger.

At first that seems to make perfect sense but upon closer inspection the numbers clearly dont work out.

 

The strong force is only 100 times as strong as electromagnetism and it only acts over a very short distance (about the size of the nucleus or 1/10,000 of an angstrom)

 

now energy = force * distance

 

so 100 * (1/10,000) = 0.01

 

even though the force is stronger it should release much less energy.

 

 

The comparison of strength of these interactions is at a common distance. But you omitted something:

 

In chemical reactions the electrons are much further away from the nucleus than the particles within the nucleus are. By about a factor of 10^5, and remember that it's a 1/r^2 dependence. So even in this gross analysis, you lose 10 orders of magnitude by having the particles be that much further away. So upon an even closer inspection, things make sense.

 

 

Here's a deeper dive into the concept

https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-forces-of-nature/the-strength-of-the-known-forces/

[granpa]

The strong force doesn't follow an inverse Square law and is virtually non-existent outside of the nucleus

[Sensei]

It's different than you think..

 

Energy released by decay of unstable particle/isotope is calculated like I showed in my signature.

 

Take for example alpha decay.

Prior decay we have parent isotope with mass m0.

Post decay we have daughter isotope with mass m1.

Mass of alpha particle is mhelium.

 

Energy released during decay is E=(m0-m1-mhelium)*c^2

 

If decay energy is less than 0, spontaneous alpha decay is prohibited!

 

Decay energy is "split" to

daughter isotope as kinetic energy of particle,

alpha particle as kinetic energy of particle,

and optionally to some gamma photons.

 

a chemical reaction might release 10 ev of energy.

a nuclear reaction releases millions of times that.

Decay energy from unstable isotopes rarely exceeds 5 MeV. It's 500,000 times more.

Edited June 28, 2016 by Sensei

[granpa]

https://en.wikipedia.org/wiki/Nuclear_fusion

Energy released in most nuclear reactions is much larger than in chemical reactions, because the binding energy that holds a nucleus together is far greater than the energy that holds electrons to a nucleus. For example, the ionization energy gained by adding an electron to a hydrogen nucleus is 13.6 eV. Less than one-millionth of the 17.6 MeV released in the deuterium-tritium (DT) reaction

The comparison of strength of these interactions is at a common distance. But you omitted something:

 

In chemical reactions the electrons are much further away from the nucleus than the particles within the nucleus are. By about a factor of 10^5, and remember that it's a 1/r^2 dependence. So even in this gross analysis, you lose 10 orders of magnitude by having the particles be that much further away. So upon an even closer inspection, things make sense.

 

 

Here's a deeper dive into the concept

https://profmattstrassler.com/articles-and-posts/particle-physics-basics/the-known-forces-of-nature/the-strength-of-the-known-forces/

So in other words the strong force is actually billions of times stronger than electromagnetism. Which is exactly what I was saying in the op.

 

So if the strong force is that strong then that could explain how the electron fits inside a neutron. What is the bohr radius for an electron orbiting around a proton under that kind of force

Edited June 28, 2016 by granpa

[Sensei]

Calculation of energy released by fusion, is the same as while calculating decay energy.

Sum masses of the all input particles,

sum masses of the all output particles,

Subtract output from input,

multiply them by c^2,

and you have energy released.

[granpa]

Calculation of energy released by fusion, is the same as while calculating decay energy.

Sum masses of the all input particles,

sum masses of the all output particles,

Subtract output from input,

multiply them by c^2,

and you have energy released.

That's true for all reactions. Not just nuclear. Even chemical reactions Edited June 28, 2016 by granpa

[swansont]

So in other words the strong force is actually billions of times stronger than electromagnetism. Which is exactly what I was saying in the op.

You said nothing of the sort in the OP.

 

So if the strong force is that strong then that could explain how the electron fits inside a neutron. What is the bohr radius for an electron orbiting around a proton under that kind of force

The electron does not "fit" inside of a neutron.

[granpa]

What else would I have been saying?

 

In order to know whether it would fit inside a neutron you first have to calculate the Bohr radius

Edited June 28, 2016 by granpa

[swansont]

https://en.wikipedia.org/wiki/Nuclear_fusion

Energy released in most nuclear reactions is much larger than in chemical reactions, because the binding energy that holds a nucleus together is far greater than the energy that holds electrons to a nucleus. For example, the ionization energy gained by adding an electron to a hydrogen nucleus is 13.6 eV. Less than one-millionth of the 17.6 MeV released in the deuterium-tritium (DT) reaction

 

Binding energy doesn't hold the nucleus together, the strong interaction does. The binding energy is a way of characterizing the interaction.

 

The electron is half an Angstrom from the nucleus (Bohr radius). You need to use the force at that distance if you calculate the work. A nucleon is a fermi away from other nucleons in a nucleus. You have to use the force at that distance for nuclear calculations. Remember, the fusion Coulomb barrier is measured in MeV

What else would I have been saying?

 

Beats me. But what you actually said was that the nuclear force should release much less energy than it does, because it's only ~100x stronger than the Coulomb force.

[granpa]

you might want to get your hearing checked someday

 

Angular momentum of electron = (planks constant)/(2pi)

Relativistic angular momentum = γmvr

Relativistic centripetal force = γmv^2/r

 

Gamma*(electron mass)*c*(10^-14 m)=(planks constant)/(2pi) solve for x

Wolfram says gamma = 38.6

 

38.6*(electron mass)*(velocity of light)^2/(10^-14 m)

Wolfram says force = 316 newtons

 

The force between 2 electrons at that distane is

( Coulomb's constant )*(electron charge)^2/(10^-14 m)^2

Wolfram says 2.3 newtons

 

According to those equations the force is 137 times stronger than electromagnetism would be at that distance (exactly as swansont said above) and that is sufficient for the electron to fit inside a neutron

Edited June 28, 2016 by granpa

[Strange]

Electrons don't "fit inside neutrons".

[granpa]

I just showed that they do

You can't argue with math

[Strange]

Do you have any evidence of electrons inside neutrons? After all, science is based on evidence...

[granpa]

https://en.wikipedia.org/wiki/Neutron#Structure_and_geometry_of_charge_distribution

 

Structure and geometry of charge distribution

An article published in 2007 featuring a model-independent analysis concluded that the neutron has a negatively charged exterior, a positively charged middle, and a negative core.^%5B67%5D In a simplified classical view, the negative "skin" of the neutron assists it to be attracted to the protons with which it interacts in the nucleus. (However, the main attraction between neutrons and protons is via the nuclear force, which does not involve charge.)

The simplified classical view of the neutron's charge distribution also "explains" the fact that the neutron magnetic dipole points in the opposite direction from its spin angular momentum vector (as compared to the proton). This gives the neutron, in effect, a magnetic moment which resembles a negatively charged particle. This can be reconciled classically with a neutral neutron composed of a charge distribution in which the negative sub-parts of the neutron have a larger average radius of distribution, and therefore contribute more to the particle's magnetic dipole moment, than do the positive parts that are, on average, nearer the core.

 

https://www.google.com/search?q=neutron+charge+distribution

Edited June 29, 2016 by granpa

[Mordred]

Great on the first link. What does that link state is the composition of the neutron.? I certainly dont see an electron on the composition list.

 

Now add up the charges of the quarks that make up the neutrons composition

[granpa]

according to this site http://www.sjsu.edu/faculty/watkins/relamomentum.html

 

Relativistic angular momentum = γ^3mvr

if so then gamma = 3.38

 

3.38*(electron mass)*(velocity of light)^2/(10^-14 m)
Wolfram says force = 27.6 newtons

and the force is 12 times stronger than electromagnetism would be at that distance

Edited June 30, 2016 by granpa

[swansont]

I just showed that they do

You can't argue with math

Do the math of the Heisenberg uncertainty principle.

https://en.wikipedia.org/wiki/Neutron#Structure_and_geometry_of_charge_distribution

 

Structure and geometry of charge distribution

An article published in 2007 featuring a model-independent analysis concluded that the neutron has a negatively charged exterior, a positively charged middle, and a negative core.^%5B67%5D In a simplified classical view, the negative "skin" of the neutron assists it to be attracted to the protons with which it interacts in the nucleus. (However, the main attraction between neutrons and protons is via the nuclear force, which does not involve charge.)

The simplified classical view of the neutron's charge distribution also "explains" the fact that the neutron magnetic dipole points in the opposite direction from its spin angular momentum vector (as compared to the proton). This gives the neutron, in effect, a magnetic moment which resembles a negatively charged particle. This can be reconciled classically with a neutral neutron composed of a charge distribution in which the negative sub-parts of the neutron have a larger average radius of distribution, and therefore contribute more to the particle's magnetic dipole moment, than do the positive parts that are, on average, nearer the core.

 

https://www.google.com/search?q=neutron+charge+distribution

So what's in a neutron that has a positive charge to make the particle neutral? How does an electron inside give a surface charge?

[granpa]

1) The model I used already accounts for the Heisenberg uncertainty principle.

 

2) proton

Edited June 30, 2016 by granpa

[imatfaal]

!

Moderator Note

 

moved to speculations

 

Take a moment to read the guidelines to posting and rules for that sub-forum

 

[swansont]

1) The model I used already accounts for the Heisenberg uncertainty principle.

No, it doesn't. Do the calculation. Confine an electron to 10^-15m. What doe the momentum uncertainty say about its speed and/or KE?

 

2) proton

The proton, neutron and electron are all spin 1/2. How do you combine two spin 1/2 particles and still end up with spin 1/2?

[granpa]

the electron is moving very fast

by adding a neutrino of course

[swansont]

the electron is moving very fast

Yes, indeed. So fast that it can't be confined inside of a neutron. (That doesn't address the spin problem, though)

[granpa]

Well I'm not going to keep repeating myself but I will say this the centrifugal force follows an inverse Cube law until the particle reaches relativistic speed at which point it then starts following an inverse Square law so at that point the centrifugal force increases exactly the same rate as the electrostatic force would increase

 

It follows an inverse Cube law if angular momentum is conserved that is

[swansont]

Well I'm not going to keep repeating myself but I will say this the centrifugal force follows an inverse Cube law until the particle reaches relativistic speed at which point it then starts following an inverse Square law so at that point the centrifugal force increases exactly the same rate as the electrostatic force would increase

 

It follows an inverse Cube law if angular momentum is conserved that is

 

 

What centrifugal force? This is the first time you've mentioned it. What is the source of this force, and why is it inverse cube? Why does it change?