Electric field, electric charge, electron and positron.

[Mr Skeptic]

Ok, but there is a problem. The definition is circular. Electron volt, the amount of energy an electron gains falling through 1 volt will depend on electron mass, hence this definition is unsuitable to be defining "mass" as it requires 'mass' to already be defined.

 

The electron's mass has nothing to do with an electron-volt. You can use any particle with a unit of charge for that.

 

Similar thing you can see with this equation for 'massless particles':

 

p= E/c

 

but...

 

E= m*c*c

 

...so, again:

 

p = m*c -> p = m*v

 

 

Without mass energy is zero, without mass momentum is zero.

 

Thus, if photons have energy and momentum they must have mass.

 

They have relativistic mass (also known as energy) but have no rest mass.

 

The equation for energy in terms of rest mass and momentum is,

[math]E^2 = m_0^2c^4 + p^2c^2[/math]

Obviously, [math]m_0 \neq m[/math] except at zero velocity. The advantage of using this formula is that it makes explicit which portion of the energy is due to the rest mass and which portion is kinetic energy.

 

Unfortunately, from all this it is still kind of unclear if 'mass' is a real property, do gravity fields really exist, or perhaps gravity force is just some side-effect due to motion of charges, effect of superposition of electric and magnetic fields and their kinematics/dynamics, their kinetic energy. But, anyhow, the real question is this: - can electrons traveling a straight line with the same velocity still have different energies due to some vibration, spin or something?

 

Well they could have different potential energy (eg one could be closer to a proton than the other), but if you mean other than that, then no.

[Sha31]

The electron's mass has nothing to do with an electron-volt. You can use any particle with a unit of charge for that.

 

Did you say:

- "An electron-volt is the amount of energy an electron gains.."?

 

To find out that amount of kinetic energy (acceleration and velocity difference) you have to know the 'mass' to start with' date=' because of: F=ma. Beside that, this acceleration will not be uniform and the result will contain error, it's approximation, so this is really terrible definition of mass (energy), it's self-referencing (circular) and it's non-exact.

 

 

They have relativistic mass (also known as energy) but have no rest mass.

 

The equation for energy in terms of rest mass and momentum is,

[math]E^2 = m_0^2c^4 + p^2c^2[/math]

Obviously, [math]m_0 \neq m[/math] except at zero velocity. The advantage of using this formula is that it makes explicit which portion of the energy is due to the rest mass and which portion is kinetic energy.

 

Some terms, like 'relativistic mass' are defined differently here:

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

 

But, that still solves nothing, it's still circular. Define "p"? Define "m0"?

 

 

How did we ever measure and conclude what is electron mass? Can someone point to a couple of different experiments that confirm these numbers for electron mass?

 

 

Well they could have different potential energy (eg one could be closer to a proton than the other), but if you mean other than that, then no.

 

Ok. From electron microscopy we know there are high-energy and low-energy electrons. It is the conclusion then, this energy difference is solely due to electron velocity? But, how is it possible then to make electrons move slower than the speed of light and then even slow them down some more? What is the velocity of an electron with energy of 1 eV and with 30 keV?

[swansont]

Did you say:

- "An electron-volt is the amount of energy an electron gains.."?

 

To find out that amount of kinetic energy (acceleration and velocity difference) you have to know the 'mass' to start with, because of: F=ma. Beside that, this acceleration will not be uniform and the result will contain error, it's approximation, so this is really terrible definition of mass (energy), it's self-referencing (circular) and it's non-exact.

 

electron-Volt works for any charge. KE = qV It measures the amount of work a field does on the charge. If a fundamental charge has moved through a 1 Volt potential difference, it will have gained 1 eV of energy. This is exact, because it's by definition. F=ma doesn't really enter into it, because if the charge has not accelerated to the correct speed, then it will not have gone through a 1 V potential difference. This is trivially derived from the definitions of potential difference and electric field.

 

It might be worth considering taking a physics class or two before tackling particle structure problems. Physics builds upon a foundation and intermediate levels, and if you are struggling with or lacking in the freshman-physics concepts, the advanced ideas are going to be pretty much impossible.

Edited December 12, 2009 by swansont
typo

[Sha31]

electron-Volt works for any charge. KE = qV

 

That definition is specifically using electron to define this particular amount of energy' date=' hence the name. Do you think you can take a proton and it will gain the same amount of energy as electron?

 

 

It measures the amount of work a field does on the charge.

 

So, we have some distance and we have some force, it also means we have some delta-time, right? What exactly is the equation you suggest here to calculate energy, work, acceleration, velocity and distance from this, without using mass?

 

 

If a fundamental charge has moved through a 1 Volt potential difference, it will have gained 1 eV of energy.

 

Yes, fundamental charge - "electron", not any charge. Gained *kinetic* energy, not any energy, since mass is constant that can mean only one thing - gained velocity. So, how exactly do you calculate this difference in energy independent of mass variable?

 

 

This is exact, because it's by definition. F=ma doesn't really enter into it, because of the charge has not accelerated to the correct speed, then it will not have gone through a 1 V potential difference. This is trivially derived from the definitions of potential difference and electric field.

 

I say it is not exact and that it very much depends on mass and F=ma. There is a circular logic for confirmation in your reasoning again, so it wold explain a lot, and it would be the most helpful, if you could just print down those trivially derived equations you're talking about?

 

 

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

- "The electron volt is not an SI unit and its value must be obtained experimentally."

[swansont]

That definition is specifically using electron to define this particular amount of energy, hence the name. Do you think you can take a proton and it will gain the same amount of energy as electron?

 

Yes.

 

So, we have some distance and we have some force, it also means we have some delta-time, right? What exactly is the equation you suggest here to calculate energy, work, acceleration, velocity and distance from this, without using mass?

 

You're moving the goalposts. I said you don't need the mass to calculate the energy. I did not say you could also calculate the other terms you have listed.

 

Yes, fundamental charge - "electron", not any charge. Gained *kinetic* energy, not any energy, since mass is constant that can mean only one thing - gained velocity. So, how exactly do you calculate this difference in energy independent of mass variable?

 

No, any charge. As we had previously discussed, the charge on a proton has the same size, as is true of many other nuclear particles. Electrons are easily obtained, though, and when the term was coined, many of those other particles were undiscovered.

 

 

I say it is not exact and that it very much depends on mass and F=ma. There is a circular logic for confirmation in your reasoning again, so it wold explain a lot, and it would be the most helpful, if you could just print down those trivially derived equations you're talking about?

 

As I said, anyone who has passed freshman physics should be able to do this:

 

If we have a constant electric field E, the potential across a distance d is Ed, i.e. V = Ed

 

E is, by definition, the force per unit charge, so the force on any charge is qE, and this is a constant, since the field is constant. W = KE = Fd for a constant force.

 

thus, KE = qV

 

(if the field is not constant you have to do a more general approach using calculus, but you get the same answer)

 

No mention of mass. Why is it not necessary? Because for a given force, a larger mass means the acceleration is smaller and it takes a longer time, and the effects cancel. And potential difference is force per unit charge, so any identical charge will feel the same amount of force.

 

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

- "The electron volt is not an SI unit and its value must be obtained experimentally."

 

That's because the fundamental charge is an experimentally determined quantity. The definition, i.e. the equation, is still exact.

[Sha31]

You're moving the goalposts. I said you don't need the mass to calculate the energy. I did not say you could also calculate the other terms you have listed.

 

The point is not to skip any derivation steps' date=' so to see if the definition is self referencing, if it is circular. When you say "work", that automatically involves force and distance, which automatically involves acceleration and velocity. Velocity is the variable we are after.

 

And while there are shortcuts when stuff gets canceled and some things end up to be equal, for the point I'm making we have to realize what came first, what means what, what is a shortcut, assumption or derivation and what is actual definition.

 

 

Circular definition and self-reference:

 

"chicken": thing that comes out an 'egg'

 

"egg": thing from which 'chicken' comes out

 

 

 

 

So, what came first, work or velocity?

 

[img']http://upload.wikimedia.org/math/9/a/e/9aeac7ca01e03ffd4b80c513dbeb1b6a.png[/img]

 

 

 

 

No mention of mass. Why is it not necessary? Because for a given force, a larger mass means the acceleration is smaller and it takes a longer time, and the effects cancel.

 

Ok, distance is constant, force is constant, so work done is the same. But, how can objects have the same velocity after traveling the same distance if acceleration was different?

 

 

In any case, my point is that it makes no sense to say electron mass is 0.511MeV if "mass" itself is involved in the definition of electron volt. The other thing is that mass is NOT energy, mass*velocity is, so how can unit of energy replace unit of mass when it involves one more variable? It's as if some velocity is assumed or suggested in the same time with this "mass".

 

 

That's because the fundamental charge is an experimentally determined quantity. The definition, i.e. the equation, is still exact.

 

Why not define mass as amount of work done by gravity force on an object free falling through the distance of 1 meter?

Edited December 12, 2009 by Sha31

[Mr Skeptic]

The point is not to skip any derivation steps, so to see if the definition is self referencing, if it is circular. When you say "work", that automatically involves force and distance, which automatically involves acceleration and velocity. Velocity is the variable we are after.

 

Why are you interested in the velocity?

 

In any case, you can get the velocity by (for non-relativistic velocities) taking [math]v=\sqrt{2 \cdot KE/m}[/math], where KE is the kinetic energy, and is equal to the work gained by an electric charge falling through 1 volt. Calculate this for 1 eV and the mass of an electron, and you get the electron's velocity at 1 eV. Calculate the same for a proton, and you get the proton's velocity at 1 eV. The proton will be moving slower.

 

So, what came first, work or velocity?

 

 

 

It doesn't matter. If you know the force and distance, you calculate work that way. If you know the mass and change in velocity, you calculate it the second way.

 

Why not define mass as amount of work done by gravity force on an object free falling through the distance of 1 meter?

 

Because that is not specific. You are not specifying a gravitational force, so the energy could be anything.

[Sha31]

Why are you interested in the velocity?

 

In any case' date=' you can get the velocity by (for non-relativistic velocities) taking [math']v=\sqrt{2 \cdot KE/m}[/math], where KE is the kinetic energy, and is equal to the work gained by an electric charge falling through 1 volt. Calculate this for 1 eV and the mass of an electron, and you get the electron's velocity at 1 eV. Calculate the same for a proton, and you get the proton's velocity at 1 eV. The proton will be moving slower.

 

That is exactly my point. I'm interested in velocity because that is one of only two variables defining kinetic energy, second being mass. Yes, the proton will be moving slower and for equation to work, for energy to come up the same, we have this larger mass so the product can give the same result: mass*velocity. And there it is 'mass' right there in the definition of electron volt, as one of the two properties defining 'kinetic energy'. The other thing is how can eV = kg? How can unit for energy substitute unite of mass, what happens to velocity?

 

 

It doesn't matter. If you know the force and distance, you calculate work that way. If you know the mass and change in velocity, you calculate it the second way.

 

Yes, but work is imaginary concept, some useful relation, while distance, force and velocity are true properties of the real world. Mass is also supposed to be one of these true and measurable properties, like charge, which is what I kind of question here.

 

 

Because that is not specific. You are not specifying a gravitational force, so the energy could be anything.

 

But, charge in coulombs is not specific too, and is experimentally determined, so naturally, the mass was involved in the measurements of these forces via F=ma. Don't you see something is strange about defining gravity force (mass) with electric force? It's like defining 'charge' with free fall.

Edited December 13, 2009 by Sha31

[Mr Skeptic]

Well, you are doing it backwards. You calculate the kinetic energy from the work done on the electron (as a charged particle, the mass never figures into the equation). Then you calculate the velocity it will be at to have that kinetic energy, since you know its mass. Why? Because it is easier that way.

 

An electron volt is not used to measure the mass of an electron. It is a unit of energy with which the mass of an electron can be expressed in, just like meters, feet, and cubits are used to measure distance.

 

An electron has a mass of 9.10938215×10⁻³¹ kg or 8.18710414 × 10^-14 joules or 510,998.903 electron volts. As you can see, an electron volt is the only of these units that is not obscenely too large to measure the mass of an electron. Actually mega electron volt is the one that is exactly right for particle masses.

[Sha31]

Well' date=' you are doing it backwards. You calculate the kinetic energy from the work done on the electron (as a charged particle, the mass never figures into the equation). Then you calculate the velocity it will be at to have that kinetic energy, since you know its mass. Why? Because it is easier that way.

[/quote']

 

Maybe I wasn't clear. Yes, I agree definition works for any charge, but I don't think it correctly describes gravity force and that is a proper definition for mass. -- Electron volt is defined as a difference in 'kinetic energy', and kinetic energy is a product of two physical properties, mass and velocity. It just so happens some imaginary relation named "work" is defined as exactly this energy difference, but that in no way avoids the fact this energy is actually "made of" mass, so 'mass' ends up being defined by using concept of "mass". There is no definition, or unit, nothing like this in the whole physics, as far as I know, it's just wrong, like chicken and egg.

 

 

An electron volt is not used to measure the mass of an electron. It is a unit of energy with which the mass of an electron can be expressed in, just like meters, feet, and cubits are used to measure distance.

 

 

Ok, after reading that article linked below, I think I can answer that question now - eV actually gets divided by c^2. So, it's just a shortcut to just write "eV" instead of "eV/c^2". Funny. Well, this surely answers one of my questions, but I'm afraid it opens a few new ones, like - just how slow electrons can go and what happens with all that mass increase stuff as they close to the speed of light, do they kind of start to look like muons with their newly gained mass?

 

 

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

- "The electronvolt (eV) is primarily a unit of energy, but because of the mass-energy equivalence it can also function as a unit of mass. In this context it is denoted eV/c^2, or simply as eV."

 

 

An electron has a mass of 9.10938215×10⁻³¹ kg or 8.18710414 × 10^-14 joules or 510,998.903 electron volts. As you can see, an electron volt is the only of these units that is not obscenely too large to measure the mass of an electron. Actually mega electron volt is the one that is exactly right for particle masses.

 

Ok. Can you tell what is the velocity of an electron with energy of 1eV, 30keV and 250MeV?

[swansont]

Ok, distance is constant, force is constant, so work done is the same. But, how can objects have the same velocity after traveling the same distance if acceleration was different?

 

They don't have the same velocity. Nobody said they had the same velocity. They have the same energy. an electron, or a proton, or any other particle with a fundamental charge, will gain 1 eV of energy when the potential difference is 1 Volt.

 

In any case, my point is that it makes no sense to say electron mass is 0.511MeV if "mass" itself is involved in the definition of electron volt. The other thing is that mass is NOT energy, mass*velocity is, so how can unit of energy replace unit of mass when it involves one more variable? It's as if some velocity is assumed or suggested in the same time with this "mass".

 

No, energy is not mass*velocity, that's momentum. E=mc^2 The electron mass is 0.511 MeV, if you express it in terms of energy (i.e. you multiply by c^2). That's a common shortcut/notation in physics.

 

These are all commonly used, defined terms. You don't get to redefine them on a whim.

[Sha31]

They don't have the same velocity. Nobody said they had the same velocity. They have the same energy. an electron' date=' or a proton, or any other particle with a fundamental charge, will gain 1 eV of energy when the potential difference is 1 Volt.

[/quote']

 

Ok. Next question. Can you explain how can electrons be slowed down and just how slow can they go? I mean without jumping from atom to atom, but moving in a beam through air or vacuum. Also, what happens with special relativity and all that mass increase stuff as electrons close to the speed of light, do they kind of start to look like muons due to their newly gained mass?

 

 

No, energy is not mass*velocity, that's momentum. E=mc^2 The electron mass is 0.511 MeV, if you express it in terms of energy (i.e. you multiply by c^2). That's a common shortcut/notation in physics.

 

These are all commonly used, defined terms. You don't get to redefine them on a whim.

 

I was talking about kinetic energy since that is how eV is defined. Anyway, what is the velocity of an electron with energy of 1eV, 30keV and 250MeV?

 

E = m/2 * v^2

30keV = m/2 * v^2

30,000 = 510,998/2 * v^2

v= sqrt(30,000 / 255,499) = 0.342662 ??

 

 

That's wrong obviously, and what happens to units?

Can someone do it properly?

[Klaynos]

How the charge of an electron was first measured:

 

http://en.wikipedia.org/wiki/Oil-drop_experiment

 

You might wish to read a bit more about that.

 

In an electrons rest frame it is moving at 0 velocity, this all becomes a bit complicated as we need to take into account quantum mechanics and we need to talk about electrons as wave-particles which don't really have a well defined position or velocity.

 

In your maths to find the velocity, write down the full units at each step (mass in eV is actually eV/c^2), I don't have the time right now to work out whether you've done it right but if you put in all the units it's far easier for someone to just scan it to tell if it's correct. Also remember that if the velocity is relativistic you need to use the relativistic equations, not Gallilain.

[swansont]

Ok. Next question. Can you explain how can electrons be slowed down and just how slow can they go? I mean without jumping from atom to atom, but moving in a beam through air or vacuum. Also, what happens with special relativity and all that mass increase stuff as electrons close to the speed of light, do they kind of start to look like muons due to their newly gained mass?

 

An individual electron can have an arbitrarily small speed. With multiple electrons you have interactions, so that's more complicated.

 

 

We'll stick with rest mass. No need to complicate things unnecessarily. The rest mass of the electron doesn't increase.

 

I was talking about kinetic energy since that is how eV is defined.

 

An electron-Volt is not mass times speed. It has units of energy (Coulombs * Volts, and a Volt is a Joule/Coulomb, so it has units of Joules, as it must)

 

 

Anyway, what is the velocity of an electron with energy of 1eV, 30keV and 250MeV?

 

E = m/2 * v^2

30keV = m/2 * v^2

30,000 = 510,998/2 * v^2

v= sqrt(30,000 / 255,499) = 0.342662 ??

 

 

That's wrong obviously, and what happens to units?

Can someone do it properly?

 

You used mass as .511 MeV/c^2, so your answer is a fraction of the speed of light. A third of the speed of light. But as Klaynos has already cautioned, you'd need to use the relativistic formula for higher energies, and this one probably has a few percent error.

[Sha31]

How the charge of an electron was first measured:

 

http://en.wikipedia.org/wiki/Oil-drop_experiment

 

You might wish to read a bit more about that.

 

In an electrons rest frame it is moving at 0 velocity, this all becomes a bit complicated as we need to take into account quantum mechanics and we need to talk about electrons as wave-particles which don't really have a well defined position or velocity.

 

In your maths to find the velocity, write down the full units at each step (mass in eV is actually eV/c^2), I don't have the time right now to work out whether you've done it right but if you put in all the units it's far easier for someone to just scan it to tell if it's correct. Also remember that if the velocity is relativistic you need to use the relativistic equations, not Gallilain.

 

 

E = 1/2 * m * v^2

30,000[eV] = 0.5 * 510,998/299,792,458^2[eV/m/s] * v^2[m/s]

v = sqrt(30,000/2.84281e-12)[m/s]

= 3,248,528 m/s

 

 

1 eV, v= 593,097 m/s

 

30 keV, v= 3,248,528 m/s

 

250 MeV, v= 93,835,378,660 m/s

 

 

Ok, this looks better, but 250 MeV is out of range, so what are the minimum/maximum electron energies we can produce/measure?

---

Merged post follows:

Consecutive posts merged

An individual electron can have an arbitrarily small speed.

 

This sounds very strange to me. For some reason, I simply can not imagine electrons move at any slow speed, like a meter per second or slower? What is the process and technology to slow down electrons? What words to google?

[Mr Skeptic]

Ok, this looks better, but 250 MeV is out of range, so what are the minimum/maximum electron energies we can produce/measure?

 

What do you mean out of range? You need to use the relativistic equation for kinetic energy if you want to go at such high energies.

[Sha31]

What do you mean out of range? You need to use the relativistic equation for kinetic energy if you want to go at such high energies.

 

 

This one? Ok, what is velocity equal to?

[Mr Skeptic]

 

This one? Ok, what is velocity equal to?

 

Well just solve the equation and plug the numbers in.

[math]v=c \sqrt{1 - \frac{1}{(E_k/m_oc^2 +1)^2}}[/math]

 

So for 250 MeV of kinetic energy, an electron would be moving at 299,791,834 m/s or 0.99999792 c

calculation

 

The same equation will work for the other speeds, just plug them into the equation I linked to.

Edited December 14, 2009 by Mr Skeptic

[Sha31]

Well just solve the equation and plug the numbers in.

[math]v=c \sqrt{1 - \frac{1}{(E_k/m_oc^2 +1)^2}}[/math]

 

So for 250 MeV of kinetic energy' date=' an electron would be moving at 299,791,834 m/s or 0.99999792 c

calculation

 

The same equation will work for the other speeds, just plug them into the equation I linked to.

 

250 MeV [93,835,378,660 m/s]

v=299,792,458 * sqrt( 1 - 1/ (250,000,000/510,998 +1)^2 )

= 299,792,458 * sqrt( 1 - 1/ 240,333.9883 )

= 299,791,834 m/s

 

 

30 keV [3,248,528 m/s]

v=299,792,458 * sqrt( 1 - 1/ (30,000/510,998 +1)^2 )

= 299,792,458 * sqrt( 1 - 1/ 1.120864 )

= 98,444,785 m/s

 

 

1 eV [593,097 m/s]

v=299,792,458 * sqrt( 1 - 1/ (1/510,998 +1)^2 )

= 299,792,458 * sqrt( 1 - 1/ 1.000003914 )

= 593,096 m/s

 

I got different result for 30 keV, but the same result for 1 eV, as before. Anyway, when you measure some particle has 250 MeV energy how do you know whether it is fast electron or slow muon?

[Mr Skeptic]

I got different result for 30 keV, but the same result for 1 eV, as before.

 

It shouldn't be that different; that's not very close to c.

I got this:((2 * 30 keV) / electron mass)^.5 = 102 727 411 m / s

which as expected is just a slight overestimate compared to using the relativistic formula.

 

Anyway, when you measure some particle has 250 MeV energy how do you know whether it is fast electron or slow muon?

 

You measure the charge-mass ratio; also, muons are unstable and will decay into other particles.

[uncool]

Also an electron is made up of a quark and anti-quark: a Down quark and an Anti Up quark. The Down quark has a charge of -1/3 (e), where e is the charge of an electron (in absolute value) and the Anti Up quark has a charge of -2/3 (e), giving you -1 (e).

You are talking about a pion, not an electron. An electron is truly an elementary particle - a lepton - while a pion is a kind of meson. A pion can decay into an electron and a neutrino, but the biggest difference between the two is that the electron is a fermion, while a pion is a boson.

=Uncool-

[Sha31]

You measure the charge-mass ratio; also' date=' muons are unstable and will decay into other particles.[/quote']

 

But how, what exactly can we measure? Can even bubble chamber trajectories tell what's effect of what force? Can even particle accelerators measure the speed of those particles?

 

And then, I could not google anything about slowing electrons down. I just can't imagine electrons floating by at low speeds, like a meter per second. I'm not sure why, but that just sounds impossible. Hmm?

 

 

Also, isn't there some "wave energy" associated with electrons or electron beams, and isn't there some "spin energy"? Then, what if electrons can spin around more than one axis, will the kinetic energy, or angular momentum, of those two spins add? In other words, could there be 3 independent spins, around 3 axes, each contributing to the total angular momentum? And finally, what if muon is just an electron with larger angular momentum than "normal" electrons, could you tell the difference?

[swansont]

This sounds very strange to me. For some reason, I simply can not imagine electrons move at any slow speed, like a meter per second or slower? What is the process and technology to slow down electrons? What words to google?

 

Electrons in a thermal distribution will always have some moving at very small speeds, just like any thermal distribution. (Though this is a classical view, and quantum effects are going to be present)

---

Merged post follows:

Consecutive posts merged

Also, isn't there some "wave energy" associated with electrons or electron beams, and isn't there some "spin energy"? Then, what if electrons can spin around more than one axis, will the kinetic energy, or angular momentum, of those two spins add? In other words, could there be 3 independent spins, around 3 axes, each contributing to the total angular momentum? And finally, what if muon is just an electron with larger angular momentum than "normal" electrons, could you tell the difference?

 

Electrons are spin 1/2. That's all the spin angular momentum they can have. It affects the energy if you put it in a magnetic field, by a predictable amount.

[Sha31]

Electrons in a thermal distribution will always have some moving at very small speeds, just like any thermal distribution. (Though this is a classical view, and quantum effects are going to be present)

 

I don't understand how that can slow down electrons. "Thermal" is just a description of kinetic state of some medium. Higher temperatures correspond to higher atom velocities or higher molecular vibrations/oscillations. Low temperature decreases velocities, but I don't think electron velocity changes, just this velocity of atoms or molecular vibrations. Electrons need their velocities to stay in their orbit, classically speaking. -- But anyhow, this is the question: how can electron microscopes produce different electron energies, how do they emit slow electrons?

 

 

Electrons are spin 1/2. That's all the spin angular momentum they can have. It affects the energy if you put it in a magnetic field, by a predictable amount.

 

Ok, so there is south and north magnetic pole, and whatever the orientation we can change it in arbitrary direction, yes? In fact, we should be able to spin this electron by influencing this magnetic dipole moment, just like electric motors do, and when we turn our spin induction magnets off, the electron should continue to spin, right?

[npts2020]

I don't understand how that can slow down electrons. "Thermal" is just a description of kinetic state of some medium. Higher temperatures correspond to higher atom velocities or higher molecular vibrations/oscillations. Low temperature decreases velocities, but I don't think electron velocity changes, just this velocity of atoms or molecular vibrations. Electrons need their velocities to stay in their orbit, classically speaking. -- But anyhow, this is the question: how can electron microscopes produce different electron energies, how do they emit slow electrons?

 

 

 

 

Ok, so there is south and north magnetic pole, and whatever the orientation we can change it in arbitrary direction, yes? In fact, we should be able to spin this electron by influencing this magnetic dipole moment, just like electric motors do, and when we turn our spin induction magnets off, the electron should continue to spin, right?

 

The word "thermal" in physics does not refer to the medium. It refers to whatever object is being discussed, in this case an electron. Nuclear reactors (nearly all of them in America anyway) are referred to as thermal because neutrons must be slowed down from their ejection during fission in order to cause more fissions and thus sustain the chain reaction. I will allow someone more versed than myself to explain the rest of what you asked.