# Relativistic Electrons: Kinetic Energy & 20keV Threshold

• IntuitioN
In summary, at what kinetic energy do electrons speeds become relativistic? It depends on the precision of your equipment/experiment.
IntuitioN
Just a general q: at what kinetic energy do electrons speeds become relativistic? I heard some people mention from 20keV onwards... is that correct?

I don't understand your question. If you are talking about relavistic speed, it is always relative (to something else). Even if its going .001 meter per second, it has more KE than a body going 0 meters per second.

Are you asking at what speed does relavistic mass (hence energy) become noticable?

"relativistic" usually means v> 0.1 c, where SR corrections are more than 1%. That would mean about 50 keV for an electron.

He means when do you need to use relativistic mechanics rather than classical mechanics to describe it.

Obviously there's no definite cut-off point, but as a general rule of thumb a particle is considered to become relativistic when its kinetic energy is roughly equal to its rest mass energy, so for an electron that would be ~500keV.

KE = 1/2 m v^2

m = 9.1*10^-31 kg

and to be a relativist electron its speed should exceed 0.1v where v is the speed of light; thus exceeds 30000 km/s

the minimum kinetic energy required for an electron to be relativist is proportional to a minimum of 30000 km/s as speed.
do the calculations and u will find the energy required in Joules unit

if you need to convert to eV
energy (J) / 1.6*10^-19 = energy (eV)

Hope this helps to get a clearer view

This reminds me of a neat calculation:
What is the speed of an electron in a (Bohr model) hydrogen atom? (Express your answer as a fraction of the speed of light.) [You don't need a calculator.]

Well, I guess that all depends on how presise you want your calculations to be. Relavistic effects always take place from a stationary reference frame (even tho its small) when v > 0.

robphy said:
This reminds me of a neat calculation:
What is the speed of an electron in a (Bohr model) hydrogen atom? (Express your answer as a fraction of the speed of light.) [You don't need a calculator.]
0 in the frame of the electron!

ropbhy - first Bohr orbit it is apha = (1/137)c

Ich said:
"relativistic" usually means v> 0.1 c, where SR corrections are more than 1%. That would mean about 50 keV for an electron.
ups, that means 2.5 keV of course (1/2 %).

yogi said:
ropbhy - first Bohr orbit it is apha = (1/137)c

Yes! So, one could use this as a rule of thumb to determine if a non-relativistic treatment is good enough.

It's interesting how important energies and lengths in atomic/quantum physics are related by powers of alpha.

Coulombs law has no intrinisc dimension if written in appropriate units as V=e^2/r.
Any physical result for the hydrogen atom must either be dimensionless or depend on some power of m and powers of alpha (which is dimensionless).

robphy said:
This reminds me of a neat calculation:
What is the speed of an electron in a (Bohr model) hydrogen atom? (Express your answer as a fraction of the speed of light.) [You don't need a calculator.]

Here's another:
An electron has a spin angular momentum of $\frac{\sqrt{3}}{2}{\hbar}$ and a classical radius $$R_0=\frac{\alpha{\hbar}}{mc}$$.
Classically, what is the maximum surface speed of the electron,
assuming the electron has spherical shape and its mass is uniformly distributed.

Creator

What; no takers?

NR, I would get 5\sqrt{3}/(4*alpha), which means it is a relativistic question. Is that what you meant? This becomes relativistic, with a bit of a messy integral involved.

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Meir Achuz said:
NR, I would get 5\sqrt{3}/(4*alpha), which means it is a relativistic question. Is that what you meant? This becomes relativistic, with a bit of a messy integral involved.

Yes; thanks for the attempt, Meir. But its more than simply 'relativistic', it becomes an 'impossibility'.

Start with the classical spin angular momentum L = I$\omega$.
Then substitute in values for electron spin angular momentum, solid spherical moment of inertia I, and angular velocity $\omega=\frac{v}{R}$, and R=classical electron radius as previously given.
Then solve for v (surface velocity) in terms of c.
What do you get?

Creator

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The model is wrong, but not impossible. Your simple formula L=I\omega is wrong in SR.
Actually angular momentum can get quite complicated in SR.
If the model had any interest, I would plow through the messy integration.

IntuitioN said:
Just a general q: at what kinetic energy do electrons speeds become relativistic? I heard some people mention from 20keV onwards... is that correct?
It depends on the precision of your equipment/experiment. Take protons as an example. There have been experiments in which the mass of low speed protons has been detectable. I think the experiment was called a "Penning Trap."

Pete

Meir Achuz said:
The model is wrong,... .

That's the whole point, Meir! Duh! Double Duh! You obviously missed the entire purpose of the exercise!

The whole point of my post to Robphy was in response to his question about trying to figure the 'orbital velocity' of an electron. Which actually one cannot use as 'correct' model either.
Well, my question was about the analogous situation involving the velocity of 'spin'.

Neither of us was trying to produce a 'correct ' model; it was merely an exercise.
The whole point of my exercise was that (if one uses the classical radius) then in order to account for the electron angular momentum classicallythen the surface velocity of the electron would have to be around 300 c !

And yes, that is an impossibility, (whether you care to agree or not).
Had you done the simple calculation I suggested you would have realized that!..MAYBE.
I'm sure you'll want to argue about it...Just forget it ...it's not worth trying to explain.

Creator

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L=I\omega is wrong. Wrong is wrong. Using the right formula would give v<c.

Hold on,the electron is a point particle.It doesn't spin.Its state vector does,however,but in "spin space",of course,not in Schrödinger's scalar theory.

Daniel.

## 1. What are relativistic electrons?

Relativistic electrons are electrons that are moving at speeds close to the speed of light. This means that they have a significant amount of kinetic energy and are subject to the effects of special relativity.

## 2. How is kinetic energy related to relativistic electrons?

The kinetic energy of a particle is directly related to its speed. As electrons approach the speed of light, their kinetic energy increases significantly, making them relativistic.

## 3. What is the 20keV threshold for relativistic electrons?

The 20keV threshold refers to the minimum amount of energy that an electron must have in order to be considered relativistic. At this energy level, the electron will have a speed close to the speed of light and will exhibit the effects of special relativity.

## 4. What are some practical applications of understanding relativistic electrons?

Understanding relativistic electrons is important in many fields, such as particle physics, astrophysics, and medical imaging. It also has practical applications in the development of technologies such as particle accelerators and high-energy lasers.

## 5. How do relativistic electrons impact our understanding of the universe?

Relativistic electrons play a significant role in many astrophysical phenomena, such as the formation of cosmic rays, the dynamics of black holes, and the acceleration of particles in space. Understanding their behavior is crucial in expanding our understanding of the universe.

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