Relativistic Electrons: Kinetic Energy & 20keV Threshold

[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?

 

[eNathan]

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?

 

[Ich]

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

 

[Miserable]

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.

 

[A_I_]

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

 

[robphy]

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.]

 

[eNathan]

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.

 

[mathlete]

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!

 

[yogi]

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

 

[Ich]

"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 %).

 

[robphy]

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.

 

[Meir Achuz]

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).

 

[Creator]

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.
Express your answer in fractions (or multiples) of c.

Creator

 

[Creator]

What; no takers?

 

[Meir Achuz]

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.

 

[Creator]

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

 

[Meir Achuz]

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.

 

[pmb_phy]

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

 

[Creator]

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. :tongue2:
I'm sure you'll want to argue about it...Just forget it ...it's not worth trying to explain.

Creator

 

[Meir Achuz]

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

 

[dextercioby]

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.