Electrons are accelerated from rest by a p.d. of 100V. What is their final velocity?

I have no idea how to begin. What is the relation between the voltage and motion of the electron?
Thanks

__________________
To each his own.

When the electrons are accelerated by a p.d., the kinetic energy acquired by them is eV, where e is the charge on the electron.
So 0.5mv^2 = eV
Now calculate the velocity v.

Hey there,
It isn't quite that simple because you have to take into account the relativistic effects that occur with the electron when nothing is in its path that slows it down taking energy away from it in the process. The mass of an electron is 0.510998910(13) MeV/c2 which when multiplied by (y - 1) yields the energy of the electron.


In other words the energy of an electron is given by:



where


and
Me = 0.510998910(13) MeV/c2 = Me already has already been normalized by the speed of light squared, so the energy of an electron is really just:

E = ( - 1)* 510998.91013 eV

E = qV = (1.60217*10^-19C)(100V) = 1.60217*10^-17 eV, so
this states that the energy obtained by an electron is the product of the charge associated with the electron and the voltage it feels.

1.60217*10^-17 = (1/sqrt(1-(v/c)^2) - 1)*510998.91013
(3.1353815*10^-23 + 1)^2 = 1/(1-(v/c)^2)
1-(v/c)^2 = (3.1353815*10^-23 + 1)^-2
(v/c) = (1-(3.1353815*10^-23 + 1)^-2)^0.5

v = c*(1-(3.1353815*10^-23 + 1)^-2)^0.5

v = 2.375644 mm/s, which may seem pretty slow, but when electrons typically travel through a conductor on the order of mm's per second than it seems more reasonable.
Of course the reason why electrons typically travel in the mm's per second is due to collisions within the conductor making their average velocity, called their drift velocity very slow indeed.
In your case, the electron is moving so slowly because at first it really started taking off, but given the relativistic speed it obtained its mass became huge which slowed it considerably. The trade off is with mass and speed.
That is, these electrons will travel at this speed under that acceleration from the E field inducing the 100 V potential difference.
Any faster than this and their mass increases such that they would slow down again decreasing their mass making them speed up again, eventually an equilibrium speed is reached.
That speed seems slow, but given the extraordinarily small mass of an electron, it doesn't have to go all that fast for its mass to significantly increase; what's a significant increase?
Well the mass of an electron is 9.10938215*10^-31 kg so simply doubling that mass would slow it considerably and doubling it adds so little mass to the electron that the increase can be made from a small speed the electron may obtain.


So the small speed increases the mass of the electron by an amount that normally would be considered insignificant for anything else, but for something that has a mass that is insignificantly small to begin with, very tiny increases in mass for that particle have much larger influences on that particles speed and acceleration.

I don't know if this helped or not, but here it is anyway.

Many Smiles,
Craig

Thank you for the explanation! the answer is 6000000 m/s and the electrons are in a vacuum in this context because i'm dealing now with their deflection in a magnetic and electric field, that is free electrons. I think the simpler explanation was good enough at my level, but of course a better understanding of it is also good.
But thanks anyway

__________________
To each his own.

Hi there,
Sorry about that, when I found the energy of the electron I thought it was in eV but it was in Joules; I shouldn't have been using it in Joules I wanted it in eV. If you go through and use my same numbers and convert the Joules to eV you get about 6*10^6 m/s (like 5.9 something, close enough to round up to 6). So I had 1.60217 *10^-17 Joules
(100V*charge of 1 electron = energy in Joules for that electron) when it should have been in eV and 1 eV = 1.602*10^-19 Joules (which should make sense since an electron volt is the charge of an electron times 1 volt, so 1 eV = 1.602*10^-19 Joules), so converting the answer in Joules to one in eV I get:

X eV = (1 eV/1.602^-19Joules)*1.60217*10^-17 Joules
1.60217*10^-17Joules = 100 eV

100 eV = (1/sqrt(1-(v/c)^2) - 1)*510998.91013
(1-(1.956951e-4 + 1)^-2) = (v/c)^2
v = c*(1-(1.956951*10^-4 +1)^-2)^0.5

v = 5.9*10^6 m/s or about 6,000,000 m/s which is what you said it was. Sorry about that I should have known by looking at it that it should have been much larger in a vacuum. In a conductor they really do move at speeds in mm's/sec, but if they traveled as slow as I had them in an old analog TV it would be hard to use them to "paint" a picture across the screen. The reason why the energy level is 100 eV is because 100eV is the 100 Volts times 1 electron, which in Joules is 100 V times the charge of 1 electron making it far smaller in Joules at about 1.602*10^-19 Joules = 1 eV.

I need to pay more attention sometimes to what I'm doing, often I solve these problems while I'm working on other things at the same time, so I do not always have my attention on the problem, I'm just going through the "motions" of solving the problem and then just post my answer without thinking about it first. If I had I would have realized it was too small. It's true that the relativistic effects have influence on the electron speed, as I get something a little over 5.9*10^6 m/s and you get 6*10^6 m/s using traditional Newtonian Physics, but if you include the relativistic effects it slows it a little to just above 5.9 *10^6 m/s, not a big difference by any means, but it is a real and known affect from the objects relative speed.

Hopefully this makes more sense than the bad answer I had given to you. Again, sorry about that, I should have just used everything in kg and Joules and not tried to mix a mass in Mev/c^2 for use with energy in eV and not Joules, and simply assumed I was using the energy in eV (when I was really using the energy in Joules). I didn't even see it.

Craig

Hello again,
I just thought I give you the function for finding the force applied upon some charge due to either or both fields. It's called the Lorentz force and is given by:



where E is the electric field in V/m, B is the magnetic field in Tesla's or W/m^2 (same thing) and v is the vector velocity of the electron. Note that for a B field to deflect an electron the electron must be moving with some velocity v at any angle other than 0 degrees with respect to the B fields direction. It applies the greatest displacement force when the velocity of the electron is at a right angle to the B field, while an E field applies a force upon a charged particle no matter if it is moving or not; and is independent of the angle between a moving electron and the direction of the E field. If a charged particle gets into an E field it will be forced in the direction of the E field every time. If it is moving in some direction perpendicular to the B field it will be forced in a direction that is at a right angle to both its initial direction and that of the B field itself.

That's it, I just thought I'd add that since you are now dealing with deflections due to these fields.

Many Smiles,
Craig