Calculating Electron Velocity: Ek = ½ MV^2

[Brooke22]

Physics help PLEASE!

How fast will the electron be travelling?

Ek = ½ MV^2
V = 2Ek/M
V = √2Ek/M
= √ 2 x 1.6x10^-19/9.1 x 10^-31
= ?
This number is too high since it implies that a larger voltage (say 10) would make the electron travel faster than light. The reason is?

Cheers for the help

 

[cepheid]

1.6x10^-19

Why did you plug in the above number for the kinetic energy in your expression? This is the charge of an electron (in coulombs).

I am not sure what you are doing, and I certainly don't see how voltage enters into it.

Keep in mind this thread should be in the HW help section.

 

[AlphaNumeric]

I think Brooke is working out the velocity an electron gets when it moves through a 1 volt potential.

$$qV = \frac{1}{2}mv^2$$
$$v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2V \times 1.6\times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{4V \times 10^{11}}$$

Brooke, the reason why it appears if you put in a large enough V you get v>c, it is because KE = 0.5mv^2 is a Newtonian expression, it assumes that the mass of an object is constant, but relativity says that energy has mass, so if you've an object which has loads of kinetic energy it's going to have a noticably increased mass.

If you plugged in q, V and the electron rest mass into relativistic equations you'd find that no matter how high you push V you'll never get a speed higher than light. Current particle accelerators can produce particle experiements with over 1 trillion electron volts in energy, and as yet nothing has been seen to move faster than light.

 

[Brooke22]

Thanks a lot, that completely answered my question CHEERS!