# Relativistic kinetic energy versus classic kinetic energy [closed]

I have a homework problem where I am getting the wrong answer and I just want advice on the route I took.

So I am told that electrons in a television set are accelerated through a potential difference of 50 000 volts. I am to find speed of the electrons using relativistic kinetic energy and classical Kinetic energy. The answer I found relativistically is correct. Since $E_K=W$ and $W=qV$, I know work, so I know $E_K$, then I solved for speed, 0.412c. All good.

So here's the question. I want to use $E_K = (1/2)mv^2$ to solve it again, and I am supposed to get 0.422c. I use the same $E_K$ as the first time, based on my $qV$. (I suspect this is the problem, but please tell me why or why not is it the correct value to be used here. If it is the incorrect value, I don't know then what I should have for $E_K$). When I solve for $v$ now, I get 0.44c.

I have ruled out sig fig problems. Thanks for your help.

Oh, and then I'm supposed discuss if this difference (0.412 vs. 0.422) makes a difference in designing television sets. I suppose it must, but I don't know how tvs work. Some guidance here would help too. Thanks.

## closed as off-topic by Brandon Enright, ACuriousMind♦, BMS, John Rennie, JamalSJan 22 '15 at 7:51

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• I get $0.442\, c$. Without more information, my hunch is that the answer you're "supposed" to get is wrong; such things happen. – Javier Jan 21 '15 at 22:25

## 1 Answer

You are definitely right with $0.44c$ I have checked the calculation and also found the exact question here http://www.phas.ubc.ca/~mcmillan/rqpdfs/1_relativity.pdf with a worked solution, all on page 11.

I honestly can't remember much about how CRT televisions work but I seem to remember that they use electromagnets to direct electrons to different parts of a phosphorus screen to produce images. So this would mean that the electrons feel a force based on the magnetic part of the Lorentz force law $$\vec{F}=q\left( \vec{E}+\vec{v}\times \vec{B} \right)\underset{\vec{E}=0}\rightarrow q\vec{v}\times \vec{B}$$

so the magnitude of this force is proportional to the velocity of the electrons so if the velocity is incorrect based on the design of the magnet etc the electron will hit the wrong part of the screen and you won't get a good image.