# Homework Help: Acceleration of electrons in lamp

1. Aug 15, 2012

### MateuszS

1. The problem statement, all variables and given/known data

Hello. I have an exam soon so I need to do many exercises. I have a problem with one. I need to get an acceleration of electrons in lamp which energy is 14 000 eV and magnetic induction is B = 5.5*10^(-5) T.

3. The attempt at a solution

The correct result is 6.4*10^14m/s^2 but i don't know how to get it. I tried with F=qVB but without any progress.

Sorry for my English I would be grateful if anybody could correct my mistakes.

Greetings,
Mateusz

2. Aug 15, 2012

### Simon Bridge

$\vec{F} = q\vec{E} + q\big ( \vec{v}\times\vec{B}\big )$

3. Aug 15, 2012

### colinven

MateuszS, you are using the correct formula: $F=qvB$. where $v$ is the velocity of the electrons coming off the lamp.

Notice, 14000eV is the kinetic energy of the electrons coming off the lamp, and this energy can be converted to Joules (J). We can write the kinetic energy as: $K=\frac{1}{2}mv^{2}=14000eV=4.4860×10^{-15} J$.

It is easy to solve for the velocity $v$ using the kinetic energy.

Your initial equation can be re-written as: $F=ma=qvB$. where $a$ is the acceleration of the electrons coming off the lamp.

Putting everything together, the final formula will look like this $a=\frac{qvB}{m}$. where $v$ can be found using the kinetic energy $K$

The final answer I received was 6.8×10$^{14}m/s^{2}$

Hope this helps,

Colinven

4. Aug 15, 2012

### MateuszS

Yes, I did it the same way, but I had a different result... thank you.

Did you multiply energy by 2? It should be 2.24
$$14 000ev = 1.60*10^{-19}*14 000 = 2.24*10^{-15}$$

Yes, we must multiplay it - but later, when we want to find $v$ in last equation. What did you substitute for $q$ and have you used $9.1*10^{-31}$ as a mass of electron?

Last edited: Aug 15, 2012
5. Aug 16, 2012

### Simon Bridge

I'll show you guys a trick using the mass-energy relation:

$\frac{1}{2}mv^2=\frac{1}{2}(mc^2)\frac{v^2}{c^2}= \frac{1}{2} (511\text{eV})\left ( \frac{v}{c}\right )^2= 14000\text{eV}$ ... so you can do everything in nice numbers!

What do you notice about the relationship between v and c here?

Last edited: Aug 16, 2012
6. Aug 16, 2012

### colinven

Yes I used the mass of the electron as 9.11 E -31 Kg. And "q" is the charge of one electron: 1.6022 E -19 C.

7. Aug 16, 2012

### colinven

Simon Bridge: How can you equate kinetic energy and rest mass energy? The electrons are moving, not at rest.

8. Aug 16, 2012

### colinven

Nevermind, I see what you did there! Haha.

9. Aug 16, 2012

### Simon Bridge

Cool huh? ;)
It's a very useful relation.
In case someone else misses it - I said that $m = (mc^2)/c^2$ so I could use the rest-mass energy in there.

But - the relationship between v and c here is very important to this question.
According this calculation - what is the speed of the electron compared to the speed of light?

Last edited: Aug 16, 2012
10. Aug 20, 2012

### Simon Bridge

That was four days ago... how did you get on?