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Electrons leaving a cathode

  • Thread starter rgo
  • Start date

rgo

7
0
I am not sure if this is the right place to post this question but here it is any way...

Electrons leave the cathode of a TV tube at essentially zero speed and are accelerated toward the front by 10,000v potential. At what speed do they strike the screen? Express this value also as a faction of the speed of light.

OK so here is what I did does it look right?

PE1 + KE1 = PE2 + KE2
KE1 = PE2 = 0
so
PE1 = KE2

PE = 1/2QV
KE = 1/2mv^2

therefore

1/2QV = 1/2mv^2

Rearrange for velocity

v = sqroot(QV/m)
v = sqroot(1.6 x 10^-19 x 10,000 V / 9.11 x 10^-31)
v = 4.2 x 10^7 m/s

c 3.00 x 10^8 m/s

Therefore the speed of the electron equals approximately 2/15c

Does this look right or am I way off base?????
thanks for your help....
 

Gokul43201

Staff Emeritus
Science Advisor
Gold Member
6,987
14
1. The correct place to post this would be the Homework and Coursework subforum (near the top of the main page). Keep that in mind for your next thread.

2. Recheck the formula you've used for PE - there's an error in it.

3. If you haven't yet had any Relativity, then this approach is correct. If you fix the error in 2 you should be good.

4. I haven't actually checked the numbers.
 

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