Electron Speed in a TV: 0.197c

[Myr73]

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

I found-
vi=0 m/s V=10 000 V q=-1.6 X 10^-19 C (not sure if it's negative)
m= 9.1 X 10^-31 kg vf= ?(speed of light)
v(light)= 3.00 X 10^8 m/s


Delta PE= qV=(-1.6 X 10^-19 C) ( 10 000 V) = -1.6 X 10^-15 J
And then --> Delta KE= -Delta PE --> 0.5mv^2=1.6 X 10^-15
vf= SqRt{ 2(1.6 X 10^-15)/9.1 X 10^-31 = 59 295 868= 5.9 x 10^7 m/s

Fraction of speed of light --> vf/v(light)= 0.197

I am unsure if I am doing this properly-Just looking for some confirmation

 

[ehild]

It looks correct.

ehild

 

[rude man]

You might have included relativistic effects. At 0.2c the error is about 2%. So admittedly it's not much.

 

[Myr73]

Awesome-

 

[HalfLostAndSearching]

or help!

I can confirm that your calculations are correct. The electrons leaving the cathode of a TV tube have essentially zero speed, but are accelerated by the potential difference of 10,000 V. This results in the electrons striking the screen with a speed of 5.9 x 10^7 m/s, which is approximately 0.197 times the speed of light. This is a significant fraction of the speed of light, indicating the high energy and velocity of these electrons in a TV.