What Speed Do Electrons Reach in a TV Tube?

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The discussion revolves around the speed of electrons in a TV tube after being accelerated by a potential difference of 10,000 V. The original poster attempts to calculate the final speed of the electrons and express it as a fraction of the speed of light.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster presents calculations involving potential energy and kinetic energy, questioning the correctness of their approach and the sign of the charge of the electron. Some participants suggest considering relativistic effects at higher speeds.

Discussion Status

The discussion includes a confirmation from one participant regarding the calculations, while another raises the point about the potential impact of relativistic effects on the results. There is an acknowledgment of the accuracy of the original poster's method, but also a suggestion for further consideration.

Contextual Notes

Participants are exploring the implications of relativistic speeds, particularly as the calculated speed approaches a significant fraction of the speed of light. The original poster expresses uncertainty about certain values and assumptions in their calculations.

Myr73
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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
 
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It looks correct.

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

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