I am not sure if this is the right place to post this question but here it is any way...(adsbygoogle = window.adsbygoogle || []).push({});

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....

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Electrons leaving a cathode

**Physics Forums | Science Articles, Homework Help, Discussion**