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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....