Question

Newton’s kinematics break down when traveling near the speed of light.

Compare the speed of an electron in the cathode-ray tube calculated from

~a =(q/m)E to the speed of light c = 3 · 108 m/s. Are relativistic corrections

necessary?

Electric Field = 522 Volts

Answer 1

using conservation of energy

0.5 m v^2 = q V

0.5* 9.11* 10^-31* v^2 = 1.6* 10^-19* 522

v = 1.354* 10^7 m/s

since velocity v < c, so no need of relativistic correction

for more learning perspective if we want to apply the relativistic effect anyway, we need to follow the following steps

m c^2 ( 1 / sqrt ( 1 - v^2/c^2) - 1) = q V

9.11* 10^-31* (3* 10^8)^2* ( 1 / srt ( 1- v^2/c^2) - 1) = 1.6* 10^-19* 522

v = 1.353* 10^7 m/s

=======

Comment before rate in case any doubt, will reply for sure.. goodluck