Question

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 33000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

Answer 1

solution

kinetic energy is given by

$KE=\frac{mv^2}{2}........eq.1$

when potential diffrence V is applied on a charge then kinetic energy

$KE=qV........eq.2$

from equation 1 and 2

$v=\sqrt{\frac{2qV}{m}}$

for electron $m=9.1\times10^{-31}kg$

$v=\sqrt{\frac{2\times1.6\times10^{-19}\times33000}{9.1\times10^{-31}}}$

$v=10.77\times10^6m/s$