Question

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27,800 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

Answer:

$v = 9.88 \times 10^7 m/s$

Explanation:

As we know that electrostatic potential energy of the electron is converting here into kinetic energy

so we will have

$qV = \frac{1}{2}mv^2$

so here we know that

$q = 1.6 \times 10^{-19} C$

$V = 27800 Volts$

$m = 9.1 \times 10^{-31} kg$

so from above equation we have

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

$v = \sqrt{\frac{2(1.6 \times 10^{-19})(27800)}{9.11 \times 10^{-31}}}$

$v = 9.88 \times 10^7 m/s$