Question

An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.

Answer 1

Answer:

$9.38\times 10^7 m/s$

Explanation:

We are given that

Potential ,V=25 kV=$25\times 10^3 V$

Distance,r =1 cm=$\frac{1}{100}=0.01 m$

1 m=100 cm

Mass of electron, m=$9.1\times 10^{-31} kg$

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

We have to find the final velocity of the electron.

Speed of electron,$v=\sqrt{\frac{2qV}{m}}$

Using the formula

$v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}$

v=$9.38\times 10^7 m/s$

Hence, the final velocity of the electron=$9.38\times 10^7 m/s$