Determine the speed of the ordered movement of electrons at the exit from the electron gun at a potential difference of 500 V.

From the condition of the problem given U = 500v – the potential difference of the gun. The electron charge e = 1.6 * 10 ^ -19Kl.

First, let’s find the work of the electron gun, while accelerating the electron. Where A is work.

A = U * e = 500v * 1.6 * 10 ^ -19Kl = 800 * 10 ^ -19Kl = 8 * 10 ^ -17J.

Now we use the formula for the kinetic energy of a body to determine its speed. Where m = 9.11 * 10 ^ -31kg is the mass of an electron.

E = m * v²: 2.

From here we find the speed of the electron.

V = √ (2 * E: m) = √ (2 * 8 * 10 ^ -17J: 9.11 * 10 ^ -31kg) = 1.3 * 10 ^ 7m / s.

The solution found the speed is approximately 130000 km / s.