Question

If the accelerating voltage va were 250 volts, what would be the speed of the electrons emerging from the gun?

Answer 1

`v=9.34*10^(6)(m)/(s)`

Step-by-step explanation:

According to the law of conservation of energy, the amount of energy an electron gains after being accelerated is equal to its kinetic energy, that is, the electrical potential energy is converted into kinetic energy :

`U=K\\eV=(mv^2)/(2)`

Solving for v and replacing the given voltage:

`v=\sqrt(2eV)/(m)\\v=\sqrt(2(1.6*10^(-19)C)250V)/(9.1*10^(-31)kg)\\v=9.34*10^(6)(m)/(s)`