# Electricity problem

electricity problem :(

so heres the problem:

through what potential difference would an electron need to be accelerated for it to achieve a speed of 42.0% of the speed of light, starting from rest? The speed of light is c = 3.00e8 m/s

so i thought the equation i would use would be Vf-Vi=deltaPE / q

so i thougth the change in potential energy would be equal to the opposite of change in kinetic energy, so delta PE would = .5mv^2, and v is 3e8 X .42, so:

.5 X 9.11e-31 X (1.26e8)^2 / 1.6e-19 = 4520V, but this is wrong, so yea, no clue. any help would be much appreciated, thanks

## Answers and Replies

Tom Mattson
Staff Emeritus
Gold Member
You've got the right idea. All of the potential energy should show up as kinetic energy of the electron. Where you went wrong is right here:

so delta PE would = .5mv^2,

The expression $K=\frac{1}{2}mv^2$ is the nonrelativistic kinetic energy. In relativity there is a different expression:

$$K=\gamma\left(mc^2-1\right)$$
$$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$$

If you use the correct expression, you should get the correct answer.