Calculating Potential Difference for Electron Acceleration

[buttterfly41]

electricity problem :(

so here's 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

 

[quantumdude]

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.