electric petntial difference.. please help me?

Question

through what potential difference would an electron need to be accelerated for it to achieve a speed of 3% of the speed of light (2.99792 * 10^8 m/s), starting from rest? an answer in units of V.

Answer 1

Kinetic energy = K =1/2mv^2

Electric Potential energy = U = qV where q = number of electrons and V = potential difference in Volts. U is units of electron-volts.

YOu are gvien v = 0.03*c = 9x10^6 m/s and m for an electron is 9.1x10^-31 kg, so K = 3.69x10^-17 J.

Now there are 1.6x10^-19 Joules in 1 electron-volt so K = 230 eV

Since U = q*V and q = 1, V =230 Volts