What potential difference would an electron have to fall through to acquire a speed of 3.00*10^6 m/sec?

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6) ^2 / (2 x 1.6 x 10^-19)

V = 25.6 V