What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kv?

Answer: v = 1.19*10^-8 m/s

Explanation: according to the work-energy theorem, the work done in accelerating the proton equals the kinetic energy that it gains.

Mathematically, we have that

qV = 1/2mv²

Where

q = magnitude of proton = 1.609*10^-16c

V = 4kv = 4000v

m = mass of proton = 9.11*10^-31 kg

v = velocity of proton = ?

By substituting the parameters, we have that

1.609*10^-19 * 4000 = 1/2 * 9.11*10^-31 * v²

1.609*10^-19 * 4000 * 2 = 9.11*10^-31 * v²

v² = (1.609*10^-19 * 4000 * 2) / 9.11*10^-31

v² = 12872*10^-19/9.11*10^-31

v² = 1412.95*10^-19

v² = 1.41*10^-16

v = √1.41*10^-16

v = 1.19*10^-8 m/s