What is the speed of a proton that has been accelerated from rest through a potential difference of - 1300 v?

For this situation, we calculate the velocity of the proton from the idea that the potential energy of the proton when at rest should be equal to the kinetic energy as it moves. We calculate as follows:

Given:

q = 1.602 x 10^-19 C

mass of proton = 1.6726 x 10^-27 kg

V = 1300 V

Solution:

Potential Energy = Kinetic Energy

Potential energy = q x V = 6.408 x 10^-16 J

Kinetic energy = (1/2) x mass of proton x v^2:

v = Sqr (2 x potential energy / mass of proton)

v = Sqr (2 x 6.408 x 10^-16 / 1.6726 x 10^-27)

v = 875341.263 m/s

The velocity of the proton would be 875341.263 m/s.

The lack of info given to us requires that we use an energy approach. We start with electrical energy, U = - qV, where U is potential electrical energy, q is charge (which for a proton is e, which equals 1.6*10⁻¹⁹ C) and V is our potential difference. We can then use conservation of energy, assuming all our energy is converted to motion, so kinetic energy: K = 1/2m*v², where K is kinetic energy, m is the mass of the proton (1.67*10⁻²⁷ kg) and v is the velocity of the proton. We can set these energy terms equal to each other to get - qV = 1/2m*v², then we can solve for v, v = sqrt (-2qV/m), plug in the numbers, v = sqrt (-2 (1.6*10⁻¹⁹) * (-1300) / (1.67*10⁻²⁷)) = 499,100 m/s, or about 1/10,000 the speed of light, which is well under the range of worrying about special relativity, so our method was correct.