What is the speed of a proton that has been accelerated through 2300 V?

Calculation of Proton Speed

Explanation: The question is asking about the resulting speed of a proton when it is accelerated through a potential difference of 2300 V. To find the speed of a proton after acceleration, we must use the principle of energy conservation, which tells us that the electrical potential energy gained by the proton is converted into kinetic energy.

The formula that relates the potential difference (V) to the kinetic energy (K.E.) of a proton with charge q (where q is 1.602 x 10^-19 C, the charge of a proton) is K.E. = qV. We can set this equal to the kinetic energy formula K.E. = ½ m^2, where m is the mass of the proton (1.67 x 10^-27 kg) and v is the velocity, to be determined.

Thus, 1.602 x 10^-19 C × 2300 V = ½ × 1.67 x 10^-27 kg × 2. Solving for v provides the speed of the proton.