Question

A proton moving from rest from one point to another under the action of an electric field acquires a kinetic energy of 10^4 keV. Determine the voltage between the two points and the final velocity of the proton.

Answer 1

The voltage between the two points is 10^7 V and the final velocity of the proton is 4.85 x 10^6 m/s.

To determine the voltage between the two points, we need to convert the kinetic energy of the proton to joules. 1 keV is equal to 1.6 x 10^-16 J. Therefore, the kinetic energy of the proton is 10^4 keV = 1.6 x 10^-12 J. The potential difference or voltage is equal to the kinetic energy of the proton divided by its charge. The charge of a proton is 1.6 x 10^-19 C. So, the voltage is (1.6 x 10^-12 J)/(1.6 x 10^-19 C) = 10^7 V.

To determine the final velocity of the proton, we can use the equation for kinetic energy: KE = 0.5mv^2, where KE is the kinetic energy, m is the mass of the proton, and v is its velocity. Rearranging the equation, we get v = sqrt((2KE)/m). The mass of a proton is 1.67 x 10^-27 kg. Plugging in the values, we find v = sqrt((2 * 1.6 x 10^-12 J)/(1.67 x 10^-27 kg)) = 4.85 x 10^6 m/s.