If protons are accelerated from rest in a Van de Graff accelerator through a potential difference of 1.00 MV, what is their resulting speed?

Question

Quizplus · Accepted Answer

The kinetic energy gained by the protons is equal to the potential energy they gain while moving through a potential difference of 1.00 MV. We can use the formula for potential energy gained by a charge q moving through a potential difference V: PE = qV.

For a proton, the charge (q) is 1.6 × 10^-19 C. Therefore, the potential energy gained by a proton is

PE = (1.6 × 10^-19 C) × (1.00 MV) = 1.6 × 10^-13 J

This potential energy gained by the proton is equal to the kinetic energy gained by the proton, since energy is conserved. We can use the equation for kinetic energy: KE = (1/2)mv^2, where m is the mass of the proton and v is its velocity.

Rearranging the equation to solve for v:

v = sqrt((2KE) / m)

Substituting in the values of KE and m for a proton:

v = sqrt((2 × 1.6 × 10^-13 J) / (1.67 × 10^-27 kg))

v = 1.38 × 10^7 m/s

Therefore, the resulting speed of the proton is 1.38 × 10^7 m/s, which corresponds to answer choice D.

If Protons Are Accelerated from Rest in a Van De