What is the potential at a distance of 0.0529 nm from a proton?

Question

Quizplus · Accepted Answer

The potential at a distance r from a proton is given by V = kq/r, where k is the Coulomb constant and q is the charge of the proton. Substituting r = 0.0529 nm (converted to meters) and q = +1.602 x 10^-19 C, we get:

V = (9 x 10^9 N*m^2/C^2)*(1.602 x 10^-19 C)/(0.0529 x 10^-9 m)

V = 2.72 x 10^-10 J/C

Converting this to volts, we get:

V = 2.72 x 10^-10 V = 0.272 eV = 27.2 V

Therefore, the correct choice is C. Note that choices A, B and D refer to the ionization energy of hydrogen, which is not directly relevant to the potential at a distance from a proton. Choice E (9.11 pC) is a charge, not a potential, and is not relevant to the problem.

What Is the Potential at a Distance of 0