Question

You cause a particle to move from point A, where the electric potential is 14.7 V, to point B, where the electric potential is −24.5 V. Calculate the change that occurs in the particle's electrostatic potential energy, when the particle is an electron, a proton, a neutral hydrogen atom, and a singly ionized helium atom (i.e., lacking one electron from its neutral state).

Answer 1

Step-by-step explanation:

If a particle to move from point A, where the electric potential is 14.7 V, to point B, where the electric potential is −24.5 V, the change in electric potential is expressed as:

∆V = VB-VA

Given VB = -24.5V

VA = 14.7V

∆V = -24.5-14.7

∆V = -39.2V

To calculate the change that occurs in the particle's electrostatic potential energy, when the particle is an electron, we will use the formula:

∆U = q∆V where q is the charge on the electron

Charge on an electron = -1.6×10^-19C

∆U = -1.6×10^-19C×-39.2

∆U = 62.72×10^-19 Joules

For a proton:

q = 1.6×10^-19C

∆U = 1.6×10^-19C×-39.2

∆U = -62.72×10^-19 J (calculation is similar as above, the only difference is the sign)

For a neutral hydrogen atom:

Charge on a neutral hydrogen atom is +1

∆U = 1×-39.2

∆U = -39.2Joules

For a singly ionized helium atom

Charge on the atom q = +1

∆U = 1×-39.2

∆U = -39.2Joules