The H+2 ion is composed of two protons, each of charge +e=1.60×1019C, and an ele

Question

The H+2 ion is composed of two protons, each of charge +e=1.60×1019C, and an electron of charge e and mass 9.11×1031kg. The separation between the protons is 1.07×1010m. The protons and the electron may be treated as point charges.

Part B

Suppose the electron in part A has a velocity of magnitude 1.60×106 m/s in a direction along the perpendicular bisector of the line connecting the two protons. How far from the point midway between the two protons can the electron move? Because the masses of the protons are much greater than the electron mass, the motions of the protons are very slow and can be ignored. (Note: A realistic description of the electron motion requires the use of quantum mechanics, not Newtonian mechanics.)

Explanation / Answer

(1)

If the repulsive force between two charges q and Q is
F = k qQ / r^2
then the electric potential energy of charge q due to the field of charge Q is
PE = k qQ / r . . . [integral of -Fdr from r = infinity, if that's more your speed]
The potential energy of the electron due to both protons is, by superposition,
PE = k (-e)(+E) / (d/2) + k (-e)(+E) / (d/2)
where d is the distance between protons. That is, you're gonna have to add energy to the electron if you want to pull it away from the protons.

(2)

First, Newtonian mechanics is perfectly adequate for "realistically" describing translational motion in this system. Your instructor should solve Schroedinger's equation and compare the solutions if s/he doesn't understand why. Now, the electron will continue to travel in a straight line because the electric forces perpendicular to its motion from the two protons cancel. As it moves away from the protons, the electron will do work on the field, increasing its potential energy at the expense of its kinetic energy. What you must do is calculate the distance R from protons to electron at which all the kinetic energy has become potential energy. If PE1 is its potential energy when it starts from the line joining the protons, then its potential energy PE2 when it's slowed to a stop is
PE2 = PE1 + KE = k qQ / R^2 + k qQ / R^2
You calculated PE1 in part (1), and you're given KE = 1/2 mv^2. [arithmetic]

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