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It depends on how far the electrons are from the atomic nucleus. You know that electrons have a negative charge = -q (1.6X10^-19 Coulombs) protons have a charge of equal magnitude but opposite sign +1.6X10^-19. Remember columbs law:

F = (k X Q1 X Q2 / r^2)

where:

F is the magnitude of the force exerted,
Q1 is the charge on one body,
Q2 \ is the charge on the other body,
r is the distance between them,
k = 1/(4 X Pi (=3.14) X Epsilon Not. "Epsilon not 8.854×10−12 C2 N−1 m−2 (also F m−1) is the permittivity of free space, approx 8.988×109 N m2 C−2.

From the equation you can immediately tell that, just like with any force that exists (ie. magnetic, gravitation) the force of attraction decreases sharply with the square of the distance between the two charges.

With this in mind, you can probably answer your own question now. Electron are located at various distance, orbiting around a positively charged nucleus. The farther out you go from the nucleus, the greater the distance between the positive nucleus and the negative electrons. The charge of all electron are the same and the charge of a proton is off-course constant, it is basically the number of protons X the charge of a proton which = +q. Consequently it is a multiple of q. Therefore electron farther away have much higher potentially energy, simply because they are farther away and hence the electrostatic forces exerted on them by the nucleus are much weaker

2007-08-09 11:02:37 · answer #1 · answered by Anonymous · 3 0

The planetary analogy works here up to a point.
The farther from the oppositly charged, attractive nucleus, the greater the potential energy.
The wavelength of the electron defines the orbits it can inhabit.
Thus the quantum jumps in energy levels instead of smooth gradation..

2007-08-10 04:05:31 · answer #2 · answered by Irv S 7 · 1 0

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