MadSci Network: Physics

Re: About Total Energy of electrons?

Date: Mon Apr 18 01:43:02 2016
Posted By: Vladimir Escalante-Ramirez, Faculty
Area of science: Physics
ID: 1454090954.Ph

You are right in that the potential energy (P.E.) increases and the kinetic energy (K.E.) decreases as the electron moves farther from the nucleus. However they do it at different rates. If you work out the equations for the Bohr atom model, you must get the result that P.E. = − 2 K.E. (Just remember F=ma, where F is the electrostatic force, m is the mass of the electron in an orbit of radius r, and a=v²/r is the centripetal acceleration for the electron moving with velocity v. Then multiply both sides of the equation by r, and remember that P.E. is negative: −ke²/r, where e is the electron charge, and k is the electrostatic constant.) Thus P.E. increases at twice the rate of decrease of K.E. when moving to an orbit of higher energy, and prevails over the decrease of K.E. The result is that the total energy P.E.+ K.E. = − K.E. increases at the same rate of the decrease of K.E.

The energy of the atom comes from the attractive force between the nucleus and the electrons. The K.E. comes from the motion of the electrons in the Bohr atom model. That model works only for hydrogen. Atoms with more than one electron require a more sophisticated quantum theory that does not consider electron orbits around the nucleus, but uses the concepts of wave mechanics. Nevertheless it is still possible to talk about the K.E. and P.E. of electrons in the atom. Using general procedures in advanced quantum mechanics, it is possible to show that the relation P.E. = − 2 K.E., called the virial theorem, still holds for the electrons in an atom. See for example problem 10 in chapter III of Problems in Quantum Mechanics by Constantinescu and Magyari (Pergamon International, 1971).

Vladimir Escalante Ramírez
Institute for Radio Astronomy and Astrophysics
Morelia, Michoacán, México

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