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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).

Greetings,

Vladimir Escalante Ramírez

Institute for Radio Astronomy and Astrophysics

Morelia, Michoacán, México

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