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m v r = n h / (2 π).Equating the centripetal force to the Coulomb force (and E = K + U) gives us a corresponding discrete spectrum of orbital radii and energy levels:
where Z is the nuclear charge, the Bohr radius
a_{0} = ε h^{2} / (π m e^{2})and the Rydberg energy= .529 * 10^{-10} m (= .529 Angstroms)
E_{0} = m e^{4} / (8 ε^{2} h^{2})These are the first 20 energy levels of a Hydrogen atom:= 13.606 electron Volts (1 eV = e J)
P ψ = - i (h / (2 π)) dψ/dx,and the energy operator is
E ψ = i (h / (2 π)) dψ/dt,where ψ(x, t) is the electron wave function. With these prescriptions, the statement of conservation of energy becomes Schrodinger's equation:
E ψ = P^{2} ψ / (2 m) + U(x) ψor
i (h / (2 π)) dψ/dt = - (h / (2 π))^{2} / (2 m) ∇^{2} ψ + U(x) ψHere ∇^{2} ψ ≡ ∇ ⋅ (∇ ψ). In general, solving this differential equation is beyond our present means. But if we restrict ourselves to spherically symmetric wave functions, where now
∇^{2} ψ = d(r^{2} dψ/dr)/dr / r^{2},and take ψ(x, t) = ψ(r) χ(t), with the Coulomb potential energy Schrodinger's equation becomes
i (h / (2 π)) dχ(t)/dt / χ(t) = - ((h / (2 π))^{2} (2 dψ/dr + r d^{2}ψ/dr^{2}) / (2 m r) + (- e^{2} / (4 π ε r)) ψ) / ψThis equation is obviously separable, and so both sides equal a constant (E). The solution for χ is
χ(t) = e^{-i E t / (h / (2 π))}and the right hand side is equivalent to the Laguerre equation. In general, it too is beyond our capabilities, but the first few solutions are instructive:
These wave functions have been normalized such that ∫_{all space} ψ^{2} dx^{3} = 1.Note the agreement with our previous analysis based on Bohr's quantization condition. While the expectation values for r are not the same, if we substitute 3 a_{0} / 2 for a_{0} in Bohr's results, they still obey the relation<r> is the expected value (the most "probable value"), and is computed using
<r> = ∫ ψ^{2} r dx^{3},which in our case is4 π ∫_{0}^{∞} ψ^{2} r^{3} dr.If we interpret <r> as the most probable value, ψ takes the role of a probability amplitude, and ψ^{2} (which in general is ψ^{*} ψ, since ψ can be complex) takes the role of a probability distribution.
r_{n} = n^{2} a_{0} / ZNote too that the second and third solutions have nodes at r = 2 a_{0} and r = 3 (3 +- √ 3) a_{0} / 2, respectively. These nodes are reminiscent of the nodes in standing waves, and lead to a similar interpretation for the electron wave functions.
Also note that as decaying exponentials, they only approach zero asymptotically, but they have significant nonzero portions far beyond the orbital radii suggested by Bohr's model.
Finally, note that for a hydride ion (H^{-}), the electron wave functions overlap.
The spin is an intrinsic angular momentum which creates a magnetic moment, but is not associated with mechanical motion.
This applet will draw three dimensional plots of several complete (not simply spherically symmetric) electron wave functions for Hydrogen atoms and Hydride ions.These quantum numbers explain the form of the periodic table.
Because of charge screening of the nuclear charge by intervening electrons, electrons in an energy level n>1 of multielectron atoms see an effective nuclear charge of approximately n.
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©2012, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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