In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave, λ=2L/n, where L is the length of its home. Given that λ = h/p, we would have p = nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argument is perfectly valid when the potential energy is 0 (inside the well) and L is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length L allowed the wave should be roughly the region classically allowed to the particle, which depends on the "height" of the total energy E relative to the potential energy U.(cf. Figure T^Tfie' "wall" is the classical turning point, where there is no kinetic energy left: E = U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (7-10). Find the location r of the classical turning point in terms of E, use twice this distance for L (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E. For the average potential, use its value at half the distance from the origin to the turning point, again in terms of E. Then write out the expected average total energy and solve for E. What do you obtain for the quantized energies?
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