Hydrogen atom: potential well and orbit radii

Question

I happened to open up an old solid-state electronics book by Sah, and in it he says:

"it is evident that the electron orbit radius is half the well radius at the energy level En"

The orbit radius is $r_n=\frac{4\pi\epsilon_0 ℏ^2 n^2}{mq^2}$ and the potential well $V(r_n)=\frac{−q^4m}{(4\pi\epsilon_0)^2ℏ^2n^2}$

Of course the orbit has to be confined in the well, but it's not obvious to me why it should be exactly half the well radius? This isn't something I recall seeing before either in any other text.

Thanks

Answer 1

Of course the orbit has to be confined in the well, but it's not obvious to me why it should be exactly half the well radius? This isn't something I recall seeing before either in any other text.

Please keep in mind that when people are speaking of orbits in the microcosm of particles and nuclei, they are speaking about average numbers over the orbitals . The quantum mechanical solution does not give orbits. We can only calculate wave functions whose magnitude squared gives us orbitals, probability distributions, not classical orbits.

Cross-section of computed hydrogen atom orbital (ψ(r, θ, φ)2) for the 6s (n = 6, ℓ = 0, m = 0) orbital. Note that s orbitals, though spherically symmetrical, have radially placed wave-nodes for n > 1. However, only s orbitals invariably have a center anti-node; the other types never do.

So of course the electron is not confined into a specific radius except in the Bohr model. It is averages that hold in the quantum mechanical calculations, which describe reality.

Answer 2

Express V and E as explict functions of r.

$$V = \frac{-q^2}{(4\pi\epsilon_0)r}$$

$$E = \frac{-q^2}{2(4\pi\epsilon_0)r}$$

Also, the previous page of Sah emphasizes that 2E = V.