Schrödinger's Equation and the Hydrogen Atom — Question 10

Question 10

Direct substitution into the radial equation recovers the hydrogen ground scale

Treat hydrogen with a fixed nucleus, a nonrelativistic Coulomb potential and no spin, fine structure or external field. Write the Bohr length as a. Let $V(r)=-\mathcal K/r$, where $\mathcal K=k_ee^2$. For an spherically symmetric state use $-\hbar^2[\psi\prime\prime+2\psi\prime/r]/(2m)-\mathcal K\psi/r=E\psi$. Try $\psi=Ne^{-r/a}$ and use $\langle1/r\rangle=1/a$ for the normalized exponential.

1. Determine a and E by requiring the equation to hold for every $r>0$.

2. Find the mean potential and kinetic energies and verify the Coulomb virial relation.

Question 10 - Solution

Direct substitution into the radial equation recovers the hydrogen ground scale

(a) Method and calculation

$$\psi\prime=-\psi/a,\quad\psi\prime\prime=\psi/a^2\Rightarrow E=-\frac{\hbar^2}{2ma^2}+\frac1r\left(\frac{\hbar^2}{ma}-\mathcal K\right).$$$$\boxed{a=\hbar^2/(m\mathcal K)},\qquad\boxed{E=-\hbar^2/(2ma^2)}.$$

(b) Interpretation and further calculation

$$\boxed{\langle V\rangle=-\mathcal K/a=-\hbar^2/(ma^2)=2E}.$$$$\boxed{\langle K\rangle=E-\langle V\rangle=-E},\qquad\boxed{2\langle K\rangle+\langle V\rangle=0}.$$

Physical check

The exponential is a spatial probability amplitude, not a circular path. The equation reproduces the Bohr ground energy while providing a different physical description of the electron.