We continue with the three dimensional equation encountered in the previous post. This time, we shall solve the radial part of the wave function with the potential provided by the Hydrogen atom nucleus(which contains single proton). This potential is nothing but the simple Coulomb potential:

V(r) = – e^{2}/(4 π ε_{0} r)

As before, the final solution is a product of Radial part R, and angular part Y(θ,Φ) (we have already solved for Y which turn out to be spherical harmonics). Note that the variable has been changed from R(r) to u(r) first, just to make the calculation neater:

u(r) = r x R(r)

The equation in u(r) can be solved with analytical methods used to solve linear ordinary differential equation. Here we quote only the results.

In the case of Hydrogen atom, energy only depends on the principal quantum number, n. The final wavefunction Ψ_{n,l,m} however depends on the three quantum numbers n(principal), l(azimuthal) and m(magnetic) as shown.