(a) Figure shows the electron and its image. Accordingly the electric energy for the system is
(b) Separating the variables by assuming solutions of the type
Note that since V(y) = \(- \frac{e^2 }{4y}\) depends on y only, px and pz are constants
of the motion. Hence
(c) The remaining boundary condition is \(\psi\)(x, y, z) = 0 for y ≤ 0.
(d) Now consider a hydrogen-like atom of nuclear charge Z. The Schrodinger equation in the radial direction is
which is identical with (1) with the replacements r → y, Z → a. Hence the solutions of (1) are simply y multiplied by the radial wave functions of the ground state of the atom. Thus
Note that the boundary condition in (c) is satisfied by this wave function. The ground-state energy due to y motion is similarly obtained:
(e) The complete energy eigenvalue for quantum state n is
where A is the normalization constant.