Find the energy of an electron fo a hydrogen atom in a stationary state for which the wave function takes the from `Psi(r )=A(1+ar)e^(-alpha r)`, where `A, a and alpha` are constant.

Explanation

The Schrodinger equation for the problem in Gaussian units
`grad^(2)Psi+(2m)/(ħ^(2))[E+(e^(2))/(r )]Psi=0`
In `MKS` units we should read `(e^(2)//4pi epsilon_(0))` for `e^(2)`
we put `Psi=(chi(r ))/(r ).` Then `chi''+(2m)/(ħ^(2))[E+(e^(2))/(r )]xho=0` ....(1)
We are given that `chi=rPsi=Ar(1+ar)e^(-ar)`
so `chi'=A(1+2ar)e^(-ar)-alphaAr(1+ar)e^(-ar)`
`chi''=alpha^(2)Ar(1+ar)e^(-ar)-2alphaA(1+2ar)+2aAe^(-ar)`
Substituting in (1) gives the condition
`alpha^(2)(r+ar^(2))=2alpha(1+2ar)+2a+(2m)/(ħ^(2))=0`....(2)
`a alpha^(2)+(2m)/(ħ^(2))Ea=0`....(3)
`a^(2)-4a alpha+(2m)/(ħ^(2))(E+e^(2)a)=0` (4)
From (3) either `a=0`,
In the first case `alpha=(me^(2))/(ħ^(2))=-(ħ^(2))/(2m)alpha^(2)=-(me^(4))/(2ħ^(2))`
This state is the ground state.
`a=alpha-(me^(4))/(8ħ^(2))` and `a= -(1)/(2)(me^(2))/(ħ^(2))`
This state is one with `n=2(2s)`.