Hydrogen Radical Factor
A final Schrodinger-equation, hydrogen atom step gives expressions for the allowed energies and the radical factor of the hydrogen atom wave function.
We have now, finally worked down to one involving the single remaining variable r. this single variable is obtained by setting the left side of equation to β = l (l + 1). We have
-1/R d/dr (r^{2} dR/dr) + 2m_{e}r^{2}/h^{2} [ε – U(r)] = l(l +1)
To proceed we must insert a function for U(r). Notice that up to now it was enough to know that U was a function only of r. the angular factors and their angular momentum implications are therefore independent of the form of U(r).
Now let us think specifically about a one electron atom H or a one-electron ion such as He^{+} or Li^{2+}. The nuclear charge is +Ze, where Z is the atomic number of the atom or ion and –e is the charge of an electron. The potential energy of an electron at a distance r from such a nucleus is –Z[e^{2}/(4∏ε_{0})/r. with this expression for U(r), the radical equation, can be rearranged to
1/r^{2} d/dr (r^{2} dR/dr) + (2m_{e}/h^{2} {ε + Z[e^{2}/(4∏ε_{0})]/r} – l(l + 1)/r^{2})R = 0
And, with the substitution σ =Zr/a_{0} and the expression for a0 from equation simplified to
1/σ^{2} 1/dσ (σ^{2} dR/dσ) + [2m_{e}/h^{2} (a_{0}/Z) 2ε + 2/σ + l(l + 1)/ σ^{2}]R = 0
Again we have an expression for which the solution functions are not easily recognized. Some examples, which can be verified to satisfy, are given here. Each of these radical functions is specified by the value fo the quantum number. Solutions to exist only for n an integer with value at least one unit greater than l. this restriction is usually expressed as
L = 0, 1, 2….. n – 1
As substitution of the solution functions would show, the energy ε is given by
ε = h^{2}/2m_{e}a_{0}^{2} n^{2} 1/n^{2 }= m_{e}Z^{2} [e^{2}/ (4∏ε_{0})] 2/2h^{2} 1/n^{2} n = 1, 2, 3,
This result is equivalent to that produced by the Bohr treatment. Thus we again arrive at an expression for the energies of the states of hydrogen gemlike atoms that agrees with spectral results, as summarized by the Rydberg equation. In so doing, however, we have developed a much more detailed and valid picture of the behaviour of the electron of the atom.
Example: by inspection of the number of hydrogen atom states, some of the hydrogen atom allowed energies suggest an expression for the degeneracies of hydrogen atom energy levels.
Solution: for any value of l there are 2l + l states. These differ in their values of m, these values suggesting different orientations of the angular momentum vector. (Notice that this degeneracy term is the counterpart of g_{1} = 2J + 1 used in the treatment of rotating linear molecules introduced.
Now we recognize that, for each n value, one or more states with different l values occur. For n = 1 there is only an l = 0, or s, state. For n = 2 there are one s state and three l = 1, or p, states, giving a total of four states. For n = 3 there are an s state, three p states, and five l = 2, or d, sates, or nine states in all.
Continuation of this calculation would lead to the recognition that for any hydrogen atom value of n there are n^{2}sates. (When, as after electron spin is recognized, the number of states with a particular n value is doubled, to 2n^{2}).
Example: verify that the radical function for n = 2, l = 1, that is, for the 2p orbital, given in and the allowed energies satisfy the radical part of the wave equation.
Solution: insertion of energy expression gives
1/σ^{2} d/dσ (σ^{2} dR/dσ) + [1/n^{2} + 2/σ + l(l + 1)/ σ^{2}]R = 0
For n = 2 and l = 1 this becomes
1/ σ^{2} d/dσ (σ^{2} dR/dσ) + (-1/4 + 2/σ + 2/ σ^{2})R = 0
Now the 2p radical function can be used. The constant factors will be present in both terms of the above equation and can be canceled out. We can proceed with the simpler function ƒ = σe^{-}^{σ/2}. The first term of the radical equation is developed in the following steps:
Dƒ/dσ = e-σ/2 (1 - σ/2)
σ^{2} dƒ/dσ = e-σ/2 (σ^{2} – σ^{3/2})
d/dσ (σ^{2} dƒ/dσ) = e^{-σ/2} (2σ - 2σ^{2} + σ^{3/4})
1/σ^{2} d/dσ (σ^{2} dƒ/dσ) = e^{-σ/2 }(2/σ – 2 + σ/4)
Now, again with ƒ = σe^{-σ/2 }in place of R, we return to the original equation and obtain
Left side = e-σ/2 in place of R, we return to the original equation and obtain
Left side = e^{-σ/2} (2/σ – 2 + σ/4) + (-1/4 + 2/σ + 2/σ^{2}) σe^{-σ/2}
= e^{-σ/2} (2/σ – 2 + σ/4 – σ/4 + 2 + 2/σ)
= 0
The R^{2}p expression is a solution of the radial part of the H-atom wave function.
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