The Full Hydrogen Atom Wavefunctions

The form taken by the 6 function depends upon the two quantum numbers, / and m,. 

When 1 = 0, must also equal zero, and the 6 function becomes a constant. Thus, there is no angular variation to the wavefunction, and we obtain an s orbital. When / = I, ntj can have the values 0, 1. For nil = 0, 0(0) = COS0 and we obtain a 2p, orbital, but we do not get the familiar 2p and 2p, orbitals when m,= 1. Instead, we obtain doughnut-shaped orbitals in which the electrons can be circulating in either a clockwise or an anti-clockwise direction, and the wavefunction now has a complex component. 

The energies are obtained by solving the Schrodinger equation (equation 6.9). It is found that only the principal quantum number n is involved in determining the energy, which is given by the equation  

The mathematics involved in obtaining this equation are quite lengthy, and will not be discussed here. A relatively simple way of obtaining the energy when n = 1 and Z =1 has already been given in Worked Problem 6.3, 

The rotational motion of the electron around the nucleus is quantized in a similar way to that of a particle on the surface of a sphere, which was described in Chapter 5. The total angular momentum of the electron is equal to + 1)] , and the component in the z direction is equal 

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