Radial Function (Hydrogen Atom) | Quantum Science Philippines
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Radial Function (Hydrogen Atom)

Derive \big[\frac{d^{2}}{d\rho^{2}}-\frac{l(l+1)}{\rho^{2}}+\frac{2}{\rho}-\lambda_{k,l}^{2}\big]U_{k,l}(\rho)=0 .

Solution:

The radial equation is given by

\Big[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+\frac{l(l+1)\hbar^{2}}{2\mu r^{2}}+V(r)\Big]U_{k,l}(\rho)=E_{k,l}U_{k,l}

But for a hydrogen atom, V(r)=-\frac{e^{2}}{r}. Substituting this into the equation above and introducing dimensionless variables \rho=\frac{r}{a_{0}} or r=\rho a_{0} and \lambda_{k,l}=\sqrt{-\frac{E_{k,l}}{E_{I}}} or E_{k,l}=\lambda_{k,l}^{2}E_{I}, where E_{I} is the ionization energy, we have

\Big[-\frac{\hbar^{2}}{2\mu}\frac{1}{a_{0}}\frac{d^{2}}{d\rho^{2}}+\frac{l(l+1)\hbar^{2}}{2\mu r^{2}}\frac{1}{a_{0}^{2}}-\frac{e^{2}}{\rho a_{0}}+\lambda_{k,l}^{2}E_{I}\Big]U_{k,l}(\rho)=0

Recall from Bohr’s model E_{I}=\frac{\mu e^{4}}{2\hbar} and a_{0}=\frac{\hbar^{2}}{\mu e^{2}} . Substituting this to the above equation, we get

\Big[-\frac{\mu e^{4}}{2\hbar^{2}} \frac{d^{2}}{d\rho^{2}} + \frac{\mu e^{4}}{2\hbar^{2}} \frac{l\big(l+1\big)}{\rho^{2}} - \frac{\mu e^{4}}{\hbar^{2}} \frac{1}{\rho} + \frac{\mu e^{4}}{2\hbar^{2}} \lambda_{k,l}^{2}\Big] U_{k,l}(\rho)=0 .

Multiplying both sides by \frac{\hbar^{2}}{\mu e^{4}}, we obtain

\Big[\frac{d^{2}}{d\rho^{2}}-\frac{l(l+1)}{\rho^{2}}+\frac{2}{\rho}-\lambda_{k,l}^{2}\Big]U_{k,l}(\rho)=0.

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