Radial Function (Hydrogen Atom) | Quantum Science Philippines

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