Radial Function For Hydrogen Atom | Quantum Science Philippines

## Radial Function For Hydrogen Atom

Problem:

Derive the radial function of hydrogen atom, $R_{20}$.

Solution:

Given the equation

$R_{kl}(r)=\frac{1}{r}\bigg(\frac{r}{a_{0}}\bigg)^{l+1}\sum_{q=0}^{k}C_{q}\bigg(\frac{r}{a_{0}}\bigg)^{q}e^{-\frac{r}{a_{0}(k+l)}}$,

For $k=2$ and $l=0$:

$R_{kl}(r)=R_{20}(r) =\frac{1}{r}\bigg(\frac{r}{a_{0}}\bigg)\sum_{0}^{2}C_{q}\bigg(\frac{r}{a_{0}}\bigg)^{q}e^{-\frac{r}{2a_{0}}}$ $R_{20}(r) = \frac{1}{a_{0}}C_{0}e^{-\frac{r}{2a_{0}}}+\frac{1}{a_{0}}C_{1}\bigg(\frac{r}{a_{0}}\bigg)e^{-\frac{r}{2a_{0}}}+\frac{1}{a_{0}}C_{2}\bigg(\frac{r}{a_{0}}\bigg)^{2}e^{-\frac{r}{2a_{0}}}$

Now, the recursion relation is given as

$C_{q}=-\frac{2(k-q)}{q(q+2l+1)(k+l)}C_{q-1};$
Hence,
$C_{1}=-\frac{C_{0}}{2}$ andÂ  $C_{2}=0.$

Thus,
$R_{20}(r)=\frac{C_{0}}{a_{0}}\Bigg(1-\frac{r}{2a_{0}}\Bigg)e^{-\frac{r}{2a_{0}}}=C\Bigg(1-\frac{r}{2a_{0}}\Bigg)e^{-\frac{r}{2a_{0}}}$

where $C=\frac{C_{0}}{a_{0}}$

Solving for the normalization constant, C:

$\int_{0}^{ \infty }r^{2}R_{20} ^{\ast} R_{20}dr=1$ $\int_{0}^{\infty }r^{2}|C|^{2}\Bigg(1-\frac{r}{2a_{0}}\Bigg)^{2}e^{-\frac{r}{a_{0}}}dr=1$ $|C|^{2}\int_{0}^{\infty }r^{2}\Bigg(1-\frac{r}{a_{0}}+\frac{r^{2}}{4a_{0}^{2}}\Bigg)e^{-\frac{r}{a_{0}}}dr=1$ $\frac{|C|^{2}}{4a_{0}^{2}}\Bigg(4a_{0}^{2}\int_{0}^{\infty }r^{2}e^{-\frac{r}{a_{0}}}dr-4a_{0}\int_{0}^{\infty }r^{3}e^{-\frac{r}{a_{0}}}dr+\int_{0}^{\infty }r^{4}e^{-\frac{r}{a_{0}}}dr\Bigg)=1$ $\frac{|C|^{2}}{4a_{0}^{2}}\bigg(8a_{0}^{5}-24a_{0}^{5}+24a_{0}^{5}\bigg)=1$ $C=\frac{1}{\sqrt{2a_{0}^{3}}}$

Therefore,

$R_{20}(r)=\frac{1}{\sqrt{2a_{0}^{3}}}\Bigg(1-\frac{r}{2a_{0}}\Bigg)e^{-\frac{r}{2a_{0}}}$
$=\frac{2}{\sqrt{2^{3}a_{0}^{3}}}\Bigg(1-\frac{r}{2a_{0}}\Bigg)e^{-\frac{r}{2a_{0}}}$
$R_{20}(r)=2\big(2a_{0}\big)^{\frac{-3}{2}}\Bigg(1-\frac{r}{2a_{0}}\Bigg)e^{-\frac{r}{2a_{0}}}$