Ground State of Hydrogen

Relevant wiki: Hydrogen Atom

The ground state of hydrogen corresponds to $n=1$ , $\ell = 0$ . The radial solutions as given in the relevant wiki are:

$R_{n \ell} (r) = \frac{1}{r} \rho^{\ell + 1} e^{-\rho} v(\rho).$

Plugging in these values of $n$ and $\ell$ gives:

$R_{10} (r) = \frac{1}{r} \rho e^{-\rho} v(\rho).$

The function $v(\rho)$ is defined by the series:

$v(\rho) = \sum_{j=0}^{\infty} c_j \rho^j, \qquad c_{j+1} = \frac{2(j+ \ell + 1 - n)}{(j+1)(j + 2\ell + 2)} c_j.$

All $c_j$ are zero after $j_{\text{max}} = n-\ell - 1$ . In this case, all $c_j$ are thus zero after $c_0$ , so $v(\rho) = 1$ up to normalization, and

$R_{10} (r) = \frac{1}{r} \rho e^{-\rho}.$

Normalizing, should have:

$\int_0^{\infty} C^2 (\frac{1}{r} \rho e^{-\rho})^2 r^2 dr = 1 \implies \frac{C^2}{4} \frac{r}{\rho} = 1 \implies C = 2\sqrt{\frac{\rho}{r}}.$

Therefore, the solution is given by:

$R_{10} (r) =2\sqrt{\frac{\rho}{r}} \frac{\rho}{r} e^{-\rho},$

which is equivalent to the listed answer.