We set up the Schrödinger equation: $$\label{ec1} \left[-\frac{\hbar^2}{2\mu}\nabla^2-\frac{Ze^2}{r}\right ]\Psi=E\Psi$$ Where $\Psi=R_{n,l}(r)Y_{l,m}(\theta,\varphi)$

Substituting the wave function into the Schödinger equation $$-\frac{\hbar^2}{2\mu}\left(R''+\frac{2}{r}R'\right)+ \frac{l(l+1)\hbar^2}{2\mu r^2}R-\frac{Ze^2}{r}R=ER \label{ec2}$$

We simplify by multiplying all the terms of the equation (\ref{ec2}) by $-\frac{2\mu}{\hbar^2}$

$$R''+\frac{2}{r}R'+\frac{2\mu ER}{\hbar^2}+\frac{2\mu Ze^2}{r\hbar^2}R-\frac{l(l+1)}{r^2}R= 0$$

Putting $a=\frac{\hbar^2}{\mu e^2}$ into the equation (8) gives:

$$\label{ec4} R''+\frac{2}{r}R'+\left[\frac{2E}{ae^2}+\frac{2Z}{ra}-\frac{l(l+1 )}{r^2}\right]R=0$$

We solve the equation (\ref{ec4}) in two cases:

• If $r\rightarrow\infty$, the electron is outside the attraction of the nucleus. Substituting $r$ for $\infty$ in the equation (\ref{ec4}) gives us: $$R''+\frac{2E}{ae^2}R=0 \label{ec5}$$ We propose a solution of the type $R=e^{\alpha r}$, deriving: $R'=\alpha e^{\alpha r}$ and $R''=\alpha^2 e^{ \alpha r}$. Substituting these derivatives into the equation (\ref{ec5}) gives us: $$\alpha^2 e^{\alpha r}+\frac{2E}{ae^2}e^{\alpha r} =0 \label{ec6}$$ Solving $\alpha$ of the equation (\ref{ec6}). $$\alpha=\sqrt{\frac{-2E}{ae^2}} \label{ec7}$$
• Now we solve the Schödinger equation for small values of $r$ (electron close to the nucleus).

In this situation the equation (\ref{ec4}) has the following solution: $$\label{ec8} R=r^se^{-\alpha r}M(r)$$ Where: $$\label{ec9} M(r)=\sum_{j=0}^{\infty}{b_j}r^j$$ Differentiating the function $R$ with respect to the variable $r$ gives us: $$\label{ec10} R'=sr^{s-1}e^{-\alpha r}M(r)+r^s(-\alpha)e^{-\ alpha r}M(r)+r^se^{-\alpha r}M'(r)$$

Grouping terms in (\ref{ec10}):

$$\label{ec11} R'=\left[(s-\alpha r)r^{s-1} M(r)+r^s M '(r)\right]e^{-\alpha r}$$

Differentiating $R'$ we get $R''$

\begin{eqnarray}\label{ec12} R''=-\alpha r^ {s-1} e^{-\alpha r} M(r)+(s-\alpha r)(s-1)r^{s-2}e^{-\alpha r} M(r)+ (s-\alpha r)r^{s-1}(-\alpha)e^{-\alpha r}M(r) \nonumber\\ +(s-\alpha r)r^{s-1} e^{-\alpha r}M'(r)+sr^{s-1}e^{-\alpha r} M'(r)+r^s (-\alpha)e^{-\alpha r } M'(r) +r^se^{-\alpha r} M''(r) \end{eqnarray}

Factoring out a common factor of $e^{-\alpha r}$ and grouping by derivatives of $M( r)$:

$$\label{ec13} R''=\left[r^s M''(r)+2(s-\alpha r)r^{s-1} M'(r )+(\alpha^2 r^2-2\alpha{rs}+s^2-s)r^{s-2} M(r)\right]e^{-\alpha r}$$

Substituting into the radial Schödinger equation (\ref{ec4}) the equations (\ref{ec7}), (\ref{ec11}) and (\ref{ec13})

\begin{eqnarray}\label{ec14} \left[r^s M''(r)+2(s-\alpha r)r^{s-1} M'(r)+(\alpha^2 r^2-2\alpha{rs}+ s^2-s)r^{s-2} M(r)\right]e^{-\alpha r} \nonumber\\ +\frac{2}{r}\left[(s-\alpha r)r^{s-1} M(r)+r^s M'(r)\right]e^{-\alpha r}+\left(\frac{2Z}{ar}-\alpha^2-\frac{l(l+1)}{r^2}\right){r^s}e^{-\alpha r} M(r)=0 \end{eqnarray}

Dividing the equation (\ref{ec14}) by $r^{s-2}$

\begin{eqnarray}\label{ec15} r^2 M''(r)+2(sr-\alpha r^2+ r)M'(r)+(\cancel{\alpha^2 r^2}-2\alpha rs+s^2-\cancel{s}+\frac{2Zr}{a}-\cancel{\alpha ^2 r^2}-l(l+1)+\cancel{2}s \nonumber\\ -2\alpha r)M(r)=0 \end{eqnarray}

Simplifying:

$$\label {ec16} r^2 M''(r)+2(sr-\alpha r^2 +r)M'(r)+\left[s(s+1)-2\alpha r(s+1) +\frac{2Zr}{a}-l(l+1)\right]M(r)=0$$

Equation (\ref{ec16}) must hold for $r=0$.

$$\label{ec17} s(s+1)-l(l+1)=0$$

The equation (\ref{ec17}) tells us that $s=l$ Substituting $s$ times $l$ in the equation (\ref{ec16}) is obtained:

$$\label{ec18} r^2 M''(r)+2(lr-\alpha r^2+r) M'(r)+\left[\cancel{l(l+1)}-2\alpha r(l+1)+\frac{2Zr}{a}-\cancel{l(l+1)}\right]M(r)=0$$

Dividing all terms by $r$ in the equation (\ref{ec18}) gives:

$$\label{ec19} rM''(r) +2(l+1-\alpha r)M'(r)+\left[\frac{2Z}{a}-2\alpha(l+1)\right]M(r)=0$$

Now we proceed to calculate the derivatives of the polynomial $M(r)$ to substitute them in the previous equation.

\begin{eqnarray} M(r) &=& \sum_{j=0}^{\infty}b_{j}r^j \\ M'(r) &=& \sum_{j=1}^{ \infty}b_{j}jr^{j-1}=\sum_{j=0}^{\infty}b_{j+1}(j+1)r^j \\ M''(r) & =& \sum_{j=2}^{\infty}b_{j+1}j(j+1)r^{j-1}=\sum_{j=0}^{\infty}b_{j+ 1}j(j+1)r^{j-1} \label{ec22} \end{eqnarray}

Substituting into equation (\ref{ec19}):

$$\label{ec23} r\sum_ {j=0}^{\infty}b_{j+1}j(j+1)r^{j-1}+2(l+1)\sum_{j=0}^{\infty}b_{ j+1}(j+1)r^j-2\alpha r\sum_{j=0}^{\infty}b_{j}jr^{j-1}+\left(\frac{2Z}{ a}-2\alpha(l+1)\right)\sum_{j=0}^{\infty}b_{j}r^{j}=0$$

Grouping terms in (\ref{ec23 })

$$\label{24} \sum_{j=0}^{\infty}\left[b_{j+1}j(j+1)+2(l+1)(j+1 )b_{j+1}-2\alpha jb_{j}+\left(\frac{2Z}{a}-2\alpha(l+1)\right)b_{j}\right]r^j= 0$$

Equating the bracket to zero and solving for $b_{j+1}$ gives the recurrence law

$$\label{ec25} b_{j+1}=\frac{2\alpha j-2Z/a+2\alpha(l+1)}{(j+1)(j+2l+2)}b_{j}$$

For $M(r)$ to be valid in mechanics quantum trun is necessary set the series in the term $j=k$, so that $b_k$ is the last non-zero term of the expansion and $b_{k+1}=0$.

Setting $j=k$ in the law of recurrence gives:

$$\label{ec26} b_{k+1}=\frac{2\alpha k-2Z/a+2\alpha(l+1)}{(k+1)(k +2l+2)}b_{k}$$

For this equation to hold, the numerator

$$\label{ec27} 2\alpha k-2Z/a+2\alpha(l+1)=0$$

Factoring out the common factor of $2\alpha$:

$$\label{ec28} 2\alpha(l+k+1)=\frac{2Z}{a}$$

Calling $n=l+k+1$ leaves us with:

$$\label{ec29} \alpha n=\frac{Z}{a}$$

Substituting $\alpha$ for the ( \ref{ec7})

\begin{eqnarray} \sqrt{\frac{-2E}{ae^2}} &=& \frac{Z}{na} \frac{-2E}{ae^2} &= & \frac{Z^2}{a^2 n^2} \\ E &=& -\frac{Z^2 e^2}{2an^2} \label{ec32} \end{eqnarray}

Substituting $a$ by its value, the following expression is obtained for the energy of the hydrogen atom.

$$\label{ec33} E=-\frac{Z^2 \mu e^4}{2\hbar^2 n^2}$$

Starting from the equation (\ref{ec8} ) gives the radial part of the wave function

$$\label{ec34} R_{n,l}(r)=r^le^{Zr/na}\sum_{j=0}^{nl -1}b_{j}r^j$$