\documentclass{article}

%% Created with wxMaxima 14.12.1

\setlength{\parskip}{\medskipamount}
\setlength{\parindent}{0pt}
\usepackage[utf8]{inputenc}
\usepackage{graphicx}
\usepackage{color}
\usepackage{amsmath}

\definecolor{labelcolor}{RGB}{100,0,0}

\begin{document}

\pagebreak{}
{\Huge {\sc hydrogen2sb.mac, corrected, from Steeb \& Hardy Ch.26}}
\setcounter{section}{0}
\setcounter{subsection}{0}
\setcounter{figure}{0}



\subsection{This script allows to find energy levels of Hydrogen atom from Schroedinger equation,
and here we illustrate how to use physical constants in Maxima}



\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i1)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
load (physical_constants);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
% \begin{math}\displaystyle
{\it
Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp.

Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp.

Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp.

Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp.

Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3 Finished compiling /tmp/gazonk\_4257\_0.lsp.

Compiling /tmp/gazonk\_4257\_0.lsp.End of Pass 1.  End of Pass 2.

OPTIMIZE levels: Safety=2, Space=3, Speed=3Finished compiling /tmp/gazonk\_4257\_0.lsp.
}
% \end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o1) }
/usr/share/maxima/5.32.1/share/ezunits/physical\_constants.mac
\end{math}
%%%%%%%%%%%%%%%

Let us use notation from S.Fluegge ``Practical Quantum Mechanic'', vol.1, problem 61.
Kepler's problem for infinetely heavy point-like nucleus of charge Z. For Z=1 we have hydrogen atom.
Denote:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i2)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
2*m0*E/hb^2 = -%gamma^2;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o2) }
\frac{2\,m0\,E}{{hb}^{2}}=-{\gamma}^{2}
\end{math}
%%%%%%%%%%%%%%%

 here hb denotes $\hbar$, i.e. Planck constant divided by $2\pi$.  Let

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i3)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
kappa = Z*e0^2*m0/(%gamma*hb^2);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o3) }
\kappa=\frac{{e0}^{2}\,m0\,Z}{\gamma\,{hb}^{2}}
\end{math}
%%%%%%%%%%%%%%%

Now for usual substitution in wavefunction $u(r,\theta,\phi)$ to separate variables:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i4)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
u = (1/r) * chi(r) * Ylm(theta,phi);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o4) }
u=\frac{\chi\left( r\right) \,\mathrm{Ylm}\left( \theta,\phi\right) }{r}
\end{math}
%%%%%%%%%%%%%%%

we write the Schroedinger equation as (with $y$ in place of $\chi$):

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i5)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
'diff(y, r, 2)  + (-%gamma^2 + 2*%gamma*kappa/r-l*(l+1)/r^2)* y = 0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o5) }
\frac{{d}^{2}}{d\,{r}^{2}}\,y+\left( \frac{2\,\gamma\,\kappa}{r}-\frac{l\,\left( l+1\right) }{{r}^{2}}-{\gamma}^{2}\right) \,y=0
\end{math}
%%%%%%%%%%%%%%%

Define orbital number $l$, magnetic $m$, and radial $k$, e.g.:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i6)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
l: 1; m: 0; k: 1;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o6) }
1
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o7) }
0
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o8) }
1
\end{math}
%%%%%%%%%%%%%%%

now the principal quantum number n is

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i9)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
n:k+l+1;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o9) }
3
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i10)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
alpha: e0^2/(hb*c);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o10) }
\frac{{e0}^{2}}{c\,hb}
\end{math}
%%%%%%%%%%%%%%%

alpha is fine structure constant.
Now substitute a polynomial $v1$ and $\exp$ as an ansatz for $y$, where

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i11)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
v1: (sum (a [j] *r^j, j, 1 ,k));
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o11) }
{a}_{1}\,r
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i12)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
v1: v1+1;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o12) }
{a}_{1}\,r+1
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i13)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
y1: r^(l+1)*exp(-sqrt(g2)*r)*v1;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o13) }
{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}
\end{math}
%%%%%%%%%%%%%%%

 $y1$ is the trial function of the radial wave function.
 Insert the ansatz into Schroedinger differential equation

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i14)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
left: diff(y1,r,2)+(-g2+2*m0*c*alpha/(hb*r)-l*(l+l)/(r*r))*y1;
left: left*exp(sqrt(g2)*r);
left: left/r;
left: ratsimp(left);
left: num(left);
left: expand(left);
left: subst(x,sqrt(g2),left);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o14) }
-2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r}-\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}-4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o15) }
\left( -2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r} \\
-\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}-4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r}\right) \,{e}^{\sqrt{g2}\,r}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o16) }
\frac{\left( -2\,{a}_{1}\,\sqrt{g2}\,{r}^{2}\,{e}^{-\sqrt{g2}\,r}+\left( \frac{2\,{e0}^{2}\,m0}{{hb}^{2}\,r}-\frac{2}{{r}^{2}}-g2\right) \,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+g2\,{r}^{2}\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r} \\
-4\,\sqrt{g2}\,r\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+2\,\left( {a}_{1}\,r+1\right) \,{e}^{-\sqrt{g2}\,r}+4\,{a}_{1}\,r\,{e}^{-\sqrt{g2}\,r}\right) \,{e}^{\sqrt{g2}\,r}}{r}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o17) }
-\frac{\sqrt{g2}\,\left( 6\,{a}_{1}\,{hb}^{2}\,r+4\,{hb}^{2}\right) -2\,{a}_{1}\,{e0}^{2}\,m0\,r-2\,{e0}^{2}\,m0-4\,{a}_{1}\,{hb}^{2}}{{hb}^{2}}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o18) }
-\sqrt{g2}\,\left( 6\,{a}_{1}\,{hb}^{2}\,r+4\,{hb}^{2}\right) +2\,{a}_{1}\,{e0}^{2}\,m0\,r+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o19) }
2\,{a}_{1}\,{e0}^{2}\,m0\,r-6\,{a}_{1}\,\sqrt{g2}\,{hb}^{2}\,r+2\,{e0}^{2}\,m0-4\,\sqrt{g2}\,{hb}^{2}+4\,{a}_{1}\,{hb}^{2}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o20) }
-6\,{a}_{1}\,{hb}^{2}\,r\,x-4\,{hb}^{2}\,x+2\,{a}_{1}\,{e0}^{2}\,m0\,r+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2}
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i21)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
h[0]: coeff(left,r,0);
h[1]: coeff(left,r,1);
h[2]: coeff(left,r,2);
h[3]: coeff(left,r,3);
h[4]: coeff(left,r,4);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o21) }
-4\,{hb}^{2}\,x+2\,{e0}^{2}\,m0+4\,{a}_{1}\,{hb}^{2}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o22) }
2\,{a}_{1}\,{e0}^{2}\,m0-6\,{a}_{1}\,{hb}^{2}\,x
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o23) }
0
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o24) }
0
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o25) }
0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i26)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
s: solve ([h[0]=0, h[1]=0, h[2]=0, h[3] =0, h[4]=0],
   [x,a[1],a[2],a[3],a[4]]);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o26) }
[[x=\frac{{e0}^{2}\,m0}{2\,{hb}^{2}},{a}_{1}=0,{a}_{2}=\%r1,{a}_{3}=\%r2,{a}_{4}=\%r3],[x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}},{a}_{1}=-\frac{{e0}^{2}\,m0}{6\,{hb}^{2}},{a}_{2}=\%r4,{a}_{3}=\%r5,{a}_{4}=\%r6]]
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i27)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
sol: last(s);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o27) }
[x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}},{a}_{1}=-\frac{{e0}^{2}\,m0}{6\,{hb}^{2}},{a}_{2}=\%r4,{a}_{3}=\%r5,{a}_{4}=\%r6]
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i28)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
sol: first(sol);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o28) }
x=\frac{{e0}^{2}\,m0}{3\,{hb}^{2}}
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i29)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
sol: rhs(sol);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o29) }
\frac{{e0}^{2}\,m0}{3\,{hb}^{2}}
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i30)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
E_kl: -hb^2*sol*sol/2/m0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o30) }
-\frac{{e0}^{4}\,m0}{18\,{hb}^{2}}
\end{math}
%%%%%%%%%%%%%%%

 energy eigenvalue for k, l

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i31)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
print ("E_kl=" ,E_kl);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
E\_kl=-\frac{{e0}^{4}\,m0}{18\,{hb}^{2}}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o31) }
-\frac{{e0}^{4}\,m0}{18\,{hb}^{2}}
\end{math}
%%%%%%%%%%%%%%%

The output is the energy eigenvalue E\_kl.
Now we may use maxima package "physical\_constant" which has a set of many dimensional constants.
It uses package ezunits which by default employs SI.
One can try
demo(ezunits);
display\_known\_unit\_conversions;

We can list the constants:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i32)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
propvars (physical_constant);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o32) }
[\%c,\%mu\_0,\%e\_0,\%Z\_0,\%G,\%h,\%h\_bar,\%m\_P,\%\%k,\%T\_P,\%l\_P,\%t\_P,\\
\%\%e,\%Phi\_0,\%G\_0,\%K\_J, \\
\%R\_K,\%mu\_B,\%mu\_N,\alpha,\%R\_inf,\%a\_0,\%E\_h,\%ratio\_h\_me,\%m\_e,\%N\_A,\%m\_u,\%F,\%R,\%V\_m,\\
\%n\_0,\%ratio\_S0\_R,\sigma,\%c\_1,\%c\_1L,\%c\_2,\%b,\%b\_prime]
\end{math}
%%%%%%%%%%%%%%%

And print some of them, e.g.

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i33)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
constvalue(%alpha);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o33) }
0.0072973525376
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i34)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
constvalue (%c);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o34) }
299792458 ` \frac{m}{s}
\end{math}
%%%%%%%%%%%%%%%

we may copy paste this SI value into CGS value for speed of light (adding two orders):

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i35)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
c0:29979245800;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o35) }
29979245800
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i36)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
constvalue (%h_bar);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o36) }
\frac{41412931}{125000000000000000000000000000000000000000\,\pi } ` s\,J
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i37)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
float(%);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o37) }
1.054571628251774\,{10}^{-34} ` s\,J
\end{math}
%%%%%%%%%%%%%%%

we may copy paste this SI value into CGS value (7 orders difference in J and erg):

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i38)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
hb:1.054571628251774e-27;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o38) }
1.054571628251774\,{10}^{-27}
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i39)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
constvalue(%%e);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o39) }
\frac{1602176487}{10000000000000000000000000000} ` C
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i40)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
float(%);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o40) }
1.602176487\,{10}^{-19} ` C
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i41)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
e0:1.602176487e-19*c0/10;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o41) }
4.80320427187535\,{10}^{-10}
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i42)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
constvalue(%m_e);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o42) }
9.10938215\,{10}^{-31} ` kg
\end{math}
%%%%%%%%%%%%%%%

In the same way

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i43)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
m0:9.10938215e-28;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o43) }
9.10938215\,{10}^{-28}
\end{math}
%%%%%%%%%%%%%%%

Now we get the energy of the level in ergs

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i44)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
E: ev(E_kl);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o44) }
-2.4220799671188296\,{10}^{-12}
\end{math}
%%%%%%%%%%%%%%%

and transform to electron-Volts (do not confuse evaluation operator ev above with eV)

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i45)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
load("physconst.mac");
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o45) }
/usr/share/maxima/5.32.1/share/physics/physconst.mac
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i46)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
float(%Ry);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o46) }
\frac{1.0973731568549\,{10}^{7}}{m}
\end{math}
%%%%%%%%%%%%%%%

Here we get Rydberg in inverse meters. Next line gives its uncertainty:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i47)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
 get(%Ry, RSU);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o47) }
7.6\,{10}^{-12}
\end{math}
%%%%%%%%%%%%%%%

We better get Ry in eV. We must multiply electron charge e0 by one Volt, expressed in CGS. The unit of voltage in CGS is
1 statvolt = 299.792458 volts.
(The conversion factor 299.792458 is simply the numerical value of the speed of light in cm/s divided by 10\verb|^|8). Hence,

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i48)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
eV0:e0/(c0*1e-8);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o48) }
1.602176487\,{10}^{-12}
\end{math}
%%%%%%%%%%%%%%%

This is the value of eV in ergs, now we have the energy of our level in eV:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i49)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
E_eV:E/eV0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o49) }
-1.511743548086928
\end{math}
%%%%%%%%%%%%%%%

Since Rydberg is

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i50)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
Ry0:m0*e0^4/hb^2;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o50) }
4.3597439408138926\,{10}^{-11}
\end{math}
%%%%%%%%%%%%%%%

in ergs, and in eV

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i51)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
Ry0eV:Ry0/eV0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o51) }
27.2113838655647
\end{math}
%%%%%%%%%%%%%%%

We obtain the energy of our level in Rydbergs:

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i52)
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
E/Ry0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o52) }
-0.055555555555555
\end{math}
%%%%%%%%%%%%%%%

It is easy to check that we get $-1/(2n^2)$ when the principal quantum number of the level is $n$.
\end{document}
