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%\subsubsection*{Notice}
%{\footnotesize In this assignment I'll refer to equations, theorems, definitions and so forth. When I do this, I don't make any references to where they're coming from, but -- unless I state anything else -- they're from the book used in the course, \cite{atomintro}. 
%
%I'll write vectors with bold fonts, to make the assignment easier to read. }


\setcounter{subsection}{1}
\subsection*{The assignment}
\ramme{ The Schrödinger equation for an electron of mass $m$ and charge $-e$, bound to an infinitely heavy nucleus of charge $e$, is
        \begin{equation}
         \left( -\frac{\hbar^2}{2m} \nabla^2 - \frac{e^2}{4\pi\epsilon_0 r} \right) \psi (\vec r) = E \psi(\vec r) .
         \label{schroedinger}
        \end{equation}\vspace{-0.2cm}}

\subsubsection*{Question a}\setcounter{thespnr}{1}
\ramme{ Indicate how this equation must be modified in order to account for singly ionised helium (\Hefireplus). Write 
        the equation in a form which is correct to first order in the ratio of the mass of the electron to the mass of the nucleus. }
I have to take into account that the nucleus has double the charge; namely $2e$. Hence the Coulomb potential, which for the nucleus of charge $e$ was $V(\vec r) = -e^2/4\pi\epsilon_0 r$, becomes instead\footnote{\label{foot1}In general, for a single electron of charge $-e$, bound to a nucleus of charge $Ze$, the Coulomb potential is $V(\vec r) = -Z e^2/4\pi\epsilon_0 r$, according to simple electrodynamics (see for example \citet{griffithsel}).} 
\begin{equation}
 V(\vec r) = -\frac{2 e^2}{4\pi\epsilon_0 r} .
 \label{potential}
\end{equation}

Apart from the changed potential, I also have to take a look at the mass $m$ (from equation \eqref{schroedinger}): Instead of only having the mass of a single electron as $m$, I have to use the reduced mass, given as
\begin{equation}
 m = \frac{m_e M_p A}{m_e + M_p A} ,
 \label{reducedmassground}
\end{equation}
where $m_e$ is the mass of an electron, $M_p$ is the mass of a proton (as well as the mass of a neutron), and $A$ is the number of protons and neutrons in the nucleus (this equation is a changed form of equation (1.13) from \cite{atomintro}). So for \Hefireplus, which has two protons and two neutrons, and hence $A = 4$, the reduced mass is
\begin{equation*}
 m = \frac{4 m_e M_p}{m_e + 4 M_p} 
   = \frac{4 m_e}{\dfrac{m_e}{M_p} + 4}
   = m_e \left( 1 + \frac{m_e}{4M_p} \right)^{-1} ,
\end{equation*}
and since $m_e \ll 4M_p$, I can expand to first order in $m_e/4M_p$ (exactly the ratio of the mass of the electron to the mass of the nucleus, as wanted from the question-text), to get that
\begin{equation}
 m \approx m_e \left( 1 - \frac{m_e}{4M_p} \right) .
 \label{reducedmass}
\end{equation}

Hence the Schrödinger equation in \eqref{schroedinger} can be modified to fit the \Hefireplus{} ion to first order in the ratio of the mass of the electron to the mass of the nucleus, by using the potential in \eqref{potential} and the reduced mass in \eqref{reducedmass}, to be
\begin{equation*}
 \left( - \left( \frac{\hbar^2}{2m_e} + \frac{2 M_p \hbar^2}{m_e^2} \right) \nabla^2 
        - \frac{2 e^2}{4\pi\epsilon_0 r} \right) \psi (\vec r) = E \psi(\vec r) .
\end{equation*}



\clearpage
\subsubsection*{Question b}\setcounter{thespnr}{2}
\ramme{ Give the expression for the energy levels of \Hefireplus{} and find the energy of the $n=3$ level in J and eV. }
To find the expression for the energy levels of \Hefireplus{}, I simply use the expression in (1.10) and (1.9) from \cite{atomintro}, with $e^2$ replaced\footnote{Just as in footnote \vref{foot1}, the general case is simply to replace $e^2$ by $Ze^2$.} by $2e^2$ (and of course I use the reduced mass $m$ in \eqref{reducedmassground} instead of the electron mass $m_e$). Then I get the expression
\begin{equation}
 E_n = -\frac{m}{2 \hbar^2} \left( \frac{2e^2}{4\pi\epsilon_0} \right)^2 \frac{1}{n^2} 
     = -\frac{8 m_e M_p}{\hbar^2 (m_e + 4M_p)} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 \frac{1}{n^2} .
 \label{energies}
\end{equation}

By looking up all the constants and inserting these along with the value $n=3$ into \eqref{energies}, I then get that the energy for the $n=3$ level in \Hefireplus{} is
\begin{align*}
 E_3 &= -\frac{8 \cdot \meValue \cdot \MpValue}{(\hbarValue)^2 \cdot ( \meValue + 4 \cdot \MpValue )} \\
     &\qquad\qquad    \cdot\left( \frac{(\eValue)^2}{4\pi\cdot \epsilonnulValue} \right)^2 \frac{1}{3^2} \\
     &= - 9.687\E{-19} \,\text{J} .
\end{align*}
To get this value in electron volts I simply use the rule that $1 \,\text{eV} = 1.602\E{-19} \,\text{J}$, so the answer in electron volts is $E_3 = -6.046 \,\text{eV}$. 


\subsubsection*{Question c}\setcounter{thespnr}{3}
\ramme{ Find the wavelength of a photon which is just capable of removing the remaining electron from the $n=3$ level. }
I know that $E = h c /\lambda$, so I simply get the wavelength in terms of the energy as the following
\begin{equation*}
 \lambda = \frac{hc}{E} .
\end{equation*}

Since I need the electron to be removed from the $n=3$ level, for which I found the energy in question b, I simply have to insert $E = 9.687\E{-19} \,\text{J}$, and hence I get that a photon just capable of removing the electron from the level $n=3$ (ionizing the \Hefireplus{} to \Hefireplusplus) has the wavelength
\begin{equation*}
 \lambda = \frac{\hValue \cdot \cValue}{9.687\E{-19} \,\text{J}} 
         = 2.051\E{-7} \,\text{m}
         = 205.1 \,\text{nm} .
\end{equation*}



\subsubsection*{Question d}\setcounter{thespnr}{4}
\ramme{ The 4\textsuperscript{th} harmonic of a laser with wavelength 808 nm is used for photo-ionizing He\textsuperscript{+} to
        He\textsuperscript{++}. Find the velocity of the liberated electron. }
The energy of the laser light is $E_L = h c /\lambda_L$, where $\lambda_L$ is the $808 \,\text{nm}$ divided by four (because it is the 4\textsuperscript{th} harmonic). When ionizing the \Hefireplus{} into \Hefireplusplus, only the energy $E_3$, found in question b, is needed. The rest of the energy goes to kinetical energy of the electron, so
\begin{equation}
 E_L - E_3 = \frac{1}{2} m_e v^2   \qquad\Rightarrow\qquad
 |v| = \sqrt{\frac{2(E_L - E_3)}{m_e}}
     = \sqrt{\frac{2 \left( \dfrac{h c}{\lambda_L} - E_3 \right) }{m_e}} ,
 \label{velocity}
\end{equation}
where $m_e$ is again the mass of the electron, and $|v|$ here is the velocity of the electron (of course without a direction to it). 

So to get the velocity of the electron, I simply insert all my given values into equation \eqref{velocity}, by which I get the value
\begin{equation*}
 |v| = \sqrt{\frac{2 \left( \dfrac{\hValue \cdot \cValue}{\frac{1}{4} \cdot 808\E{-9} \,\text{m}} - 9.687\E{-19} \,\text{J} \right) }{\meValue}}
     = 1.7944\E{5} \,\text{m/s} .
\end{equation*}









\subsubsection*{Question e}\setcounter{thespnr}{5}
\ramme{ List in the notation $\ket{nlm}$ all the states with $n=3$, and give the total degeneracy, excluding spin. }
The azimuthal quantum number $l$ must obey the rule that it is in the set 
\begin{equation*}
 \{ 0,1,2,\ldots ,n-1 \} ,
\end{equation*} 
so because I here have $n=3$, $l$ can take the values $0$, $1$ and $2$. For all these $l$-values, the magnetic quantum number $m$ must obey the rule that it is in the set 
\begin{equation*}
 \{ -l, -l+1, \ldots, -1, 0, 1, \ldots, l-1, l \} .
\end{equation*}
So for $l=0$, $m$ can be only $0$; for $l=1$, $m$ can be $-1$, $0$ and $1$; and for $l=2$, $m$ can be $-2$, $-1$, $0$, $1$ and $2$. 

Concluding, I hence for $n=3$ have the nine states
\begin{equation*}
 \ket{300}, \;
 \ket{31,-1}, \;
 \ket{310}, \;
 \ket{311}, \;
 \ket{32,-2}, \;
 \ket{32,-1}, \;
 \ket{320}, \;
 \ket{321}, \;
 \ket{322} ,
\end{equation*}
which means that I have $9$-fold degeneracy\footnote{This can be checked by noting that $n^2 = 9$, which is exactly the degeneracy of the $n$\textsuperscript{th} state according to the calculation on page 28 of \citet{dias090421}.} (all of the nine states above have the same energy, since the energy is only dependent on the principal quantum number $n$). 



\subsubsection*{Question f}\setcounter{thespnr}{6}
\ramme{ Sketch for all states with $m=0$ the angular and the radial dependence of the wave function in the $xz$-plane, taking the $z$-axis as the
        quantisation axis. }
For the first of the three states with $m=0$ ($\ket{300}$, $\ket{310}$ and $\ket{320}$), $\ket{300}$, I know that the wave function is $\psi_{300}(r, \theta, \phi) = R_{30}(r) Y_{00}(\theta, \phi)$, where the radial wave function $R_{30}(r)$ is given by\footnote{I have found all the radial functions in table 2.2 in \cite{atomintro}.}
\begin{equation*}
 R_{30}(r) = \left( \frac{Z}{3 a_0} \right)^{3/2} 2 \left( 1 - 2 \rho + \frac{2}{3} \rho^2 \right) e^{-\rho} ,
\end{equation*}
where $\rho = Z r / n a_0$ and $Z = 2$ for this atom, so
\begin{equation}
 R_{30}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} 2 \left( 1 - \frac{4}{3} \frac{r}{a_0} + \frac{8}{27} \frac{r^2}{a_0^2} \right) e^{-2r/3a_0} ,
 \label{radial300}
\end{equation}
and the angular wave function $Y_{00}(\theta, \phi)$ is given by\footnote{I have found all the angular functions (the spherical harmonics) in table 2.1 in \cite{atomintro}.}
\begin{equation}
 Y_{00}(\theta, \phi) = \sqrt{\frac{1}{4\pi}} .
 \label{angular300}
\end{equation}
I now want to sketch these two functions, \eqref{radial300} and \eqref{angular300}. I sketch the radial function as a function of $r/a_0$, and ignore the constant factor $2(2/3a_0)^{3/2}$. Hence I get the plot shown in figure \vref{fig:R300}. The angular function is easier to sketch, since it is simply a constant everywhere, and hence it will simply be a circle in the $xz$-plane -- shown in figure \vref{fig:Y300}. 

\begin{figure}[htb]
 \centering
  \subfigure[$R_{30}(r)$\label{fig:R300}]{%
   \psset{xunit=5mm,yunit=25mm,plotpoints=1000,polarplot=false}
   \begin{pspicture}(-0.3,-0.5)(16,1.1)  %\psgrid
    %Koordinatsystem:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(15,0)   \uput[0](15,0){\small$r/a_0$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.3)(0,1.1)    
    %Koordinatakselabels:
    \multido{\i=1+1}{14}{\psline[linecolor=black,linewidth=0.3pt](\i,0)(\i,-0.03)
                         \uput[-90](\i,0){\tiny$\i$}}
    %Plot:
    \psPlot[linecolor=ForestGreen]{0}{15}{( 1 - (4/3)*x + (8/27)*x^2 ) * 2.718^(-(2/3)*x)}
    %Labels:
    %\rput(10,0.8){\tiny$R_{30}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} 2 \left( 1 - \frac{4}{3} \frac{r}{a_0} 
    %                    + \frac{8}{27} \frac{r^2}{a_0^2} \right) e^{-2r/3a_0}$}
   \end{pspicture}}
  \hspace{1cm}
  \subfigure[$Y_{00}(\theta,\phi)$\label{fig:Y300}]{%
   \psset{plotpoints=200,unit=2.5cm,polarplot=true}
   \begin{pspicture}(-0.5,-0.9)(0.5,0.7)  %\psgrid
    %Axes:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(0.5,0)   \uput[-90](0.5,0){\small$z$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.7)(0,0.7)   \uput[180](0,0.7){\small$x$}
    %Plots:
    \rput{90}{\psPlot[linecolor=ForestGreen,linewidth=.35mm]{0}{360}{sqrt(1/(4*3.14))}}
    %Labels:
    %\rput(0.4,0.6){$Y_{00}$}
   \end{pspicture}}
 \caption{Sketches of the radial and angular part of the wave function $\psi_{300}(r,\theta,\phi) = R_{30}(r) Y_{00}(\theta,\phi)$ (also denoted by
          $\ket{300}$) of \Hefireplus. The radial part is given in equation \eqref{radial300}, while the angular part is given in equation 
          \eqref{angular300}.}
 \label{fig:300}
\end{figure}

For the second of the three states with $m=0$, $\ket{310}$, the wave function is given by $\psi_{310}(r, \theta, \phi) = R_{31}(r) Y_{10}(\theta, \phi)$, where the radial wave function $R_{31}(r)$ is given by
\begin{equation}
 R_{31}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} \frac{8\sqrt2}{9} \frac{r}{a_0} \left( 1 - \frac{1}{3} \frac{r}{a_0} \right) e^{-2r/3 a_0} ,
 \label{radial310}
\end{equation}
and the angular wave function $Y_{10}(\theta, \phi)$ is given by
\begin{equation}
 Y_{10}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos\theta .
 \label{angular310}
\end{equation}
The radial wave function in \eqref{radial310} is sketched in \vref{fig:R310}, again as a function of $r/a_0$ and ignoring the constant factor $(2/3a_0)^{3/2} 8\sqrt2 / 9$. The angular wave function in \eqref{angular310} is sketched in \vref{fig:Y310}.

\begin{figure}[htb]
 \centering
  \subfigure[$R_{31}(r)$\label{fig:R310}]{%
   \psset{xunit=5mm,yunit=50mm,plotpoints=1000,polarplot=false}
   \begin{pspicture}(-0.3,-0.25)(16,0.55)  %\psgrid
    %Koordinatsystem:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(15,0)   \uput[0](15,0){\small$r/a_0$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.15)(0,0.55)    
    %Koordinatakselabels:
    \multido{\i=1+1}{14}{\psline[linecolor=black,linewidth=0.3pt](\i,0)(\i,-0.015)
                         \uput[-90](\i,0){\tiny$\i$}}
    %Plot:
    \psPlot[linecolor=ForestGreen]{0}{15}{x*(1-(1/3)*x) * 2.718^(-(2/3)*x)}
    %Labels:
    %\rput(10,0.4){\tiny$R_{31}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} \frac{8\sqrt2}{9} \frac{r}{a_0} \left( 1 
    %                    - \frac{1}{3} \frac{r}{a_0} \right) e^{-2r/3 a_0}$}
   \end{pspicture}}
  \hspace{1cm}
  \subfigure[$Y_{10}(\theta,\phi)$\label{fig:Y310}]{%
   \psset{plotpoints=200,unit=2.5cm,polarplot=true}
   \begin{pspicture}(-0.5,-0.9)(0.5,0.7)  %\psgrid
    %Axes:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(0.5,0)   \uput[-90](0.5,0){\small$z$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.7)(0,0.7)   \uput[180](0,0.7){\small$x$}
    %Plots:
    \rput{90}{\psPlot[linecolor=ForestGreen,linewidth=.35mm]{0}{360}{sqrt(3/(4*3.14))*cos(x)}}
    \rput{90}{\psPlot[linecolor=ForestGreen,linewidth=.35mm]{0}{360}{-sqrt(3/(4*3.14))*cos(x)}}
    %Labels:
    %\rput(0.4,0.6){$Y_{10}$}
   \end{pspicture}}
 \caption{Sketches of the radial and angular part of the wave function $\psi_{310}(r,\theta,\phi) = R_{31}(r) Y_{10}(\theta,\phi)$ (also denoted by
          $\ket{310}$) of \Hefireplus. The radial part is given in equation \eqref{radial310}, while the angular part is given in equation 
          \eqref{angular310}.}
 \label{fig:310}
\end{figure}

For the third of the three states with $m=0$, $\ket{320}$, the wave function is given by $\psi_{320}(r, \theta, \phi) = R_{32}(r) Y_{20}(\theta, \phi)$, where the radial wave function $R_{32}(r)$ is given by
\begin{equation}
 R_{32}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} \frac{8\sqrt2}{27\sqrt5} \frac{r^2}{a_0^2} e^{-2r/3a_0} ,
 \label{radial320}
\end{equation}
and the angular wave function $Y_{20}(\theta, \phi)$ is given by
\begin{equation}
 Y_{20}(\theta, \phi) = \sqrt{\frac{5}{16\pi}} \left( 3\cos^2\theta - 1 \right) .
 \label{angular320}
\end{equation}
The radial wave function in \eqref{radial320} is sketched in \vref{fig:R320}, again as a function of $r/a_0$ and ignoring the constant factor $(2/3a_0)^{3/2} 8\sqrt2 / 27\sqrt5$. The angular wave function in \eqref{angular320} is sketched in \vref{fig:Y320}.

\begin{figure}[htb]
 \centering
  \subfigure[$R_{32}(r)$\label{fig:R320}]{%
   \psset{xunit=5mm,yunit=22mm,plotpoints=1000,polarplot=false}
   \begin{pspicture}(-0.3,-0.15)(16,1.4)  %\psgrid
    %Koordinatsystem:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(15,0)   \uput[0](15,0){\small$r/a_0$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.15)(0,1.4)    
    %Koordinatakselabels:
    \multido{\i=1+1}{14}{\psline[linecolor=black,linewidth=0.3pt](\i,0)(\i,-0.03)
                         \uput[-90](\i,0){\tiny$\i$}}
    %Plot:
    \psPlot[linecolor=ForestGreen]{0}{15}{x^2 * 2.718^(-(2/3)*x)}
    %Labels:
    %\rput(11,1.2){\tiny$R_{32}(r) = \left( \frac{2}{3 a_0} \right)^{3/2} \frac{8\sqrt2}{27\sqrt5} \frac{r^2}{a_0^2} e^{-2r/3a_0}$}
   \end{pspicture}}
  \hspace{1cm}
  \subfigure[$Y_{20}(\theta,\phi)$\label{fig:Y320}]{%
   \psset{plotpoints=200,unit=2.5cm,polarplot=true}
   \begin{pspicture}(-0.5,-0.7)(0.5,0.9)  %\psgrid
    %Axes:
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(0.5,0)   \uput[-90](0.5,0){\small$z$}
    \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-0.7)(0,0.7)   \uput[180](0,0.7){\small$x$}
    %Plots:
    \rput{90}{\psPlot[linecolor=ForestGreen,linewidth=.35mm]{0}{360}{sqrt(5/(16*3.14))*(3*cos(x)^2 -1)}}
    %Labels:
    %\rput(0.4,0.6){$Y_{20}$}
   \end{pspicture}}
 \caption{Sketches of the radial and angular part of the wave function $\psi_{320}(r,\theta,\phi) = R_{32}(r) Y_{20}(\theta,\phi)$ (also denoted by
          $\ket{320}$) of \Hefireplus. The radial part is given in equation \eqref{radial320}, while the angular part is given in equation 
          \eqref{angular320}.}
 \label{fig:320}
\end{figure}










\subsubsection*{Question g}\setcounter{thespnr}{7}
\ramme{ Draw the radial probability density function for the state $n=3$, $l=2$, and find the value of $r$ for which it is maximum. }
I first find the radial probability density function $P(r) = r^2 |R_{32}(r)|^2$ (where $R_{32}(r)$ is given in \eqref{radial320}) by simple calculation, to be
\begin{equation}
 P(r) = r^2 |R_{32}(r)|^2 
      = r^2 \left( \frac{2}{3 a_0} \right)^3 \frac{128}{3645} \frac{r^4}{a_0^4} e^{-4r/3a_0} 
      = \frac{1024}{98415 a_0 } \frac{r^6}{a_0^6} e^{-4r/3a_0} .
 \label{R320squared}
\end{equation}
I have drawn this function in figure \vref{fig:R320squared}, as a function of $r/a_0$, again ignoring the constant factor $1024/98415 a_0$. 

To find the maximum of the function, I first differentiate with respect to $r$
\begin{align*}
 \dd{}{r} \big( P(r) \big) 
    &= \frac{1024}{98415 a_0} 6 \frac{r^5}{a_0^6} e^{-4r/3a_0} - \frac{1024}{98415a_0} \frac{4}{3a_0} \frac{r^6}{a_0^6} e^{-4r/3a_0} \\ 
    &= \frac{1024}{98415 a_0} \left( 6 \frac{r^5}{a_0^6} e^{-4r/3a_0} - \frac{4}{3a_0} \frac{r^6}{a_0^6} e^{-4r/3a_0} \right) \\
    &= \frac{4096}{98415 a_0^7} \left( 6 r^5 - \frac{4}{3 a_0} r^6 \right) e^{-4r/3a_0} , 
\end{align*}
this expression I then set this equal to zero, to find the value of $r$ where the maximum happens
\begin{gather*}
 0 = \frac{4096}{98415 a_0^7} \left( 6 r^5 - \frac{4}{3 a_0} r^6 \right) e^{-4r/3a_0}   \qquad\Rightarrow\qquad
 0 = \left( 6 r^5 - \frac{4}{3 a_0} r^6 \right) e^{-4r/3a_0}  \qquad\Rightarrow \\
 0 = 6 r^5 - \frac{4}{3 a_0} r^6  \qquad\Rightarrow\qquad
 0 = r^5 \left( 6 - \frac{4}{3 a_0} r \right) , 
\end{gather*}
so an extremum happens for $r = 0$ and for $r = 9 a_0 / 2$. By looking at figure \vref{fig:R320squared} I easily see that the maximum I am looking for hence must be at $r = 9 a_0 / 2$. This place is shown in the figure by a dashed line from the $r/a_0$-axis to the graph.

\begin{figure}[htb]
 \centering
  \psset{xunit=5mm,yunit=1.5mm,plotpoints=1000,polarplot=false}
  \begin{pspicture}(-0.3,-2)(16,22)  %\psgrid
   %Linie:
   \psline[linecolor=Gray,linewidth=0.5pt,linestyle=dashed](4.5,0)(4.5,20.5958)
   %Koordinatsystem:
   \psline[linecolor=black,linewidth=0.5pt,arrows=->](-0.5,0)(15,0)   \uput[0](15,0){\small$r/a_0$}
   \psline[linecolor=black,linewidth=0.5pt,arrows=->](0,-2)(0,22)    
   %Koordinatakselabels:
   \multido{\i=1+1}{14}{\psline[linecolor=black,linewidth=0.3pt](\i,0)(\i,-0.5)
                        \uput[-90](\i,0){\tiny$\i$}}
   %Plot:
   \psPlot[linecolor=ForestGreen]{0}{15}{x^6*2.718^(-(4/3)*x)}
   %Labels:
   %\rput(12,1.3){\tiny$r^2 |R_{32}(r)|^2 = \frac{1024}{98415 a_0} \frac{r^6}{a_0^6} e^{-4r/3a_0}$}
  \end{pspicture}
 \caption{The radial probability density function for the state with $n=3$ and $l=2$, $P(r) = r^2 |R_{32}(r)|^2$. This function is given in
          \eqref{R320squared}. The dashed line indicate the place of the maximum, at $r/a_0 = 9/2$.}
 \label{fig:R320squared}
\end{figure}




\subsubsection*{Question h}\setcounter{thespnr}{8}
\ramme{ Give the selection rules for electric dipole transitions, and draw a Grotrian diagram with all allowed transitions from $n=3$ to lower lying
        states. }
The change in the principal quantum number, $n$, is arbitrary, and the change in the azimuthal quantum number, $l$, must be either one up or one down ($\Delta l = \pm 1$). %And at last the change in the magnetic quantum number, $m$, must be either one up, no change, or one down ($\Delta m = 0, \pm 1$). 
These selection rules helps me make the Grotrian diagram shown in figure \vref{fig:grotrian}. 

\begin{figure}[htb]
 \centering
  \psset{xunit=5mm,yunit=0.5mm}
  \begin{pspicture}(-2,-90)(22,5)  %\psgrid
   %Energiakse
   \psline[linecolor=black,linewidth=0.5pt,arrows=->](15,-90)(15,0)      \uput[0](15,0){$E$}
   \psline[linecolor=black,linewidth=0.5pt](14.8,-9.682)(15.2,-9.682)    \uput[0](15,-9.682){\small$E_3 = -9.687\E{-19} \,\text{J}$}
   \psline[linecolor=black,linewidth=0.5pt](14.8,-21.785)(15.2,-21.785)  \uput[0](15,-21.785){\small$E_2 = -21.80\E{-19} \,\text{J}$}
   \psline[linecolor=black,linewidth=0.5pt](14.8,-87.14)(15.2,-87.14)    \uput[0](15,-87.14){\small$E_1 = -87.18\E{-19} \,\text{J}$}
   %Niveauer
   \psline[linecolor=black,linewidth=0.5pt](0,-9.682)(3,-9.682)    \uput[0](2.75,-9.682){\small $3s$}   %3s
   \psline[linecolor=black,linewidth=0.5pt](5,-9.682)(8,-9.682)    \uput[0](7.75,-9.682){\small $3p$}   %3p
   \psline[linecolor=black,linewidth=0.5pt](10,-9.682)(13,-9.682)  \uput[0](12.75,-9.682){\small $3d$}  %3d
   \psline[linecolor=black,linewidth=0.5pt](0,-21.785)(3,-21.785)  \uput[0](2.75,-21.785){\small $2s$}  %2s
   \psline[linecolor=black,linewidth=0.5pt](5,-21.785)(8,-21.785)  \uput[0](7.75,-21.785){\small $2p$}  %2p
   \psline[linecolor=black,linewidth=0.5pt](0,-87.074)(3,-87.14)   \uput[0](2.75,-87.14){\small $1s$}  %1s
   %n-labels:
   \uput[180](-0.5,-9.682){$n=3$}
   \uput[180](-0.5,-21.785){$n=2$}
   \uput[180](-0.5,-87.14){$n=1$}
   %l-labels:
   \uput[90](1.5,-5){$l=0$}
   \uput[90](6.5,-5){$l=1$}
   \uput[90](11.5,-5){$l=2$}
   %Henfaldslinier:
   \psline[linecolor=ForestGreen]{c-c}(1.5,-9.682)(6.5,-21.785)
   \psline[linecolor=ForestGreen]{c-c}(6.5,-9.682)(1.5,-21.785)
   \psline[linecolor=ForestGreen]{c-c}(11.5,-9.682)(6.5,-21.785)
   \psline[linecolor=ForestGreen]{c-c}(6.5,-9.682)(1.35,-87.14)
   \psline[linecolor=ForestGreen]{c-c}(6.5,-21.785)(1.75,-87.14)
  \end{pspicture}
 \caption{The Grotrian diagram for decay from the $n=3$ state of \Hefireplus. The energies are calculated by using equation \eqref{energies}. Note that
          I have not drawn lines between levels of same $n$, although there is in principle allowed transitions here -- these transitions have such 
          little energy difference, that their lifetime is way larger than the lifetime of the drawn transitions, and therefore does not really happen.}
 \label{fig:grotrian}
\end{figure}


%\clearpage
\subsubsection*{Question i}\setcounter{thespnr}{9}
\ramme{ What is the lifetime of the $2p$ level? }
The $2p$ level can only decay in one way, and that is directly to the $1s$ level (as seen on the Grotrian diagram in figure \vref{fig:grotrian}). First of all I have to note that the $2p$ level is three times degenerate (there's a level with $m = -1$ ($\ket{21,-1}$), $m = 0$ ($\ket{210}$) and $m = 1$  ($\ket{211}$)), and hence I should look at lifetimes for all these three levels. 

The lifetime of a level is given by the reciprocal spontaneous decay rate, $\tau = 1/A_{21}$, where $A_{21}$ is given by equation (7.23) in \cite{atomintro} as
\begin{equation}
 A_{21} = \frac{g_1}{g_2} \frac{4\alpha}{3c^2} \omega^3 |\vec D|^2 ,
 \label{Akoefficient}
\end{equation}
where the dipole moment vector $\vec D$ is given by equation (7.22) in \cite{atomintro} to be
\begin{equation*}
 \vec D = \braket{1}{\vec r}{2} ,
        %= \int \psi_1^\ast(\vec r) \,\vec r \,\psi_2(\vec r) \,\di^3 \vec r ,
\end{equation*}
where $\vec r = (x,y,z)$ is the position vector, $\ket{1}$ is the state the transition is from, and $\ket{2}$ is the state the transition is to\footnote{I have been told that the transition is actually {\slshape from} $\ket{2}$ {\slshape to} $\ket{1}$, and not the opposite as I stated -- but since I am only going to use the modulus square of $\vec D$, and $\braket{1}{\vec r}{2} = \braket{2}{\vec r}{1}^\ast$, I keep on going as I did before the notice, simply because it doesn't make any difference.}. 

Now I have to calculate the $\vec D$ coefficients for the three different transitions, from $\ket{21,-1}$ to $\ket{100}$, from $\ket{210}$ to $\ket{100}$, and from $\ket{211}$ to $\ket{100}$. I note that the vector $\vec r = (x,y,z)$ can be rewritten in terms of spherical coordinates and from there the spherical harmonics by
\begin{align*}
 x = r \sin\theta \cos\phi &= -r \sqrt{\frac{2\pi}{3}} \,\big( Y_{11}(\theta,\phi) - Y_{1,-1}(\theta,\phi) \big) , \\
 y = r \sin\theta \sin\phi &= ir \sqrt{\frac{2\pi}{3}} \,\big( Y_{11}(\theta,\phi) + Y_{1,-1}(\theta,\phi) \big) , \\
 z = r \cos\theta          &= r \sqrt{\frac{4\pi}{3}} \,Y_{10}(\theta,\phi) ,
\end{align*} 
which I can use to calculate the inner products without having to integrate. 
\\ \\
\noindent $\bullet$ For the transition from the $m=-1$ state, I have $\ket{1} = \ket{21,-1} = \ket{R_{21}} \ket{Y_{1,-1}}$, and will always have that $\ket{2} = \ket{100} = \ket{R_{10}} \ket{Y_{00}}$. So for the first coordinate, $x$, I get that
\begin{equation}
 \braket{1}{x}{2} = \bra{Y_{1,-1}} \braket{R_{21}}{x}{R_{10}} \ket{Y_{00}} 
                  = - \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{1,-1}}{Y_{11} - Y_{1,-1}}{Y_{00}} ,
 \label{eq:1x2:m-1:continue}
\end{equation}
where the angular inner product (with this I mean the inner product containing the angular parts of the wave functions) is
\begin{multline*}
 \braket{Y_{1,-1}}{Y_{11} - Y_{1,-1}}{Y_{00}} 
   = \braket{Y_{1,-1}}{Y_{11}}{Y_{00}} - \braket{Y_{1,-1}}{Y_{1,-1}}{Y_{00}} \\
   = Y_{00} \underbrace{\twobraket{Y_{1,-1}}{Y_{11}}}_{=0} - Y_{00} \underbrace{\twobraket{Y_{1,-1}}{Y_{1,-1}}}_{=1}
   = - Y_{00}
   = - \sqrt{\frac{1}{4\pi}} ,
\end{multline*}
where I used that $Y_{00}$ is simply a constant (and hence can be pulled out of the inner product because of its linearity) and that the spherical harmonics are orthonormal. The radial inner product is found by first noting that
\begin{align*}
 R_{10}(r) &= \left( \frac{Z}{2 a_0} \right)^{3/2} \frac{2}{\sqrt3} \,\rho \,e^{-\rho}
            = \left( \frac{1}{a_0} \right)^{3/2} \frac{2}{\sqrt3} \frac{r}{a_0} \,e^{-r/a_0} , \\
 R_{21}(r) &= \left( \frac{Z}{a_0} \right)^{3/2} 2 \,e^{-\rho}
            = \left( \frac{2}{a_0} \right)^{3/2} 2 \,e^{-2r/a_0} ,
\end{align*}
according to table 2.2 and equation (2.21) in \cite{atomintro} (and the fact that $Z=2$ and $n=1$ for the first and $Z=2$ and $n=2$ for the second), and hence
\begin{multline*}
 \braket{R_{21}}{r}{R_{10}} 
   = \int_0^\infty R_{21}^\ast(r) r R_{10}(r) r^2 \,\di r
   = \int_0^\infty \frac{2}{\sqrt3 a_0^{5/2}} r e^{-r/a_0} r \frac{2^{5/2}}{a_0^{3/2}} e^{-2r/a_0} r^2 \,\di r \\
   = \frac{8 \sqrt2}{\sqrt3 a_0^4} \int_0^\infty r^4 e^{-3r/a_0} \,\di r
   = \frac{8 \sqrt2}{\sqrt3 a_0^4} \frac{\Gamma(5)}{(3/a_0)^5}
   = \frac{8 \sqrt2}{\sqrt3 a_0^4} \frac{24 a_0^5}{243}
   = \frac{64 \sqrt2 a_0}{81 \sqrt3} .
\end{multline*}
So, continuing equation \eqref{eq:1x2:m-1:continue}, I have that the first coordinate of the dipole moment vector for the transition with $m=-1$ is
\begin{equation*}
 D_x^{m=-1} \equiv \braket{1}{x}{2}
            = \sqrt{\frac{2\pi}{3}} \frac{64 \sqrt2 a_0}{81 \sqrt3} \sqrt{\frac{1}{4\pi}}
            = \frac{64}{243} a_0 .
 %\label{Dm-1x}
\end{equation*}
For the second coordinate, $y$, I get that
\begin{equation*}
 \braket{1}{y}{2} = \bra{Y_{1,-1}} \braket{R_{21}}{y}{R_{10}} \ket{Y_{00}} 
                  = i \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{1,-1}}{Y_{11} + Y_{1,-1}}{Y_{00}} ,
\end{equation*}
where the angular inner product is simply the same as found for the $x$-component except for a different sign, and the radial inner product is the same as found before, and hence
\begin{equation*}
 D_y^{m=-1} \equiv \braket{1}{y}{2}
            = - i \sqrt{\frac{2\pi}{3}} \frac{64 \sqrt2 a_0}{81 \sqrt3} \sqrt{\frac{1}{4\pi}}
            = - \frac{64 i}{243} a_0 .
 %\label{Dm-1y}
\end{equation*}
For the last coordinate, $z$, I get that
\begin{equation*}
 \braket{1}{z}{2} = \bra{Y_{1,-1}} \braket{R_{21}}{z}{R_{10}} \ket{Y_{00}} 
                  = \sqrt{\frac{4\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{1,-1}}{Y_{10}}{Y_{00}} ,
\end{equation*}
where the angular inner product is
\begin{equation*}
 \braket{Y_{1,-1}}{Y_{10}}{Y_{00}} 
   = Y_{00} \underbrace{\twobraket{Y_{1,-1}}{Y_{10}}}_{=0} 
\end{equation*}
because of the orthonomality of the spherical harmonics, and hence
\begin{equation*}
 D_z^{m=-1} \equiv \braket{1}{z}{2}
            = 0 .
 %\label{Dm-1z}
\end{equation*}

$\bullet$ For the transition from the $m=0$ state, I have $\ket{1} = \ket{210} = \ket{R_{21}} \ket{Y_{10}}$, and I still have that $\ket{2} = \ket{100} = \ket{R_{10}} \ket{Y_{00}}$. So for the first coordinate, $x$, I get that
\begin{equation*}
 \braket{1}{x}{2} = \bra{Y_{10}} \braket{R_{21}}{x}{R_{10}} \ket{Y_{00}} 
                  = - \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{10}}{Y_{11} - Y_{1,-1}}{Y_{00}} ,
\end{equation*}
where the angular inner product is
\begin{equation*}
 \braket{Y_{10}}{Y_{11} - Y_{1,-1}}{Y_{00}} 
   = \braket{Y_{10}}{Y_{11}}{Y_{00}} - \braket{Y_{10}}{Y_{1,-1}}{Y_{00}} 
   = Y_{00} \underbrace{\twobraket{Y_{10}}{Y_{11}}}_{=0} - Y_{00} \underbrace{\twobraket{Y_{10}}{Y_{1,-1}}}_{=0} .
\end{equation*}
Hence
\begin{equation*}
 D_x^{m=0} \equiv \braket{1}{x}{2} = 0 . 
 %\label{Dm0x}
\end{equation*}
For the second coordinate, $y$, I get that
\begin{equation*}
 \braket{1}{y}{2} = \bra{Y_{10}} \braket{R_{21}}{y}{R_{10}} \ket{Y_{00}} 
                  = i \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{10}}{Y_{11} + Y_{1,-1}}{Y_{00}} ,
\end{equation*}
where the angular inner product is zero because of the same argument as for the $x$-component, and hence
\begin{equation*}
 D_y^{m=0} \equiv \braket{1}{y}{2} = 0 . 
 %\label{Dm0y}
\end{equation*}
For the last coordinate, $z$, I get that
\begin{equation}
 \braket{1}{z}{2} = \bra{Y_{10}} \braket{R_{21}}{z}{R_{10}} \ket{Y_{00}} 
                  = \sqrt{\frac{4\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{10}}{Y_{10}}{Y_{00}} ,
 \label{eq:1z2:m0:continue}
\end{equation}
where the angular inner product is
\begin{equation*}
 \braket{Y_{10}}{Y_{10}}{Y_{00}} 
   = Y_{00} \twobraket{Y_{10}}{Y_{10}}
   = Y_{00}
   = \sqrt{\frac{1}{4\pi}} ,
\end{equation*}
and the radial inner product is as found earlier. Hence continuing equation \eqref{eq:1z2:m0:continue} yields
\begin{equation*}
 D_z^{m=0} \equiv \braket{1}{z}{2} 
           = \sqrt{\frac{4\pi}{3}} \frac{64 \sqrt2 a_0}{81 \sqrt3} \sqrt{\frac{1}{4\pi}} 
           = \frac{64 \sqrt2}{243} a_0 .
 %\label{Dm0z}
\end{equation*}

$\bullet$ For the transition from the $m=1$ state, I have $\ket{1} = \ket{211} = \ket{R_{21}} \ket{Y_{11}}$, and, as before, I have that $\ket{2} = \ket{100} = \ket{R_{10}} \ket{Y_{00}}$. So for the first coordinate, $x$, I get that
\begin{equation}
 \braket{1}{x}{2} = \bra{Y_{11}} \braket{R_{21}}{x}{R_{10}} \ket{Y_{00}} 
                  = - \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{11}}{Y_{11} - Y_{1,-1}}{Y_{00}} ,
 \label{eq:1z2:m1:continue}
\end{equation}
where the angular inner product is
\begin{multline*}
 \braket{Y_{11}}{Y_{11} - Y_{1,-1}}{Y_{00}} 
   = \braket{Y_{11}}{Y_{11}}{Y_{00}} - \braket{Y_{11}}{Y_{1,-1}}{Y_{00}} \\
   = Y_{00} \underbrace{\twobraket{Y_{11}}{Y_{11}}}_{=1} - Y_{00} \underbrace{\twobraket{Y_{11}}{Y_{1,-1}}}_{=0} 
   = Y_{00}
   = \sqrt{\frac{1}{4\pi}} ,
\end{multline*}
and the radial inner product is the same as found earlier, and hence continuing equation \eqref{eq:1z2:m1:continue} gives
\begin{equation*}
 D_x^{m=1} \equiv \braket{1}{x}{2} 
           = - \sqrt{\frac{2\pi}{3}} \frac{64 \sqrt2 a_0}{81 \sqrt3} \sqrt{\frac{1}{4\pi}}
           = - \frac{64}{243} a_0 .
 %\label{Dm1x}
\end{equation*}
For the second coordinate, $y$, I get that
\begin{equation*}
 \braket{1}{y}{2} = \bra{Y_{11}} \braket{R_{21}}{y}{R_{10}} \ket{Y_{00}} 
                  = i \sqrt{\frac{2\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{11}}{Y_{11} + Y_{1,-1}}{Y_{00}} ,
\end{equation*}
where the angular inner product is the same as the one for the $x$-component (except the irrelevant sign of the zero-term), and the radial inner product again is as found earlier. Hence
\begin{equation*}
 D_y^{m=1} \equiv \braket{1}{y}{2} 
           = i \sqrt{\frac{2\pi}{3}} \frac{64 \sqrt2 a_0}{81 \sqrt3} \sqrt{\frac{1}{4\pi}}
           = \frac{64 i}{243} a_0 .
 %\label{Dm1y}
\end{equation*}
For the last coordinate, $z$, I get that
\begin{equation*}
 \braket{1}{z}{2} = \bra{Y_{11}} \braket{R_{21}}{z}{R_{10}} \ket{Y_{00}} 
                  = \sqrt{\frac{4\pi}{3}} \braket{R_{21}}{r}{R_{10}} \braket{Y_{11}}{Y_{10}}{Y_{00}} ,
\end{equation*}
where the angular inner product is
\begin{equation*}
 \braket{Y_{11}}{Y_{10}}{Y_{00}} 
   = Y_{00} \underbrace{\twobraket{Y_{11}}{Y_{10}}}_{=0} ,
\end{equation*}
and hence
\begin{equation*}
 D_z^{m=1} \equiv \braket{1}{z}{2}
           = 0
 %\label{Dm1z}
\end{equation*}

Now I can look at the $\vec D$-vectors for the three transitions, and see that they are of equal length:
\begin{align*}
 |\vec D^{m=-1}| &= \sqrt{|D_x^{m=-1}|^2 + |D_y^{m=-1}|^2 + |D_z^{m=-1}|^2} \\
                 &= \sqrt{\left( \frac{64}{243} a_0 \right)^2 + \left( -\frac{64i}{243} a_0 \right)\left( \frac{64i}{243} a_0 \right)}
                  = \frac{64 \sqrt2}{243} a_0 , \\
 |\vec D^{m=0}|  &= \sqrt{|D_x^{m=0}|^2 + |D_y^{m=0}|^2 + |D_z^{m=0}|^2}  \\
                 &= \sqrt{\left( \frac{64 \sqrt2}{243} a_0 \right)^2}
                  = \frac{64 \sqrt2}{243} a_0 , \\
 |\vec D^{m=1}|  &= \sqrt{|D_x^{m=1}|^2 + |D_y^{m=1}|^2 + |D_z^{m=1}|^2}  \\
                 &= \sqrt{\left( - \frac{64}{243} a_0 \right)^2 + \left( \frac{64 i}{243} a_0 \right)\left( -\frac{64 i}{243} a_0 \right)}
                  = \frac{64 \sqrt2}{243} a_0 ,
\end{align*}
and hence the lifetimes of the three transitions are also the same (assuming that the three levels have the same energy, and hence use the same $\omega$ to find it). To find the lifetime of the $2p$ level, I hence only have do one calculation of $A_{21}$ from \eqref{Akoefficient}, using
\begin{equation*}
 |\vec D|^2 = 2 \left(\frac{64}{243}\right)^2 a_0^2 
            = 2 \left(\frac{64}{243}\right)^2 (\anulValue)^2 
            = 3.885\E{-22}\,\text{m}^2 ,
\end{equation*}
and the angular frequency $\omega$ given by
\begin{align*}
 \omega &= \frac{1}{\hbar} (E_2 - E_1)
         = \frac{8 m_e M_p}{\hbar^3 (m_e + 4 M_p)} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 \left( 1 - \frac{1}{4} \right) \\
        &= \frac{8 \cdot \meValue \cdot \MpValue}{(\hbarValue)^3 (\meValue + 4 \cdot \MpValue)} 
           \left( \frac{(\eValue)^2}{4\pi \cdot \epsilonnulValue} \right)^2 \frac{3}{4} \\
        &= 6.200\E{16}\,\text{s}^{-1} ,
\end{align*}
where I used equation \eqref{energies} for the energies. I also note that the states (for the different $m$), are non-degenerate\footnote{Here I ignore spin -- If spin were taken account of, both $g_1$ and $g_2$ would instead be equal to $2$, so no change would happen.}, so both $g_1$ and $g_2$ are simply equal to $1$. So finally I get that, inserting into equation \eqref{Akoefficient},
\begin{align*}
 A_{21} %&= \frac{4\alpha}{3c^2} \omega^3 |\vec D|^2 \\
        &= \frac{1}{1} \frac{4\cdot \alphaValue}{3\cdot(\cValue)^2} (6.200\E{16}\,\text{s}^{-1})^3 \cdot 3.885\E{-40}\,\text{m}^2
         = 1.0025\E{10} \,\text{s}^{-1} ,
\end{align*}
and hence the lifetime for the $2p$ level is
\begin{equation*}
 \tau = \frac{1}{A_{21}}
      = \frac{1}{1.0025\E{10} \,\text{s}^{-1}}
      = 9.975\E{-11} \,\text{s} .
\end{equation*}











\subsubsection*{Question j}\setcounter{thespnr}{10}
\ramme{ Find the frequency difference between the Balmer-$\alpha$ line of \Hefireplus{} and \Hetreplus. }
The Balmer-$\alpha$ line is the decay from a level with $n=3$ to a level with $n=2$. The only difference between \Hefireplus{} and \Hetreplus{} is the nuclear mass, and hence the reduced mass for \Hetreplus{} is, following equation \eqref{reducedmassground},
\begin{equation*}
 m_{\text{\Hetreplus}} = \frac{3 m_e M_p}{m_e + 3 M_p} ,
\end{equation*}
because it has only a single neutron (and hence $A = 3$). The reduced mass of \Hefireplus was earlier found to be
\begin{equation*}
 m_{\text{\Hefireplus}} = \frac{4 m_e M_p}{m_e + 4 M_p} .
\end{equation*}
The energy-expression in the first part of equation \eqref{energies} is still correct, and hence the Balmer-$\alpha$ lines of \Hefireplus{} and \Hetreplus{} has the energies
\begin{equation*}
 E_i = E_3 - E_2 
     = \frac{m_i}{2 \hbar^2} \left( \frac{2e^2}{4\pi\epsilon_0} \right)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
     = \frac{5m_i}{18\hbar^2} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2  ,
\end{equation*}
where $i$ is either \Hetreplus{} or \Hefireplus. By using the known connection that $\omega = 2\pi\nu$ and $E = \hbar \omega$, I get that the frequency $\nu_i$ of the two lines are
\begin{equation*}
 \nu_i = \frac{E_i}{2\pi\hbar}
       = \frac{5m_i}{36\pi\hbar^3} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 ,
\end{equation*}
so the frequency difference of the two Balmer-$\alpha$ lines must be
\begin{align*}
 \nu_{\text{\Hefireplus}} - \nu_{\text{\Hetreplus}}
   &= \frac{5}{36\pi\hbar^3} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 \left( m_{\text{\Hefireplus}} - m_{\text{\Hetreplus}} \right) \\
   &= \frac{5}{36\pi\hbar^3} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 \left( \frac{4 m_e M_p}{m_e + 4 M_p} - \frac{3 m_e M_p}{m_e + 3 M_p} \right) \\
   &= \frac{5}{36\pi\hbar^3} \left( \frac{e^2}{4\pi\epsilon_0} \right)^2 \frac{m_e^2 M_p}{(m_e + 4 M_p)(m_e + 3 M_p)} ,
\end{align*}
which, with inserted values, is
\begin{align*}
 \nu_{\text{\Hefireplus}} - \nu_{\text{\Hetreplus}}
   &= \frac{5}{36\pi\cdot(\hbarValue)^3} \left( \frac{(\eValue)^2}{4\pi\cdot \epsilonnulValue} \right)^2 \\
   \quad&\cdot  \frac{(\meValue)^2 \cdot \MpValue}{(\meValue + 4 \cdot \MpValue)(\meValue + 3 \cdot \MpValue)} \\
   &= 8.292\E{10} \,\text{Hz} . 
\end{align*}




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