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Notebook[{

Cell[CellGroupData[{
Cell["Answers to Exam IIa (Nov.  2005)", "Subtitle"],

Cell[CellGroupData[{

Cell["Prob 1", "Subsection"],

Cell[CellGroupData[{

Cell["LS coupling", "Subsubsection"],

Cell[TextData[{
  "For LS coupling of identical particles, we have the restriction L+S is \
even.  As a first step, couple two particles\n(2s2s) \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`\^1\)]],
  "S   (even parity)\n(2s2p) \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`\^1\)]],
  "P,  ",
  Cell[BoxData[
      \(TraditionalForm\`\^3\)]],
  "P   (odd-parity)                          \n",
  Cell[BoxData[
      \(TraditionalForm\`\(\((2  p2p)\)\( \[RightArrow] \^1\)\)\)]],
  "S,  ",
  Cell[BoxData[
      \(TraditionalForm\`\^3\)]],
  "P,  ",
  Cell[BoxData[
      \(TraditionalForm\`\^1\)]],
  "D   (even-parity)   \n                \nFor an even parity 3-particle \
state  couple two 2p states (2p2p)\nthen couple the resulting state with the \
2s  (2p2p)[LS]2s \nThe possible combinations are\n(2p2p) [ ",
  Cell[BoxData[
      \(TraditionalForm\`\^1\)]],
  "S]2s  ",
  Cell[BoxData[
      \(TraditionalForm\`\^2\)]],
  "S    2\[Times]1 substates\n(2p2p) [ ",
  Cell[BoxData[
      \(TraditionalForm\`\^3\)]],
  "P]2s  ",
  Cell[BoxData[
      \(TraditionalForm\`\^2\)]],
  "P     2\[Times]3 substates\n(2p2p) [ ",
  Cell[BoxData[
      \(TraditionalForm\`\^3\)]],
  "P]2s  ",
  Cell[BoxData[
      \(TraditionalForm\`\^4\)]],
  "P     4\[Times]3 substates\n(2p2p) [ ",
  Cell[BoxData[
      \(TraditionalForm\`\^1\)]],
  "D]2s  ",
  Cell[BoxData[
      \(TraditionalForm\`\^2\)]],
  "D   2\[Times]5 substates\nTotal number of substates = 2+6+12+10=30"
}], "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell["jj coupling", "Subsubsection"],

Cell[TextData[{
  "For jj coupling of identical particles J is even. The (2p2p) states are\n\
",
  Cell[BoxData[
      \(TraditionalForm\`\((3  p\_\(1/2\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(1/2\)\)]],
  ") \[RightArrow] [0]                            \n",
  Cell[BoxData[
      \(TraditionalForm\`\((3  p\_\(1/2\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") \[RightArrow] [1], [2]                     \n",
  Cell[BoxData[
      \(TraditionalForm\`\((3  p\_\(3/2\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") \[RightArrow] [0], [2]\n      \nCombining these with a 2s state  leads \
to\n\n [(",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(3  p\_\(1/2\)\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(1/2\)\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\([0]\) 2  s\_\(1/2\)\)]],
  "]    [1/2]                          2                  substates\n [(",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(3  p\_\(1/2\)\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\([1]\) 2  s\_\(1/2\)\)]],
  "]    [1/2]  & [3/2]           2 + 4  = 6    substates\n [(",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(3  p\_\(1/2\)\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\([2]\) 2  s\_\(1/2\)\)]],
  "]    [3/2]  & [5/2]           4 + 6  = 10  substates \n [(",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(3  p\_\(1/2\)\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\([0]\) 2  s\_\(1/2\)\)]],
  "]    [1/2]                          2                  substates\n [(",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(3  p\_\(1/2\)\)\)\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`3  p\_\(3/2\)\)]],
  ") ",
  Cell[BoxData[
      \(TraditionalForm\`\([2]\) 2  s\_\(1/2\)\)]],
  "]    [3/2]  & [5/2]           4 + 6  = 10  substates \n  Total number of \
substates = 2+6+10+2+10=30"
}], "Text"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Prob 2", "Subsection"],

Cell[TextData[{
  "The  energy of the (1s3p) state  is \nE[ (1s3p) ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\^3\),
            "TraditionalForm"], "P"}], TraditionalForm]]],
  "] =  ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\[Epsilon]\_\(1  s\)\),
            "TraditionalForm"], "+", \(\[Epsilon]\_\(3  p\)\)}], 
        TraditionalForm]]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`\(+\ R\_0\)\)]],
  "(1s3p1s3p) - ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(1\/3\),
            "TraditionalForm"], \(R\_1\)}], TraditionalForm]]],
  "(1s3p3p1s) - ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(U\_\(3  p3p\)\)\)\)]],
  " \nwhere U = ",
  Cell[BoxData[
      \(TraditionalForm\`1\/r\)]],
  " accounts for the screening of the 3p (or 2s) electron. \nThe energy of \
the (1s2s) state is\n E[ (1s2s) ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\^3\)\(S\)\)\)]],
  "] =  ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\[Epsilon]\_\(1  s\)\),
            "TraditionalForm"], "+", \(\[Epsilon]\_\(2  s\)\)}], 
        TraditionalForm]]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`\(+\ R\_0\)\)]],
  "(1s2s1s2s) - ",
  Cell[BoxData[
      \(TraditionalForm\`R\_0\)]],
  "(1s2s2s1s) - ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(U\_\(2  s2s\)\)\)\)]],
  ".\nAs a first step we need lowest order energies and Coulomb wave \
functions:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Epsilon]1s\  = \ \(-Z^2\)/2\)], "Input",
  CellLabel->"In[1]:="],

Cell[BoxData[
    \(\(-\(Z\^2\/2\)\)\)], "Output",
  CellLabel->"Out[1]="]
}, Open  ]],

Cell["Let Zs = Z-1 be the screened charge.", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Epsilon]2s\  = \ \(-Zs^2\)/\((2\ 2^2)\)\)], "Input",
  CellLabel->"In[3]:="],

Cell[BoxData[
    \(\(-\(Zs\^2\/8\)\)\)], "Output",
  CellLabel->"Out[3]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Epsilon]3p\  = \ \(-Zs^2\)/\((2\ 3^2)\)\)], "Input",
  CellLabel->"In[40]:="],

Cell[BoxData[
    \(\(-\(Zs\^2\/18\)\)\)], "Output",
  CellLabel->"Out[40]="]
}, Open  ]],

Cell[BoxData[
    \(P1s[r_]\  := \ 2\ Z^\((3/2)\)\ r\ Exp[\(-Z\)\ r]\)], "Input",
  CellLabel->"In[6]:="],

Cell[BoxData[
    \(P2s[
        r_]\  := \ \((Zs^\((3/2)\)/Sqrt[2])\)\ r\ \((1 - 
            Zs\ r/2)\)\ Exp[\(-Zs\)\ r/2]\)], "Input",
  CellLabel->"In[5]:="],

Cell[BoxData[
    \(P3p[r_]\  := \ 
      8  Zs^\((5/2)\)/\((27\ Sqrt[6])\)\ r^2\ \((1 - 
            Zs\ r/6)\)\ Exp[\(-Zs\)\ r/3]\)], "Input",
  CellLabel->"In[11]:="],

Cell["Now evaluate the screening potentials and Slater Integrals", "Text"],

Cell[BoxData[
    \(v01s[
        r_]\  := \ \ Integrate[P1s[x]^2, {x, 0, r}, 
            Assumptions \[Rule] \ Z > 0]/r\  + 
        Integrate[P1s[x]^2/x, {x, r, Infinity}, 
          Assumptions \[Rule] \ Z > 0]\)], "Input",
  CellLabel->"In[23]:="],

Cell[BoxData[
    \(v01s2s[r_]\  := \ 
      Integrate[P1s[x]\ P2s[x], {x, 0, r}, 
            Assumptions \[Rule] \ {Z > 0, Zs > 0}]/r\  + 
        Integrate[P1s[x] P2s[x]/x, {x, r, Infinity}, 
          Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)], "Input",
  CellLabel->"In[24]:="],

Cell[BoxData[
    \(v11s3p[r_] := 
      Integrate[P1s[x]\ P3p[x]\ x, {x, 0, r}, 
            Assumptions \[Rule] \ {Z > 0, Zs > 0}]/
          r^2\  + \(\(r\)\(\ \)\(Integrate[
            P1s[x] P3p[x]/x^2, {x, r, Infinity}, 
            Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)\(\ \)\)\)], "Input",
  CellLabel->"In[25]:="],

Cell[BoxData[
    \(G01s2s\  := \ 
      Integrate[P2s[r]^2\ v01s[r], {r, 0, Infinity}, 
        Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)], "Input",
  CellLabel->"In[49]:="],

Cell[BoxData[
    \(G01s3p\  := \ 
      Integrate[P3p[r]^2\ v01s[r], {r, 0, Infinity}, 
        Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)], "Input",
  CellLabel->"In[50]:="],

Cell[BoxData[
    \(F01s2s\  := \ 
      Integrate[P2s[r]\ P1s[r]\ v01s2s[r], {r, 0, Infinity}, 
        Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)], "Input",
  CellLabel->"In[51]:="],

Cell[BoxData[
    \(F11s3p\  := \ 
      Integrate[P3p[r]\ P1s[r]\ v11s3p[r], {r, 0, Infinity}, 
        Assumptions \[Rule] \ {Z > 0, Zs > 0}]\)], "Input",
  CellLabel->"In[52]:="],

Cell["Now evaluate the background potentials", "Text"],

Cell[BoxData[
    \(\(\(U2s\)\(\ \)\(:=\)\(\ \)\(Integrate[
        P2s[r]^2\ /r\ , {r, 0, Infinity}, 
        Assumptions \[Rule] \ Zs > 0]\)\(\ \)\)\)], "Input",
  CellLabel->"In[53]:="],

Cell[BoxData[
    \(\(\(U3p\)\(\ \)\(:=\)\(\ \)\(Integrate[
        P3p[r]^2\ /r\ , {r, 0, Infinity}, 
        Assumptions \[Rule] \ Zs > 0]\)\(\ \)\)\)], "Input",
  CellLabel->"In[54]:="],

Cell[TextData[{
  "Put things together: ",
  "E[ (1s3p) ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\^3\),
            "TraditionalForm"], "P"}], TraditionalForm]]],
  "] =  ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\[Epsilon]\_\(1  s\)\),
            "TraditionalForm"], "+", \(\[Epsilon]\_\(3  p\)\)}], 
        TraditionalForm]]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`\(+\ R\_0\)\)]],
  "(1s3p1s3p) - ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(1\/3\),
            "TraditionalForm"], \(R\_1\)}], TraditionalForm]]],
  "(1s3p3p1s) - ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(U\_\(3  p3p\)\)\)\)]],
  " "
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(e1s3p\  = \ \[Epsilon]1s + \ \[Epsilon]3p\  + \ G01s3p\  - \ 
        F11s3p/3\  - U3p\)], "Input",
  CellLabel->"In[55]:="],

Cell[BoxData[
    \(\(-\(Z\^2\/2\)\) - Zs\/9 - 
      Zs\^2\/18 - \(36\ Z\^3\ Zs\^5\ \((56\ Z\^2 - 28\ Z\ Zs + 5\ \
Zs\^2)\)\)\/\((3\ Z + Zs)\)\^9 + \(Z\ Zs\ \((243\ Z\^6 + 567\ Z\^5\ Zs + 567\ \
Z\^4\ Zs\^2 + 315\ Z\^3\ Zs\^3 + 69\ Z\^2\ Zs\^4 + 21\ Z\ Zs\^5 + \
Zs\^6)\)\)\/\((3\ Z + Zs)\)\^7\)], "Output",
  CellLabel->"Out[55]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(e1s3p\  = \ 
      e1s3p /. \ {Z \[Rule] \ 5, \ Zs \[Rule] \ 4} // N\)], "Input",
  CellLabel->"In[56]:="],

Cell[BoxData[
    \(\(-13.407311988341895`\)\)], "Output",
  CellLabel->"Out[56]="]
}, Open  ]],

Cell[TextData[{
  "E[ (1s2s) ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\^3\)\(S\)\)\)]],
  "] =  ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          FormBox[\(\[Epsilon]\_\(1  s\)\),
            "TraditionalForm"], "+", \(\[Epsilon]\_\(2  s\)\)}], 
        TraditionalForm]]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`\(+\ R\_0\)\)]],
  "(1s2s1s2s) - ",
  Cell[BoxData[
      \(TraditionalForm\`R\_0\)]],
  "(1s2s2s1s) - ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(U\_\(2  s2s\)\)\)\)]],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(e1s2s\  = \ \[Epsilon]1s + \ \[Epsilon]2s\  + \ G01s2s\  - \ F01s2s\  - 
        U2s\)], "Input",
  CellLabel->"In[57]:="],

Cell[BoxData[
    \(\(-\(Z\^2\/2\)\) - Zs\/4 - 
      Zs\^2\/8 - \(16\ Z\^3\ Zs\^3\ \((20\ Z\^2 - 30\ Z\ Zs + 13\ \
Zs\^2)\)\)\/\((2\ Z + Zs)\)\^7 + \(Z\ Zs\ \((8\ Z\^4 + 20\ Z\^3\ Zs + 12\ \
Z\^2\ Zs\^2 + 10\ Z\ Zs\^3 + Zs\^4)\)\)\/\((2\ Z + Zs)\)\^5\)], "Output",
  CellLabel->"Out[57]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(e1s2s\  = \ 
      e1s2s\  /. \ {Z \[Rule] \ 5, \ Zs \[Rule] \ 4} // N\)], "Input",
  CellLabel->"In[58]:="],

Cell[BoxData[
    \(\(-14.76632246282222`\)\)], "Output",
  CellLabel->"Out[58]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[CapitalDelta]E\  = \ \(\(e1s3p\)\(-\)\(e1s2s\)\(\ \)\)\)], "Input",
  CellLabel->"In[59]:="],

Cell[BoxData[
    \(1.3590104744803249`\)], "Output",
  CellLabel->"Out[59]="]
}, Open  ]],

Cell[TextData[{
  "Energy in ",
  Cell[BoxData[
      \(TraditionalForm\`cm\^\(-1\)\)]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(decm\  = \ 219474.62\ \[CapitalDelta]E\)], "Input",
  CellLabel->"In[60]:="],

Cell[BoxData[
    \(298268.307462589`\)], "Output",
  CellLabel->"Out[60]="]
}, Open  ]],

Cell["Wavelength in Angstrom", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Lambda]\  = \ 10\^8/decm\)], "Input",
  CellLabel->"In[61]:="],

Cell[BoxData[
    \(335.2686071501`\)], "Output",
  CellLabel->"Out[61]="]
}, Open  ]],

Cell["\<\
The NIST table  gives 337.07 Angstrom so the  difference with NIST \
is only 0.5 %\
\>", "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Prob 3", "Subsection"],

Cell[TextData[{
  "The reduced matrix element ",
  Cell[BoxData[
      \(TraditionalForm\`R\_E1\)]],
  " =",
  Cell[BoxData[
      \(TraditionalForm\`\(\(<\)\((1  s2s)\)\)\)]],
  Cell[BoxData[
      \(TraditionalForm\`\(\ \^2\)\)]],
  "S || r || (1s3p) ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \^3\)\(\(P\)\(>\)\)\)\)]],
  " =",
  Cell[BoxData[
      \(TraditionalForm\`\(\(\ \)\(\@3\)\)\)]],
  " <0 ||C1||1> ",
  Cell[BoxData[
      \(TraditionalForm\`R\_\(1  s\ 3  p\)\)]],
  " , where \n ",
  Cell[BoxData[
      \(TraditionalForm\`R\_\(1  s\ 2  p\)\)]],
  " = \[Integral] ",
  Cell[BoxData[
      \(TraditionalForm\`P\_\(2  p\)\)]],
  "(r) r ",
  Cell[BoxData[
      \(TraditionalForm\`P\_\(1  s\)\)]],
  "(r) dr  is the radial dipole integral."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Rsp\  = \ 
      Integrate[P3p[r] r\ P2s[r], {r, 0, Infinity}, 
        Assumptions \[Rule] \ Zs > 0]\)], "Input",
  CellLabel->"In[62]:="],

Cell[BoxData[
    \(\(27648\ \@3\)\/\(15625\ Zs\)\)], "Output",
  CellLabel->"Out[62]="]
}, Open  ]],

Cell["Let cps =  <0 ||C1||1>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(cps\  = \ Sqrt[3]\ ThreeJSymbol[{0, 0}, {1, 0}, {1, 0}]\)], "Input",
  CellLabel->"In[63]:="],

Cell[BoxData[
    \(\(-1\)\)], "Output",
  CellLabel->"Out[63]="]
}, Open  ]],

Cell["Then, we find that the reduced matrix element is:", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(RE1\  = \ Sqrt[3]\ cps\ Rsp\)], "Input",
  CellLabel->"In[65]:="],

Cell[BoxData[
    \(\(-\(82944\/\(15625\ Zs\)\)\)\)], "Output",
  CellLabel->"Out[65]="]
}, Open  ]],

Cell["The associated line strength is", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(SE1\  = \ RE1^2\)], "Input",
  CellLabel->"In[66]:="],

Cell[BoxData[
    \(6879707136\/\(244140625\ Zs\^2\)\)], "Output",
  CellLabel->"Out[66]="]
}, Open  ]],

Cell[TextData[{
  "The transition rate is A = ",
  Cell[BoxData[
      \(TraditionalForm\`\(2.02613\ 10\^18\)\/\[Lambda]\^3\)]],
  " ",
  Cell[BoxData[
      \(TraditionalForm\`SE1\/g\_a\)]],
  " in 1/s, where ",
  Cell[BoxData[
      \(TraditionalForm\`g\_a\)]],
  " is the initial state degeneracy (9 for a triplet P state) and \[Lambda]= \
335.269 from previous problem"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    RowBox[{"A", " ", "=", " ", 
      RowBox[{
        RowBox[{
          FormBox[\(\(2.02613\ 10\^18\)\/\[Lambda]\^3\),
            "TraditionalForm"], " ", 
          FormBox[\(SE1\/g\_a\),
            "TraditionalForm"]}], " ", "/.", 
        " ", \({Zs \[Rule] 2, \ g\_a \[Rule] \ 9}\)}]}]], "Input",
  CellLabel->"In[67]:="],

Cell[BoxData[
    \(4.208390529806159`*^10\)], "Output",
  CellLabel->"Out[67]="]
}, Open  ]],

Cell["", "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Prob 4", "Subsection"],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`5  d\_\(5/2\)\)]]], "Subsubsection"],

Cell["\<\
This state decays to the ground state by M1 emission 
\[Lambda] = 62374 Angstrom\
\>", "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`4  f\_\(5/2\)\)]],
  " &  ",
  Cell[BoxData[
      \(TraditionalForm\`4  f\_\(7/2\)\)]]
}], "Subsubsection"],

Cell[TextData[{
  "These states decay by E1 emission to the 5d levels\n",
  Cell[BoxData[
      \(TraditionalForm\`4  f\_\(5/2\)\)]],
  " \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`5  d\_\(3/2\)\)]],
  "  \[Lambda] = 13898 Angstrom\n",
  Cell[BoxData[
      \(TraditionalForm\`4  f\_\(5/2\)\)]],
  " \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`5  d\_\(5/2\)\)]],
  "  \[Lambda] = 17883 Angstrom\n",
  Cell[BoxData[
      \(TraditionalForm\`4  f\_\(7/2\)\)]],
  " \[RightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`5  d\_\(5/2\)\)]],
  "  \[Lambda] = 14100 Angstrom"
}], "Text"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`6  s\_\(1/2\)\)]]], "Subsection"],

Cell[TextData[{
  "The 6s decays to the lower 5d levels b E2 emission\n",
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