Midterm (221b), Problem #2

a)

The cross section is computed in the lecture notes "Scattering Theory II":

\(TraditionalForm\`dσ\/dΩ = \(dσ\/dΩ\( | \_pointlike\) | 
            F(q)\( | \^2\) = \(\((\(\(2  m\)\/\^2\) 
                    Ze\^2\/q\^2)\)\^2\) \((3\ \(sin\ aq\  - \ aq\ cos\ aq\)\/\((aq)\)\^3)\)\^2\)\),

where Z = ((4  π)/3) (a^3) (Subscript[ρ, 0]) gives the relation between charge, radius, and constant charge density of the nucleus, and (q^2) = 4  (k^2) (sin^2) (θ/2) is the momentum tranfer.

b)

We can't be too careful here because our model of the nucleus as a ball of constant charge density is very crude.  However, the most salient features of the cross sections--the  periodic dips--are  enough to extract some information about the size of the nucleus.  In other words, we're going to use the cheapo method.  In our model, the dips correspond to the zeros in the numerator of the form factor.  (We get zeros instead of dips because our model is simplistic.)   We need to find radius 'a' so that the zeros of (sin aq  -  aq cos aq) match the dips in the experimental cross sections.

There is some question about whether to use relativistic or non-relativistic  formulas for the energy-momentum  relations.  It turns out we need both, and really the way to see this is to compute the cross section using relativistic formalism from the beginning.  Here's what happens:  The relativistic form factor is still the Fourier transform of the charge density (now w/ respect to relativistic momentum), so in the form factor the k in q = 2  k sin (θ/2)should be related to the incident energy as E=kc.  On the other hand, the overall dependence of the relativistic cross section is 1/(E^2), not 1/(E^4), so we need to substitute \(TraditionalForm\`\(\(\^2\) k\^2\)\/\(2  m\) -> \ E\) in the pointlike cross section.  In other words, my quick and careless relativistic calculation suggests (in units where \(TraditionalForm\` = \(c = 1\)\))
\(TraditionalForm\`dσ\/dΩ -> \ \(\(\(Z\^2\) 
                e\^4\)\/\(4 \( E\^2\) \(sin\^4\) θ/
                  2\)\) \((3\ \(sin\ aq\  - \ aq\ cos\ aq\)\/\((aq)\)\^3)\)\^2\), 
where the remaining q's are given by q = 2  E sin (θ/2).  With this formula, the cross section more or less gives the right order of magnitude, though the difference between peaks and valleys is much exaggerated compared to the data.  See bottom for a plot.

We only need to worry about the gross shape, i.e. the minima, to extract reasonable radii.  For 750 MeV electrons, k =  3.81 (fm^(-1)).

ff[θ_] = 
        Sin[2 k a Sin[(θ/2)]] - 
            2 k a Sin[(θ/2)] Cos[2 k a Sin[(θ/2)]] /. 
          k -> 3.81;

Ca40dips = {{19, 0},  {31, 0},  {47, 
              0}} (π/180);
Ca48dips  =  {{18, 0},  {30, 0},  {45, 
              0}} (π/180);

I played around until I found values for 'a' s.t. the zeros in the form factor piece (ff) corresponded roughly with the zeros from the experiemental cross section (Ca40dips, plotted as the dark black dots).  For Ca40, a ≈ 3.6 fm.  For Ca48, a ≈ 3.8 fm.

Show[Plot[ff[θ] /. a -> 3.6, {θ, 0, (π/2)}], 
      ListPlot[Ca40dips,  PlotStyle -> PointSize[ .03] ]]

Show[Plot[ff[θ] /. a -> 3.8, {θ, 0, (π/2)}], 
      ListPlot[Ca48dips,  PlotStyle -> PointSize[ .03] ]]

c)

Two dips are apparent only for the 250 MeV electrons, for which k =  1.27 (fm^(-1)).   I find for Pb208, a ≈  7.0 fm.  It appears from the graph below that the first dip wasn't resolved in the experimental data.  How do we know we shouldn't be taking a ≈ 5 and fitting the two dips to the first two zeros of ffPb instead of the second and third zeros?  The spacing between zeros would be off.  (Also, we know about what the correct answer should be--see  part (d).)

Pb208dips = {{52, 0},  {76, 0}} (π/180);

ffPb[θ_] = 
        Sin[2 k a Sin[(θ/2)]] - 
            2 k a Sin[(θ/2)] Cos[2 k a Sin[(θ/2)]] /. 
          k -> 1.27;

Show[Plot[ffPb[θ] /. a -> 6.95, {θ, 0, (π/2)}], 
      ListPlot[Pb208dips,  PlotStyle -> PointSize[ .03] ]]

d)

From more careful electron scattering experiments, the A dependence of the radius is known to be a ≈  (Subscript[a, 0]) (A^(1/3)) fm, with an average value (Subscript[a, 0]) ≈  1.1.  If we assume (A^(1/3))dependence, which at least makes the intuitive sense of constant volume per nucleon, we have found:
A		a		(Subscript[a, 0])
40		3.6		1.05
48		3.8		1.05
208		7.0		1.18
in surprisingly good agreement with the better experimental average.

e)

For fun, here is a plot of the theoretical cross section for Ca40.  I write (e^2) ->  α the fine structure constant and energy E-> En  given in GeV.  'convert' is the conversion factor from (GeV^(-2)) to (cm^2).  'k' and 'a' can still be given in (fm^(-1)) and fm since they only appear in the product ka.

<< Graphics`Graphics`

crosssection[θ_] := 
      convert (( (Z^2) (α^2))/(4 (En^2) 
                Sin[(θ/2)]^4)) (((3 ff[θ]/(((2 k a Sin[(θ/2)]))^3)))^2) /. {a -> 3.6,  k -> 3.81,  
          Z -> 40,  α -> 1/137,  En ->  .75,  
          convert ->  .389 (10^(-27))}

LogPlot[
      crosssection[θ],  {θ,  15 π/180,  
        60  π/180},  
      Ticks -> {{{20  π/180, 20}, {30  π/180, 
              30},  {40  π/180, 40}, {50  π/180, 50}},  
          Automatic}, AspectRatio -> 1.6,  
      AxesLabel -> {degrees,   (cm^2)/sr}]