Actually, people forget the 1/r term because it is gone by the end of
Chapter 6 of Woolfson.
Yes, it is true that, for the single "reference electron" the scattered
intensity falls off with the inverse square law of distance (r) and,
hence, the amplitude falls off with 1/r. However, the units of
"intensity" here is photons/area. This is not the same as the units of
"intensity" for an integrated diffraction spot (photons). That is,
spots are an integral over solid angle, so the 1/r^2 term goes away. A
shame, really, that we use the word "intensity" to describe so many
different things. Leads to a lot of confusion like this! It is also a
real shame that certain individuals are so draconianly eristic about
"units" because this discourages the use of colloquial "units" (like
photons/um^2 or electrons) as an educational tool.
The loss of the 1/r^2 term arises because diffraction from a crystal
is "compressed" into very sharp peaks. That is, as the crystal gets
larger, the interference fringes (spots) get smaller, but the total
number of scattered photons must remain constant. The photons/area at
the tippy-top of this "transform-limited peak" is (theoretically) very
large, but difficult to measure directly as it only exists over an
exquisitely tiny solid angle at a very precise "still" crystal
orientation. In real experiments, one does not see this
transform-limited peak intensity because it is "blurred" by other
effects, like the finite size of a pixel (usually very much larger than
the peak), the detector point-spread function, the mosaicity of the
crystal, unit cell inhomogeneity (Nave disorder) and the spread of
angles in the incident beam (often called "divergence" or "crossfire").
It is this last effect that often tricks people into thinking that spot
intensity falls off with 1/r. However, if you do the experiment of
chopping down the beam to a very low divergence, choosing a wavelength
where air absorption is negligible, and then measuring the same
diffraction spot at several different detector distances you really do
find that the pixel intensity is the same: independent of distance.
Yes, I have actually done this experiment!
-James Holton
MAD Scientist
On 10/14/2010 8:33 PM, William G. Scott wrote:
On Oct 14, 2010, at 2:28 PM, Jacob Keller wrote:
I have always found this angle independence difficult. Why, if the anomalous scattering
is truly angle-independent, don't we just put the detector at 90 or 180deg and solve the
HA substructure by Patterson or direct methods using the pure anomalous scattering
intensities? Or why don't we see pure "anomalous spots" at really high
resolution? I think Bart Hazes' B-factor idea is right, perhaps, but I think the lack of
pure anomalous intensities needs to be explained before understanding the
angle-independence argument.
JPK
Yo Jacob:
I think one thing that got ignored as I followed the other irrelevant tangent
is what f and F are.
f is the atomic scattering factor, and F is the corresponding Fourier sum of all of the
scattering centers. This holds for f_0 vs. F_0, f' vs. F' and f" vs. F". The
spots we are measure correspond to the capital Fs. Just like we add the f_o for each
scatterer together and we get a sum (F) that has a non-zero phase angle, this also holds for
F" (that is the part I missed when I posted the original question my student asked me).
The full scattered wave isn't given by f by the way. It is (1/r) * f(r) *
exp(ikr) so the intensity of the scattered wave will still tail off due to the that
denominator term (which is squared for the intensity). That holds for f_o, f' and
f" unless I missed something fundamental.
People tend to forget that (1/r) term because we are always focusing on just
the f(r) scattering factor.
-- Bill
