Some ways to calculate the radiation dose that a crystal has absorbed: Difference between revisions

Some ways to calculate the radiation dose that a crystal has absorbed (edit)

1,355 bytes added , 7 May 2020

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 -James Holton
 MAD Scientist
+</pre>
+and he posted two corrections, the first at Thu, 7 May 2020 09:06:29 -0700:
+<pre>
+Ah!  I did that last formula wrong.  Never do algebra in your head
+without checking. It should be:
+The equation then becomes:
+f = P/t/L/1.2e-5
+Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux
+and t=exposure (as above).
+For example, if you see an average pixel value of 20 photons on a
+Pilatus 6M, then that is P=12e6 photons.  If that was a t=0.1 s exposure
+from a sample 100 microns thick, then the beamline flux was about 1e12
+photons/s.  Note that this is the flux after any attenuation, not before.
+Oh, and if you want a reference for that 2000 ph/um^2 = 1 Gy rule, it is
+here:
+https://doi.org/10.1107/S0909049509004361
+And, of course, if you are lucky enough to have accurate flux, size and
+shape information for the beam and sample, plus chemical composition the
+most accurate dose you'll get from raddose-3D: https://www.raddo.se/
+-James Holton
+MAD Scientist
+</pre>
+and at Thu, 7 May 2020 09:33:03 -0700
+<pre>
+One more correction:
+For example, if you see an average pixel value of 20 photons on a
+Pilatus 6M, then that is P=120e6 photons.  If that was a t=0.1 s
+exposure from a sample 100 microns thick, then the beamline flux was
+about 1e12 photons/s.  Note that this is the flux after any attenuation,
+not before.
 </pre>