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

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

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'''This is James Holton's explanation - too good to be buried in the [https://www.jiscmail.ac.uk/cgi-bin/webadmin?A2=CCP4BB;674b03b0.2005 CCP4BB] archives:'''
'''This is James Holton's corrected explanation - too good to be buried in the [https://www.jiscmail.ac.uk/cgi-bin/webadmin?A2=CCP4BB;674b03b0.2005 CCP4BB] archives:'''


''Subject: Re: Dose in diffraction patterns?''
''Subject: Re: Dose in diffraction patterns?''
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In a pinch, you can estimate the flux by the total number of photons on the image (P).  This is assuming that you know the sample thickness (L) in microns.  You must also assume that the total scattering cross section of the atoms in the sample is close to that of oxygen (0.2 cm^2/g), that the sample density is 1.2 g/cm^3 and that about 50% of the scattered photons reach the detector.  None of these are terrible assumptions. The equation then becomes:
In a pinch, you can estimate the flux by the total number of photons on the image (P).  This is assuming that you know the sample thickness (L) in microns.  You must also assume that the total scattering cross section of the atoms in the sample is close to that of oxygen (0.2 cm^2/g), that the sample density is 1.2 g/cm^3 and that about 50% of the scattered photons reach the detector.  None of these are terrible assumptions. The equation then becomes:


f = P/t/L*1.2e-5
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).
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).
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Getting P from a pixel array is easy: you just add up all the pixel values.  From a CCD you want to be careful to subtract the baseline value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for ADSC Q315 (swbin) and 1.8 for Q315r (hwbin).  A few considerations, yes, but it can be a good sanity check.
Getting P from a pixel array is easy: you just add up all the pixel values.  From a CCD you want to be careful to subtract the baseline value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for ADSC Q315 (swbin) and 1.8 for Q315r (hwbin).  A few considerations, yes, but it can be a good sanity check.


-James Holton
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  
MAD Scientist
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>
 
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  
Oh, and if you want a reference for that 2000 ph/um^2 = 1 Gy rule, it is  
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https://doi.org/10.1107/S0909049509004361
https://doi.org/10.1107/S0909049509004361


And, of course, if you are lucky enough to have accurate flux, size and  
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/
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
-James Holton
MAD Scientist
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>
</pre>
JMHolton
4

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