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Q:
Take 3 imaginary isotopes.
#1 has a peak at 100 keV only, # 2 has a peak at 662 keV only and #3 has a single peak at 3 MeV.
Each sample has the same number of photons output per minute, let's call it 1000 to make it ultra simple.
What counts would be expected in peak 100 keV, 662 keV and 3 MeV respectively, assuming a 60 second count time.
Givens:
All samples are the same size, density and distance form the probe.
The geometry of all samples is equal and the probe housing passes all energies equally well.
A:
The way to answer this question would be:
100 keV = 500% N
662 keV= 100% N
3 MeV = 20% N
Whatever number of counts would be in the 662 keV peak would equal 100 %. The 100 keV peak would have 5 times that may counts while the 3 MeV peak would contain 25% as many counts.
Per this chart:
Notice that the chart is centered on 662 keV, all of the detector charts are. Detectors are "normalized" to 662 keV.This implies that whatever it reads at Cs-137 is correct, and whatever it reads at any other energy needs to be compensated.Meters and MCA's do not compensate for this factor.
The response to equal numbers of photons of different energies is different. Far different tin fact. The same thing applies to a meter unit with a dial on it. Geiger Muller tubes do the same thing, with even more wiggles in the curve due to gas factors.
The reason I beat on this is that it is a fundamental principle, against which all our measurements are eventually made.
If this is not clear to all members, I will rethink how to explain it. It's that important.
Geo
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