Light Collection Efficiency for Liquid Scintillation Detecto

AI Thread Summary
To calculate the light yield and light collection efficiency of a liquid scintillation detector, one must consider the geometric efficiency and attenuation of the scintillator material. The formula involves assessing the number of photomultiplier tubes (PMTs) used and their placement relative to the scintillator. If the source is immersed in the scintillator, the geometric efficiency approaches 1, though edge losses may occur unless the scintillator is fully surrounded by PMTs. Accurate calculations require data on the geometry of the scintillator and the configuration of the PMTs. Understanding these factors is crucial for determining the detector's performance.
Priyo
Messages
6
Reaction score
0
How do I calculate the light yield and light collection efficiency of a liquid scintillation detector? Couls someone give me a formula for that? What data do I need to calculate it?

THanks a bunch!
 
Engineering news on Phys.org
Priyo said:
How do I calculate the light yield and light collection efficiency of a liquid scintillation detector? Couls someone give me a formula for that? What data do I need to calculate it?

THanks a bunch!
How many photomultiplier tubes is one using. What is the geometry of the scintillator and the location of the PM tube or tubes?

One has to look at the geometric efficiency and the attenuation within the scintillator material.

For the system, one has to consider the geometric efficiency of the source with respect to the scintillator. If the source is immersed/dissolved within the scintillator, the geometric efficiency is close to 1, but there will be some edge losses, unless the scintillator is completely surrounded by PM detectors.
 
Hello, I'm currently trying to compare theoretical results with an MCNP simulation. I'm using two discrete sets of data, intensity (probability) and linear attenuation coefficient, both functions of energy, to produce an attenuated energy spectrum after x-rays have passed through a thin layer of lead. I've been running through the calculations and I'm getting a higher average attenuated energy (~74 keV) than initial average energy (~33 keV). My guess is I'm doing something wrong somewhere...
Back
Top