Geometric efficiency of a detector

In summary, the geometric efficiency of the setup is 15%. Approximately 15% of the gamma rays emitted by the source make it to the detector.
  • #1
Soilwork
102
0
I'm not to sure how to do this question.

Q. A 4MBq gamma source emitting 5 KeV photons is held at a distance of 5 cm from the end window of a detector. The diameter of the detector window is 3.5 cm, and the quantum detection efficiency of the detector is 15%.

i)What is the geometric efficiency of the setup?
ii) What fraction of the source activity is recorded by the detector?

Well I dont' particularly have a clue about the first part. I know that if it is point source you can neglect the term
y(gamma)= (radius of source)^2/(distance to detector)^2
But the question doesn't say if it is a point source or not.
If it isn't then I don't really know because you don't have the radius of the source.
Obviously I will have to use all the information given and I just can't see how to use it all.
I'm sorry I'm not too good at this stuff, but I'm trying to study for my exams in a couple of weeks and would appreciate any help :)
 
Physics news on Phys.org
  • #2
Soilwork said:
I'm not to sure how to do this question.

Q. A 4MBq gamma source emitting 5 KeV photons is held at a distance of 5 cm from the end window of a detector. The diameter of the detector window is 3.5 cm, and the quantum detection efficiency of the detector is 15%.

i)What is the geometric efficiency of the setup?

Hint: If the detector window COMPLETELY surrounds the source, you will have 100% geometric efficiency.

It is asking for how much of the photons actually made it into the detector at that distance, with that opening area.

Zz.
 
  • #3
To find the number of photons make it to the detector you would need to know the radius of the source which you don't know.
so would it be reasonable to assume that it's a point source then?
 
  • #4
Soilwork said:
To find the number of photons make it to the detector you would need to know the radius of the source which you don't know.
so would it be reasonable to assume that it's a point source then?

One can only assume, after reading the nature of the problem, that the source emits those photons isotropically.

Zz
 
  • #5
Thank you for that :)
I think I can do the second part using the fact that Ro=kR
where R is the activity of the source and Ro is the detected activity.
k = (number of particles produced)*G*E
where G and E are the geometric and detection efficiencies.
 
  • #6
haha ok maybe I can't do the second part :(
I assume I'd have to use the quantum detection efficiency along with the initial activity and the energy of the gamma rays.
So damn frustrating because the lecturer's notes barely cover this and we don't actually have a book.
 

Related to Geometric efficiency of a detector

1. What is the geometric efficiency of a detector?

The geometric efficiency of a detector is a measure of how well the detector can capture and detect particles or radiation coming from a specific direction. It takes into account the size, shape, and positioning of the detector, as well as the properties of the particles or radiation being detected.

2. How is the geometric efficiency of a detector calculated?

The geometric efficiency of a detector is calculated by dividing the number of particles or radiation detected by the number of particles or radiation emitted in a given direction. This calculation takes into account the solid angle subtended by the detector and the total solid angle in which particles or radiation could have been emitted in that direction.

3. Why is the geometric efficiency of a detector important?

The geometric efficiency of a detector is important because it directly affects the accuracy and sensitivity of the detector. A higher geometric efficiency means more particles or radiation can be detected, leading to more accurate measurements and a lower detection limit. It also helps in optimizing the design and placement of detectors for specific experiments or applications.

4. How can the geometric efficiency of a detector be improved?

The geometric efficiency of a detector can be improved by increasing the size and/or the sensitivity of the detector, as well as by optimizing its shape and positioning. Additionally, using materials with high atomic numbers and reducing the distance between the detector and the source can also improve the geometric efficiency.

5. Are there any limitations to the geometric efficiency of a detector?

Yes, there are limitations to the geometric efficiency of a detector. These include physical constraints, such as the size and shape of the detector, as well as the properties of the particles or radiation being detected. Additionally, external factors such as background noise and interference can also affect the geometric efficiency of a detector.

Similar threads

Replies
1
Views
192
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
877
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
16
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
34
Views
2K
  • Special and General Relativity
Replies
13
Views
1K
  • Electrical Engineering
Replies
9
Views
954
  • Quantum Interpretations and Foundations
3
Replies
79
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top