Geometric efficiency of a detector

[Soilwork]

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 :)

 

[ZapperZ]

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.

 

[Soilwork]

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?

 

[ZapperZ]

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

 

[Soilwork]

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.

 

[Soilwork]

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.