Question about detectors and solid angles

[yxgao]

There's a square detector of area A. When a point source is placed at the center of this detector, half of the emitted particles are detected. What fraction of the particles are detected when the point source is placed a distance x away?

I figured that at x=0, 50% of the particles are detected because the solid angle is 2\pi, half of the full solid angle. I also know that the amount of gamma rays detectors falls with the radius as \frac{1}{r^2}. Should I try to calculate the solid angle at a distance x?

There is an equation \Omega=\frac{A}{r^2} that relates the solid angle, area, and r, but the area here is the area on the sphere, not of the detector, so I don't know how to proceed from here.

Any help is appreciated.
Thanks!

 

[Astronuc]

You simply want the area of the detector subtending the solid angle at a distance, r, (or x as the case may be) from the source.

For a point source, the radiation goes in all directions, and this is a solid angle (4\pi) through which 100% of the radiation passes.

At a distance r, the surface area of the sphere is 4\pi\,r^2. A detector of area A placed at r would intercept a fraction of the total radiation given by

\frac{A}{4\pi\,r^2} - or simply the ratio of the areas.

This excludes scattering and reflection from other surfaces and the detector itself.

 

[yxgao]

Thanks for your reply. It's not giving me the correct answer to the problem for some reason.

The area is not along the area of the sphere, but is of a flat square detector with side s, where A = s*s. It's not part of the surface area of the sphere - how does this change your answer?

Thanks!

 

[Astronuc]

Given a source strength (particle/sec), can you calculate the flux, Q" (particle/sec-unit area), at some distance 'r' or 'x'.

Then the efficiency is the product of the flux and detector area, divided by the source strength, S.

or Q"A/S, where in this case the detector area is s².

 

[yxgao]

Here are the numbers used in the problems. We aren't given the source strength or the flux in this case.

The detector is a square 8cm by 8 cm.

What is the fraction of detected particles at a distance of 1 meter?

The answer is 4*10^-4. I have no idea how they get that. There is no factor or pi, which would come in if I used \frac{A}{4\pi\,r^2}

 

[Astronuc]

Well, the detector area is 64 cm²

and at a distance of 100 cm (=1 m), the detector would cover approximately 4\pi100².

So fraction is \frac{64}{4\pi 10000} ~5E-4.

However, the detector does not cover exactly 64 cm², because 1 pt must be tangent to the sphere at radius, r. So there is a geometric correction to the solid angle. One has to project the area of the detector onto the surface of the sphere. But I don't imagine the difference would be 20%, but I could be wrong.

On the other hand, it may be that the author(s) of the question is taking an 'effective' diameter or radius.

If one selects an effective radius of the detector as, r = s/2 = 8 cm/2 = 4 cm, then the problem becomes -

\frac{\pi 4^2}{4\pi 10000}

The pi's cancel so it is 16/(4*10000) = 4E-4, which is the number that you mention.

I would recommend working through the geometry of the square detector tangent to a sphere of radius 100 cm, and see if the projected area is approximately 16 pi or 50.265 cm².