Compton Scattering: detector is 60 degrees from the symmetry axis

[Graham87]

Homework Statement

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Relevant Equations

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It says that the detector is 60 degrees from the symmetry axis. How can angles other than 60 degrees reach the detector? Wouldn't it be only the scattering polar angle of 60 degrees that reach the detector?
Why specifically a range between 60 to 120?
I can't see it geometrically and compton scattering ranges.

Thanks

 

[vela]

Where do the photons come from? What's their direction of propagation before they're scattered into the detector?

 

[Graham87]

Where do the photons come from? What's their direction of propagation before they're scattered into the detector?

The photons come from the Am source in the cylinder. Their directions are isometric, 360 degrees?

 

[vela]

Does the approximation that $h \ll r$—the source is flat—affect your answer?

 

[Graham87]

Does the approximation that $h \ll r$—the source is flat—affect your answer?

I'm not sure how $h \ll r$ affects the propagation. It just means that the photons only propagate in 2D instead of 3D?
The polar angle is the angle between incident photon and scattered photon.

In flat source the propagation direction before scatter would be perpendicular to the surface, ie 0 or 180 degrees? And that would mean 60 or 120 degrees to the detector?
But it says 60 to 120 degrees, not 60 or 120 degrees. That would mean that the scatter propagation is not perpendicular to the surface?

I can see that if the photon propagates in perpendicular at top surface +z direction it has to scatter at 60 degrees. And in bottom propagation -z it has to scatter 120 degrees to reach detector.

 

[TSny]

In flat source the propagation direction before scatter would be perpendicular to the surface, ie 0 or 180 degrees? And that would mean 60 or 120 degrees to the detector?

If a $\gamma$ ray is emitted from a radioactive nucleus in a direction parallel to the z-axis, does the photon have much chance of being scattered?

For which initial directions of propagation is scattering most likely to occur?

 

[Graham87]

If a $\gamma$ ray is emitted from a radioactive nucleus in a direction parallel to the z-axis, does the photon have much chance of being scattered?

For which initial directions of propagation is scattering most likely to occur?

The Compton cross section probability is dependent on incoming photon energy. So the likelihood of scattering has nothing to do with direction. Am I missing crucial information about scattering?

I don't get how the thickness of the source material relates to the scattering properties and probabilities.

 

[TSny]

The Compton cross section probability is dependent on incoming photon energy. So the likelihood of scattering has nothing to do with direction. Am I missing crucial information about scattering?

I don't get how the thickness of the source material relates to the scattering properties and probabilities.

In order for a photon to be Compton scattered, it must encounter an electron.

 

[TSny]

In order for the scattered photon to reach the detector, I find the range of scattering angle of the photon to be $30^{\circ} \leq \theta \leq 150^{\circ}$ instead of $60^{\circ} \leq \theta \leq 120^{\circ}$. So, I disagree with the solution given in the OP.

@vela: have you worked it out?

 

[vela]

In flat source the propagation direction before scatter would be perpendicular to the surface, ie 0 or 180 degrees? And that would mean 60 or 120 degrees to the detector?
But it says 60 to 120 degrees, not 60 or 120 degrees. That would mean that the scatter propagation is not perpendicular to the surface?

You have to take into account the source is a flat cylinder—it extends into and out of the plane of the figure.

@vela: have you worked it out?

I agree with your answer.

 

[Graham87]

You have to take into account the source is a flat cylinder—it extends into and out of the plane of the figure.


I agree with your answer.

So the propagation of the photons when $h <<r$ is perpendicularly straight out of the flat surface at +z and -z?

In that case wouldn't the only angles that reach the detector be either 60 degrees or 120 degrees?
How did they get a min-max range?

I'm not sure how the photons propagate. Do they all either propagate at -z or +z perpendicularly to the surface, or can they propagate non-perpendicularly from the surface?

I'm not yet following the answer $30^{\circ} \leq \theta \leq 150^{\circ}$, let alone $60^{\circ} \leq \theta \leq 120^{\circ}$.

 

[TSny]

So the propagation of the photons when $h <<r$ is perpendicularly straight out of the flat surface at +z and -z?

No. The source is emitting gamma rays in all directions. But only in certain directions do the gamma rays have a significant chance of undergoing a Compton scattering. (Imagine the height $h$ of the source to be infinitesimal compared to the radius $r$.)

In that case wouldn't the only angles that reach the detector be either 60 degrees or 120 degrees?
How did they get a min-max range?

Yes, but the chance that a gamma ray emitted parallel to the z -axis encountering an electron is negligible. So, these scatterings can be neglected.

I'm not sure how the photons propagate. Do they all either propagate at -z or +z perpendicularly to the surface, or can they propagate non-perpendicularly from the surface?

The gamma rays are emitted in all directions.

I'm not yet following the answer $30^{\circ} \leq \theta \leq 150^{\circ}$, let alone $60^{\circ} \leq \theta \leq 120^{\circ}$.

Let the detector lie in the y-z plane as shown. If a gamma ray is emitted inside the source in the +y-direction, at what angle would it need to scatter in order to reach the detector?

If a gamma ray is emitted in the +x-direction, what is the scattering angle to reach the detector?

How can you get a scattering angle of 150^o?

 

[Graham87]

No. The source is emitting gamma rays in all directions. But only in certain directions do the gamma rays have a significant chance of undergoing a Compton scattering. (Imagine the height of the source to be infinitesimal compared to the radius .)

What are the certain directions the gamma rays have a significant chance for Compton scattering when h is negligible?
I asked how being a flat surface would affect the direction of the gamma rays, and I get replied that the gamma rays emit in all directions. So I assume being flat does not affect the rays direction? Now you say that they do?

My question is I'm not sure how the photons propagate, and I get the answer that they propagate in all directions. How should I interpret that answer if the photons have a range between 60 to 120 degrees that they scatter in? If they can propagate in any direction, then how can I know the likeliest angle it scatters?
If they can propagate in any direction, what does infiniteseimal h have anything to do with it?
The scatter would in that case be in any direction too?







Update:
Wait a minute, I think I get it.
When the material becomes flat there is no attenuation on z-axis, hence no possibility to scatter.
The photons will propagate along x-y-axis, and from there scatter towards the detector, which should indeed be in the range 30-150 degrees.

 

[TSny]

And why can't I get a scattering angle below 30 or above 150 degrees?

Strictly speaking, you could have a scattering angle less than 30 degrees into the detector (or greater than 150 degrees). But, it is very unlikely when $h \ll r$.

An example is shown in the diagram below. A nucleus at $e$ emits a photon in the y-z plane in the direction of the arrow $E$. The arrow $D$ also lies in the y-z plane and shows the direction toward the detector. Then it would be possible for the photon to scatter off an electron at some point between $e$ and $a$.

If the angle between the arrow $E$ and the y-axis is 10 degrees, you can see that the scattering angle $\theta_s$ into the detector would be 20 degrees (thus, less than 30 degrees). But note that the distance $ea$ for which scattering can take place is pretty small. The smaller this distance, the less likely that a Compton scattering will happen. A photon emitted from $e$ along the y-axis, for example, can travel a greater distance inside the material. So the chance of Compton scattering is increased. Then, the scattering angle $\theta_s$ into the detector would be essentially 30 degrees.

If $h \ll r$, the picture might look like this:

The distance $ea$ is greatly reduced and there is little chance of scattering into the detector. Photons emitted along the y-axis can travel much farther inside the material; therefore, they have a much greater chance of scattering.

I leave it to you to think about emission points $e$ that are not on the y-axis.

 

[TSny]

Update:
Wait a minute, I think I get it.
When the material becomes flat there is no attenuation on z-axis, hence no possibility to scatter.
The photons will propagate along x-y-axis, and from there scatter towards the detector, which should indeed be in the range 30-150 degrees.

Yes, I think you have it.