Fraction of Radiation Detected

In summary: The approximation in the solution is the assumption that the source is far away so that the solid angle seen by the detector is small. In this case, the solid angle is ~area of detector / distance^2, and the area of the detector is pi*r^2, and the area of a unit sphere is 4*pi, so the fraction seen is pi*r^2 / (4*pi*distance^2). At distance=1m, this is pi*r^2 / 4, giving the value in the solution. The approximation is that the solid angle seen is the area of the detector divided by the area of the sphere.In summary, the conversation discussed a Physics GRE practice test question involving a Na
  • #1
darkchild
155
0

Homework Statement


This is from a Physics GRE practice test:

An 8-cm-diameter by 8-cm-long NaI(Tl) detector detects gamma rays of a specific energy from a point source of radioactivity. When the source is placed just next to the detector at the center of the circular face, 50% of all emitted gamma rays at that energy are detected. If the detector is moved to 1 meter away, the fraction of detected gamma rays drops to

(A)[tex]10^{-4}[/tex]
(B)2 X [tex]10^{-4}[/tex]
(C)4 X [tex]10^{-4}[/tex]
(D)[tex]8 \pi[/tex] X [tex]10^{-4}[/tex]
(E)[tex]16 \pi[/tex] X [tex]10^{-4}[/tex]

Homework Equations



None are given, and I don't know if there are any in particular I'm supposed to know.

The Attempt at a Solution



Please see the attached jpeg image for how I set my figure up. It shows the detector from the side (basically an 8cm long circular cone). I assumed that the source was still lined up with the center of the detector, i.e., just moved 1 m straight backwards. First, I decided that, when immediately adjacent to the source, the detector was privy to gamma rays emitted at any angle such that [tex] - \pi < \theta < \pi[/tex] (for a total range of 180 degrees) where the angle is measured from the straight line runs from the source to the center of the detector. At 1 m distance, the magnitude of this unknown range is the variable [tex] \theta [/tex].

Using lines from the source to the edges of the detector, I constructed the triangle that is in blue. Using the right triangle,

[tex] 2 \arctan (\frac{4 cm}{100 cm}) = \theta [/tex].

I expressed this as a fraction of the total angular range of emitted gamma rays using a proportion: A 180 degree range of gamma rays corresponded to 50% of the total emitted, and so:

[tex]\frac{\pi}{.5} = \frac{\theta}{x}[/tex]

where x is my final answer: 0.01273 which is way off; the answer is C.

I've never even seen a problem like this before, so I'm not sure if I'm doing it correctly.
 

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  • #3
i guess you messed up..also your derivation seems to be very complex..actually it is simple..nothing to do with angle!
Do you know about the inverse law for the illumination of light/radioactive stuff..
if you apply it the answer will be 'B'?
So try with inverse square law and let me know..
 
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  • #4
Rajini said:
if you apply it the answer will be 'B'?
So try with inverse square law and let me know..

I have the solution set. The answer is supposed to be C.
 
  • #5
zachzach said:
http://grephysics.net/ans/0177/14. Most helpful website for me while studying for the GRE (along with practice tests).

Thanks a lot. Unfortunately, there seems to be a lot of disagreement about how to find the correct answer on that post.
 
  • #6
Yes there is. I personally think the question is poorly written. Great site though. Good luck.
 
  • #7
This is my derivation.
For a point source we have the illumination formula:
[tex]\frac{\pi r^2}{4\pi l^2}[/tex]
Take [tex]r=4[/tex] and [tex]l=100[/tex]. [r is radius of detector and l is distance between detector and source].
Now you get [tex]4\times 10^{-4}[/tex].
Then for 50% you need to divide the above value by 2 so the solution would be [tex]2\times 10^{-4}[/tex].
good luck
 
  • #8
Rajini said:
This is my derivation.
so the solution would be [tex]2\times 10^{-4}[/tex].
good luck

Again, the answer is supposed to be C. I don't know whether you're saying that is a mistake/misprint, or just didn't read when I typed the answer previously.
 
  • #9
What is wrong in my solution?
 
  • #10
Rajini said:
What is wrong in my solution?

I don't know. I don't know how to do the problem, that's why I'm trying to get help. I only have the answer letter.
 
  • #11
Go to the link I posted and scroll down, there are a few posts that discuss the problem. Since it is a point source it is radiating in all directions. When the detector is just next to the source it captures 50% of the photons. This corresponds to an efficiency of 100%;the detector cannot detect the photons which are going the opposite direction but detects all the photons that arrive at the detector.
 
  • #12
Rajini said:
What is wrong in my solution?

This is correct except you divided by two at the end. The question asks what fraction of the total radiation you detect. To do this absolutely correctly you are just talking about the solid angle that detector covers with the point source at the origin. At first it covers 2pi solid angle, so 2pi/4pi -> 50%

Afterwards, Rajini has shown that a great approximation for a large enough distance is to just divide from the active area of the detector the surface area of a sphere centered on the point source with radius extending to the detector. There is no reason to divide by two though, the first part was all you need to get the correct answer.
 
  • #13
Oh I agree guys...
Obviously i took 50% as efficiency.
 
  • #14
Rajini said:
Oh I agree guys...
Obviously i took 50% as efficiency.

Yep, you assume 100% detection efficiency in the detector (since you detect 50% right next to the source this is implied), which is basically never going to happen in real life due to charge leakage or readout noise and many other effects.
 

What is "Fraction of Radiation Detected"?

"Fraction of Radiation Detected" refers to the proportion or percentage of radiation that is detected by a particular instrument or detector. It is a measure of the sensitivity and efficiency of the instrument in detecting radiation.

How is "Fraction of Radiation Detected" calculated?

The "Fraction of Radiation Detected" is calculated by dividing the amount of radiation detected by the total amount of radiation emitted by a source. This can also be expressed as a percentage by multiplying the result by 100.

What factors can affect the "Fraction of Radiation Detected"?

The "Fraction of Radiation Detected" can be affected by various factors such as the type and energy of the radiation, the sensitivity and efficiency of the detector, the distance between the source and detector, and any shielding or barriers between them.

Why is "Fraction of Radiation Detected" an important concept in radiation detection?

The "Fraction of Radiation Detected" is an important concept because it helps in evaluating the performance and accuracy of radiation detection instruments. It also allows scientists to compare different instruments and determine which one is more effective in detecting radiation.

How can the "Fraction of Radiation Detected" be maximized?

The "Fraction of Radiation Detected" can be maximized by using more sensitive and efficient detectors, reducing the distance between the source and detector, and minimizing any barriers or shielding between them. Regular calibration and maintenance of the instruments can also help in maximizing the fraction of radiation detected.

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