- #1
darkchild
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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.