GREPhysics.NET: GR0177 #14

GR0177 #14

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }Cross Section Formula

When the detector is placed near the source, the particles enter it directly from the circle at the end of the cylinder. The area is $A=\pi d^2/4$

When the detector is placed 1 m away, the particles enter it through a sphere of radius 1 m, thus $A=4\pi R^2 = 4\pi$ .

The ratio of areas gives the fraction of detected gamma rays, $d^2/16=8^2E-4/16=4E-4$ , as in choice (C).

This is due to Jon Jockers.

Alternate Solutions

backyard
2006-10-24 06:19:27

I think the real answer should be 50 percent of that given by choice(c)

julio
2006-11-03 02:10:04

backyard : No I don't think so.

50 percents of all emitted gamma rays are detected in the first case because de detector doesn't dectect the rays that go in the opposit direction.

But for the second case, Jon Jockers' solution is rigorous.

dirichlet
2006-11-14 11:49:55

The solution was absolutely faulty. As the distance from the source i.e. the radius of the sphere is large compared to the diameter it produces a good result. In fact we should have worked with a particular zone of the sphere whose sagitta will be given by the formula

h=\0.5{sqrt{R^2+d^2/4}-R}/\sqrt{R^2+d^2/4}.

From this we can find the area of the required spherical zone via A=2\pi*R*h. And then dividing it by 4\pi*R^2 we get the required result which is now as I can see 3.992*10^(-4). mathbeautymobius@yahoo.com

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Comments

BerkeleyEric
2010-07-11 18:45:57

From geometry we know that the percentage of gamma rays detected is linearly proportional to the solid angle.

The solid angle for the far-away source is A/R^2 and the solid angle for the close-source is (by symmetry) 2 pi. Taking the ratio and multiplying by 50% gives choice (C).

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MuffinSpawn
2009-07-16 09:56:35

I'm thinking of this as simply a diminishing intensity problem. The 50% detected is a red herring I think. It just says that the detector is 100% efficient, but initially approximating the detector is an infinite plane only half of the particles will hit it.

So the fractional intensity is 1 ( $I_0=1$ ), and then the intensity at $R=1m$ is just $I_f=\frac{I_0}{4 \pi R^2}=\frac{1}{4 \pi}$ .

The fraction of particles that hit the front of the detector is then $I_f A=(\frac{1}{4 \pi}) (16\pi \times 10^{-4} m)=4 \times 10^{-4}$ .

chemicalsoul
2009-10-28 08:41:04

This is a the best answer!

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Ethan
2008-10-10 15:22:06

I don't know if this adds, but it clarifies the disagreement:

The exact answer is given by the solid angle subtended on the sphere ie $4\pi\sin^{2}(\frac{\theta}{2})$ over the total area $4\pi$ . For our case this is approximatelly the area of the detector and so the approximation is fine.

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Anastomosis
2008-04-10 12:33:33

I guess this really isn't an "alternate" solution, but I just set up a quick ratio, as ratios are among the fastest methods of solving problems, especially on this test.

This is going to be basically the same exact thing as the solution above, just a little easier to follow, I think.

Quick ratios:

$x=\frac{A_{detector}}{A_{sphereFar}}$

and:

$\0.5 = \frac{A_{detector}}{A_{sphereClose}}$

These ratios just say that the fraction of gamma-rays that are detected is going to be the ratio of area of the detector to the total area (of a sphere with the same radius as the distance to the detector).

So, quickly divide the first by the second, the $A_{detector}$ cancels out, and you are left with:

$\frac{x}{0.5} = \frac{A_{sphereClose}}{A_{sphereFar}}$

And what is the area of the close sphere? We know that it's just double the area of the detector $2 \times \pi(4\times 10^{-2})^2$ , since when it is at that distance, it catches 50% of the gammas. (I just thought of the detector at this distance as a hemisphere, half enveloping the radioactive sample).

So:

$x = \frac{0.5\times 2\pi (4\times 10^{-2})^2}{4\pi (1)^2}$

$= \frac{16}{4} \times 10^{-4}$

$= 4 \times 10^{-4}$

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cedric_tsui
2007-10-30 14:39:33

I agree that the solution is flawed. The math is correct, but for the wrong reasons.rnrnFirst. One must realize that 50% of the gamma rays are detected when the source is placed against the side of the detector tells us that the detector efficiency is 100%. That is, that every ray that so much as glances the face of the detector is counted. Thus the length of the detector is not important, as the radiation source can only 'see' the circular face.rnrnSecond. The detector is now 1m away from the source. Thus the fraction of the rays it detects is equal to the fraction of rays striking it, which is equal to the surface area of the circular face divided by the surface area of a sphere with a 1m radius.

cedric_tsui
2007-10-30 14:40:52

Ung. It killed my formatting.

I agree that the solution is flawed. The math is correct, but for the wrong reasons.

First. One must realize that 50% of the gamma rays are detected when the source is placed against the side of the detector tells us that the detector efficiency is 100%. That is, that every ray that so much as glances the face of the detector is counted. Thus the length of the detector is not important, as the radiation source can only 'see' the circular face.

Second. The detector is now 1m away from the source. Thus the fraction of the rays it detects is equal to the fraction of rays striking it, which is equal to the surface area of the circular face divided by the surface area of a sphere with a 1m radius.

sidharthsp
2008-10-21 01:14:46

I second that, that is the best explanation. that 50 % thing is a distraction or as "cedric_tsui" says a measure of efficiency of detector...

-Sid

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kostas_zarganis
2007-01-11 05:18:46

À =3.14...

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kostas_zarganis
2007-01-11 05:16:26

I have to show an easier way to find the solution.... Initially for a distance of the source s=1m we have a sphere's volume (4Às^2)/3=4À/3 m^3

The radiation that is detectable in that case is related with the volyme of a cone with base:the cross section of the detector and height h the distance between the source and the detector (h = s = 1m). The volume of the come is:1/3*Àr^2*h (r = 0.04 m)=(16/3) À 10^-4 m^3.

The fraction is the: (volume of the cone)/ (volume of the sphere)=4*10^-4

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freeform
2006-12-01 23:32:48

It's all about solid angles. (see wikipedia:steradians). Think of the source as an isotropically emitting pt source. With source directly in front of detector, half of all emissions captured by detector as stated in problem. 1 meter away, the fraction is simply area of disk / area of sphere of radius R.

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backyard
2006-10-24 06:19:27

I think the real answer should be 50 percent of that given by choice(c)

julio
2006-11-03 02:10:04

dirichlet
2006-11-14 11:49:55

KarstenChu
2007-03-22 13:36:37

Yes, Jon Jocker's solution *happens* to work out because the distance the detector is away from the source is large compared to the diameter of the source. In this limit we can take the area of the opening to be almost equal to the fraction of the surface area of the sphere that represents where all the radiation is going.

My point is just to take his solution with a grain of salt.

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