# Points close to the source not behaving like inverse-square

1. May 25, 2014

### steamyoshi

1. The problem statement, all variables and given/known data
The problem - confirm that the amount of radiation measured from a source is related by inverse-square to the distance.

Here is a graph of my data fitted to y = a1 * x^a2.
Every point is an average of 3-8 measurements (as many as needed to have a statistical error of <3%)
a1 = 2.7*10^4; a2 = -2.2; so it seems we have a good fit to the theory, and all the points are on the graph,
except the two closest points which are much lower than expected, and I have no explanation for this strange behavior.
What could cause those two points to be measured so low?

2. Relevant equations

3. The attempt at a solution
Tried looking it up on the internet and asking my lab instructor, and I'm stumped.

Edit: Residuals plot might be more helpful

2. May 25, 2014

### verty

Perhaps your geiger counter has a resolution of about 10ms and starts to miscount if you go much beyond 1000 per 10s. I think it's just measurement error.

3. May 25, 2014

### AlephZero

You might want to find out exactly how the least squares fitting works, if you are using a spreadsheet etc rather than doing it by hand. It is possible the different points are given different weighting factors that bias the fitting to one end of the curve more than the other.

Another possibility is the geometry of your measurement system. Maybe you should really be measuring the distance from the source to the (average) position inside the counter where the radiation is detected, not the distance to the front of the counter. That will have more effect on small distances than on large ones. You could try a fit of the form $y = a/(x + d)^2$ where $d$ is a constant depending on the shape of the counter.

4. May 26, 2014

### steamyoshi

Thank you for the replies. I'm trying to obtain the technical specs. for the counter, which will allow me to know both the resolution and tube length, and then I will check your suggestions.
Another thing a friend suggested is that because the loss from inverse-square at this distance is very low, the absorption by the air is no longer negligible in comparison.

5. May 26, 2014

### haruspex

I'm not sure your problem is with those points. After all, the exponent should be -2, not -2.2. What do you get if you plot inverse intensity against square of distance and take a straight line fit through the origin? Which points look off now?

6. May 26, 2014

### AlephZero

Ah... I missed that in the OP.

If the experiment was to measure the intensity assuming an unknown power law, -2.2 is what you measured, and you can't arbitrarily cook the results to make them closer to what you expected. If you refine the measurements, you might get closer to -2 or you might not, depending on what was causing the difference.

But if you want to verify the "correct" value of -2, then you should do what Haruspex said. There may still be a systematic deviation from the expected results, which you might be able to explain.

7. May 26, 2014

### haruspex

Actually, I'm saying a bit more than that. Even if the aim is to deduce the exponent, the set of points which look 'off' can be an artefact of what you chose to plot against against what. If you do the same curve fit but plotting inverse intensity against distance squared you might not get an exponent of 1.1, or the same set of ill-fitting points.

8. May 26, 2014

### SammyS

Staff Emeritus
Yes, I also think this may be part of the problem with the nearby points, although a power of -2.2 is quite disturbing.

What is the cross-sectional area of the active part of the detector ?

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9. May 26, 2014

### rcgldr

The formula for inverse square law needs to take into account the size and shape of the source of the radiation and the size and shape of the detector in the geiger counter. The simple formula for intensity = 1 / r^2 assumes a point source and a point detector. Spherical sources will also follow the rule as long as the spheres don't touch or intersect. If the distance is large enough, then the formula for point source and detector will closely approximate the actual intensity, but at close distances, you'll need to use a formula based on a single or double integral.