# Can this diagram make any sense? [counts of alpha decay]

1. Mar 31, 2014

### MortalWombat

Hello all you physics folks,

this is my first post, so if I screw this up, go easy on me :)

Here's the problem I'm working on and that I simply can't get my head wrapped around:

I have a table of 10 values of counts per second (cps) of alpha decay of Americium-241, depending on the pressure $p$ inside a chamber, where at a distance of $x_0 = 6$ cm a detector is mounted.

I have to plot the cps...but not against the pressure, but a distance corresponding to that pressure at normal pressure levels ($p = 1 bar$). I am to use Boyle-Marriots law
$$p \cdot V = p \cdot A \cdot x = const.$$

Since the cross-section of the chamber $A$ can be considered constant, we have
$$p \cdot x = const.$$

But this means that for decreasing pressure the distances get longer (that makes sense), but when I plot $p_i$ vs $x_i$, I get an increasing cps count for longer distances, which is pretty much the opposite of what we'd want...any idea where I made a mistake here? Do I have to modify the cps counts in any way?

EDIT: my bad, I wanted to post a screenshot of my current graph
http://imgur.com/vegvPSu

Last edited: Mar 31, 2014
2. Mar 31, 2014

### nasu

So what is the meaning of that x in the equation?
How do you change the pressure in the chamber? You have a piston that you move in the chamber?

3. Mar 31, 2014

### kurros

You mean you plot pressure "p" vs effective distance "x"... where do the counts per second appear in such a plot? Isn't this just a plot of y=A/x, for some constant A?

4. Apr 1, 2014

### 256bits

let's see.
You have a chamber with a radioactive isotope, and a detector at a set distance.
With increasing pressure in the chamber , you count the number of detections.

I imagine that the number of detections should vary inversely with the pressure. Is that what happens?

Then you want to plot the number of detections, versus an effective distance, with regards to pressure.

Your problem is how to convert the pressure into an effective distance?

Since, by increasing the pressure, the density of the gas also increases, and thereby the number of gas particles the radiation has a chance of encountering also increase.
Would not the effective distance would then just be Px/Po times Xo, where Po and Xo are the base values at 1 bar, and Px is the values of the prescribed pressure readings.
Or have I interpreted the experiment incorrectly.

5. Apr 1, 2014

### MortalWombat

Yep, that's pretty much it. I figured it out yesterday night, the plot now looks like this

So after a short distance nearly all alpha particles are absorbed or stopped by air molecules, and then the counts are pretty much 0 for higher distances.

6. Apr 2, 2014

Excellent.