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Calculating the number of alpha particles incident on a detector

  1. Nov 26, 2011 #1
    Hi, firstly I'm not sure if I'm posting in the correct forum, if not
    could I request that the moderators move it to the right one.

    Ok, so my problem is pretty much as I stated in the title.
    The background to this is that I have been asked to roughly calculate the
    number of alpha particles that would be incident on a detector surface
    in an environment where there is 147 Bq m-3 of alpha
    particles from decaying Rn222.

    So, I have taken the maximum range in air for an α-particle to be
    3.4cm, so I am considering the decays in a hemisphere around the
    detector which for simplicity's sake I am assuming to be a square of
    side 1cm, and only has radiation incident on one side.
    I have worked out the total activity in the hemisphere to be 1.22x10
    -2 Bq, but I'm not sure how to proceed to work out how many
    actually hit the sensor.

    I believe it will probably involve some kind of volume integral, but
    I'm not sure what to integrate!

    I'm guessing that I plug something into the following equation

    [itex]\int_0^π[/itex][itex]\int_0^π[/itex][itex]\int_0^R[/itex] r2sinθ dr dθ d[itex]\phi[/itex]

    (I am also assuming the radiation to be isotropic)

    If anyone could point me in the right direction it would be greatly

    Many thanks
  2. jcsd
  3. Nov 27, 2011 #2
    Well if you forget about the effect of air for a moment, the number of particles which hit a detector with cross-sectional area A (pointing straight at the source) per second is just going to be the total activity times the ratio of A to the area of a sphere with radius of the distance to the detector. i.e. [itex] 147\frac{A}{4\pi r^2} [/itex]. Assuming isotropic radiation as you say.
    If you want to take the air into account you can figure out the total number of particles that make it the distance r and then multiply that by the same ratio. Of course your detector is probably not 100% efficient so there is an efficiency factor also. Although you asked about the number incident on the surface not the number actually detected so I guess you can forget about that.

    Although reading your question again I see perhaps you are not interested in a point source but some kind of dispersed radioactivity. In which case you can forget most of what I wrote :p.
    Last edited: Nov 27, 2011
  4. Nov 27, 2011 #3


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    It depends on the source strength, the geometry of the source with respect to the detector (including attenuation over distance), and any shielding on the part of the source and detector.

    Normally, for an alpha source, one might want to put the source inside the detector gas chamber.
  5. Nov 27, 2011 #4
    lol yes, which is why I am talking about integrating over the volume of a hemisphere which has the radius of the range of the α-particles in air and centred on the detector.
    Would it be a valid approach to integrate the [itex] 147\frac{A}{4\pi r^2} [/itex] equation?

    so it would be [itex]\int_0^π[/itex][itex]\int_0^π[/itex][itex]\int_0^R[/itex] [itex] 147\frac{A}{4\pi r^2} [/itex]r2sinθ dr dθ d[itex]\phi[/itex] ?

    As I mentioned, it is radon gas in air, the activity is 147 Bq m-3, and I am considering a hemisphere of the radon/air mixture with a radius equal to the range of the alpha particles in air, and there is no shielding.
    Also it is a semiconductor detector.
  6. Nov 27, 2011 #5
    Hmm, yes I think that should be ok as a rough approximation. You are just integrating over the activity at the detector due to each volume element of gas within the hemisphere. It assumes no loss of alpha particles due to air within the volume and 100% loss of those originating outside, and that the detector area looks the same from every angle (which will sort of suck as an assumption at the edge of the hemisphere), but if you are just after a quick estimate it is probably good enough.

    You could improve it by making A a function of the angles and figuring out the function describing alpha loss with distance and putting that in there too.
  7. Nov 27, 2011 #6
    That's exactly what I'm after (at the moment) thank you! :biggrin:

    I'm not quite sure what you mean by making 'A' a function of the angles though? I'll more than likely need to make it more precise, so it would be good to know!

    Incidentally, I used the above equation and got 1.25 x 10-4Bq actually falling on a square 1x1cm detector, it would be nice if someone could check that for me if they feel like it! (just looking for a 'yes', 'no', or 'that looks reasonable' lol)
  8. Nov 27, 2011 #7
    Oops, just realise that it's actually
    [itex]\int_0^π[/itex][itex]\int_0^π[/itex][itex]\int_0^R[/itex] 1.22 x 10-2[itex] \frac{A}{4\pi r^2} [/itex]r2sinθ dr dθ d[itex]\phi[/itex]

    Since it's 147 Bq m-3 but only 1.22 x 10-2Bq within the actual hemisphere, which takes into account the range of the alphas in air
  9. Nov 27, 2011 #8
    Well, from the viewpoint of some volume element of gas, the "target" the various alpha particles have to hit has a different cross-sectional area (say it is flat and pointing to the middle of the hemisphere), i.e. it will be the projection of A in the direction of the viewing angle, [itex]A.n=A\cos\theta[/itex]. So you could replace A with this factor in your integral. This assumes the distance from the detector to the bit of gas is large compared to the detector area so it will suck for the "close" bits of gas, but it is better than just A. You are still assuming no alpha particles get in through the sides of your detector or anything sneaky like that.

    Err actually the angle there is the angle between the surface normal and the vector to the gas element, which is a different theta to your spherical polar coordinate theta. You can make them the same though if you switch the cos to sin.
  10. Nov 27, 2011 #9
    Ok, that's very helpful, thank you :smile:
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