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  • MICHAEL SHORT: OK.

  • I think things have been getting pretty derivy lately,

  • so I wanted to shift gears to something a little bit more

  • practical.

  • So I started alluding to this hypothetical radiation source

  • I might have right here, and things

  • like if you have a source of known activity, which

  • we calculated yesterday, and you have

  • a detector of unknown efficiency,

  • how do you know what the efficiency is?

  • How do you know what, let's say, your dose distance relationship

  • is?

  • And how do you calculate all this stuff?

  • So let's take the general situation that we're

  • starting to work out.

  • Let's say we have a Geiger counter right here.

  • That's our GM tube.

  • And we have a point source that's

  • emitting things in all directions.

  • Let's go with the stuff from yesterday.

  • Let's say it's a cobalt 60 source.

  • It's now 0.52 microcurie.

  • The question is, how many counts do you expect in this detector

  • when it's a certain distance away?

  • So I've actually laser-cut out a little Geiger counter jig

  • from a previous class.

  • And you guys can all do this too.

  • Who here has been to the IDC before?

  • A couple.

  • The international design center--

  • so they've got a laser cutter that you can sign up to use,

  • which is where I did this.

  • And it's set to just take a Geiger counter

  • and put your sources at some fixed distance

  • away so you can discover the dose distance

  • relationship with things.

  • Speaking of, does anybody know what the relationship

  • is between dose and distance or measured activity and distance?

  • Yeah, Luke.

  • AUDIENCE: [INAUDIBLE] r cubed.

  • MICHAEL SHORT: Close.

  • It's, let's say, the measured activity

  • would be proportional to 1 over r squared.

  • Who knows where this comes from?

  • I'll move the source a bit away to lessen the beeping.

  • Yeah.

  • AUDIENCE: Well, the flux of particles coming out

  • is just [INAUDIBLE] over the surface area of [INAUDIBLE]

  • and the [INAUDIBLE] is 4 pi r squared.

  • MICHAEL SHORT: Yeah, exactly.

  • If you were to draw a hypothetical sphere

  • around the source right here, then you've got,

  • let's say, a detector that's roughly rectangular

  • with a fixed area.

  • Let's say it's got a half length L and a half width

  • W. Then the area--

  • I'm sorry, let's just say length L, width W--

  • would be just L times W. And actually,

  • what Chris mentioned as the solid angle subtended

  • by this detector right here--

  • in other words, at a certain distance r away,

  • how much of this sphere--

  • how much does the area of this sphere-- does this detector

  • take up?

  • In other words, how many of these gamma rays

  • are going to go in a different direction than the detector,

  • versus how many we'll actually enter the detector?

  • And a simple formula for the solid angle

  • is just the surface area of whatever

  • you've got over r squared.

  • It's a pretty good approximation to the solid angle of something

  • for very long distances, and it's probably the one

  • that you'll see in the reading.

  • But I wanted to show you the actual formula, in this case,

  • for a rectangle--

  • solid angle comparison.

  • Good, that's up there.

  • So let's say on the x-axis, right here,

  • this would be distance from the source

  • to the detector in meters.

  • And I've said that we've got some sort of a detector that

  • is 2.5 by 10 meters in size.

  • That's an enormous detector.

  • Let's actually switch it to the units right here.

  • So this is roughly 10 centimeters long.

  • So let's change our length to 0.1.

  • And what do you think the width of this Geiger counter

  • is in meters?

  • AUDIENCE: A centimeter

  • MICHAEL SHORT: A centimeter.

  • 0.01.

  • We're going to have to change our axes so we can actually

  • see the graph.

  • So instead of looking all the way out to 15 meters away,

  • let's look one meter away, maybe less.

  • This whole thing is probably 50 centimeters.

  • And we'll take a look there.

  • And what we notice is that except for extremely

  • short distances, this approximate formula

  • for the solid angle-- or in other words,

  • if I were to draw a sphere around the source that's

  • the radius of the distance between the source

  • and the detector, how much of that

  • sphere's area does the detector take up?

  • This approximate formula-- the blue curve--

  • is a pretty good approximation of the red curve

  • until you get really, really close to 5 centimeters

  • away, or about this distance right here.

  • Does anyone know why this formula would break down?

  • What happens as r goes to 0?

  • What happens to our solid angle or our approximation

  • for our solid angle?

  • AUDIENCE: Goes to Infinity

  • MICHAEL SHORT: It goes to infinity, right?

  • Can a detector actually take up infinity area

  • on, well, anything?

  • Never mind that unit sphere.

  • Not quite.

  • If you were to take this detector and bring the radius

  • down to 0 so that the source and the detector,

  • if not counting for the thickness of the plastic,

  • were right upside each other, if that solid angle went to, well,

  • infinity , then the count should go to infinity,

  • and it does not compute.

  • Does anyone know how many--

  • first of all, who here has heard of solid angle before?

  • So a little more than half of you.

  • That's getting clicky.

  • I'm going to turn that off.

  • Solid angle is kind of the analog to regular old angle,

  • except in 3D.

  • So instead of looking at things in radians,

  • this has the unit of what's called steradians--

  • steradians-- with a full sphere taking up 4pi steradians.

  • Interestingly enough, 4pi is also

  • the surface area of a unit sphere with radius of 1.

  • So that's where this comes from.

  • If something were to completely cover a unit sphere--

  • like, if you were to, let's say, encase a light source in tin

  • foil completely, and say, how much of that solid angle

  • does the tin foil encase?

  • It would be 4pi steradians, regardless

  • of the size of the sphere or how much tin foil you had to use.

  • So this pretty simple formula isn't the best approximation

  • for it.

  • And I'm not going to go through the derivation,

  • because like I said, today is going

  • to be a more practical nature.

  • There is a more complex and rigorous formula

  • for the solid angle of something,

  • let's say, in this case, a rectangle of length L

  • and with W, from a certain distance r, or, in this case,

  • on our graph, x away from the sphere.

  • And you can actually see that red curve right there.

  • Once you get to a few centimeters away,

  • it's pretty close.

  • Anyone want to guess what the maximum value of the red curve

  • is?

  • If I take this source and slam it right up next

  • to the detector, how much of sphere

  • is the detector subtending?

  • AUDIENCE: 2pi

  • MICHAEL SHORT: 2pi-- half the sphere.

  • Because let's say this whole side of the source

  • is completely obscured by the detector and this whole side

  • is free to move.

  • And if you look really closely, yep, at 0, the correct formula

  • does give you 2pi steradians.

  • Which is to say that half the gamma rays leaving the source

  • would enter the detector.