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MICHAEL SHORT: So since I know series decay is

a difficult topic to jump into, I wanted to quickly re-go over

the derivation today and then specifically go over

the case of nuclear activation analysis, which reminds me,

did you guys bring in your skin flakes and food pieces?

We have time.

So if you didn't remember, start thinking about what you

want to bring in, what you got.

AUDIENCE: Aluminum foil.

MICHAEL SHORT: OK, so you've got aluminum foil.

You want to see what in it is not aluminum-- excellent.

Well, what else did folks bring in?

AUDIENCE: [INAUDIBLE]

MICHAEL SHORT: OK, rubber stopper--

sound perfect.

Anyone else bring something in?

AUDIENCE: [INAUDIBLE]

MICHAEL SHORT: OK, so tell you what, when you bring stuff in,

bring it in a little plastic baggie.

I can supply those if you don't have them

with your name on them just so we know whose samples are what

because that's going to be the basis for another one

of your homeworks where are you going

to use the stuff that we're learning today

to determine which impurities and how much are in whatever

thing that you looked at.

And, of course, you're not going to get all the impurities

because in order to do that, we'd

have to do a long nuclear activation analysis,

irradiate for days, and count for a longer time.

So you'll just be responsible for the isotopes

on the shortlist, which we've posted on the learning module

site.

So again, bring in your whatever,

as long as it's not hair because, apparently, that's

a pain to deal with or salty because the sodium activates

like crazy or fissionable, which you shouldn't have, anyway.

I hope none of you have fissionable material at home.

So let's get back into series decay.

We very quickly went over the definition of activity

which is just the decay constant times the amount of stuff

that there is, the decay constants,

and units of 1 over second.

The amount of stuff-- let's call it

a number density-- could be like an atoms per centimeter cubed,

for example.

So the activity would give you the amount of,

let's say decays, per centimeter cubed per second.

If you wanted to do this for an absolute amount of a substance,

like you knew how much of the substance there was,

you just ditched the volume.

And you end up with the activity in decays per second.

That unit is better known as becquerels or BQ,

named after Henri Becquerel, though I don't

know if I'm saying that right.

But my wife's probably going to yell

at me when she sees this video.

But so becquerel is simple.

It's simply 1 decay per second, and there's

another unit called the curie, which is just a whole lot more

decays per second.

It's a more manageable unit of the case

because becquerel-- the activity of many things in becquerels

tends to be in the millions or billions or trillions

or much, much more for something that's really radioactive.

And it gets annoying writing all the zeros

or all the scientific notation.

And so last time we looked at a simple situation--

let's say you have some isotope N1 which

decays with the k constant lambda 1

to isotope N2, which decays with the k constant lambda 2 and N3.

And we decided to set up our equations

in the form of change.

Everyone is just a change equals a production

minus a destruction for all cases.

So let's forget the activation part.

For now, we're just going to assume that we have

some amount of isotope, N1.

We'll say we have N1 0 at t equals 0.

And it decays to N2 and decays N3.

So what are the differential equations

describing the rate of change of each of these isotopes?

So how about N1?

Is there any method of production

of isotope N1 in this scenario?

No, we just started off with some N1,

but we do have destruction of N1 via radioactive decay.

And so the amount of changes is going

to be equal to negative the activity.

So for every decay of N1, we lose an N1 atom.

So we just put minus lambda 1 N1.

For every N1 atom that decays, it produces an N2.

So N2 has an equal but different sign production term

and has a similar looking destruction term.

Meanwhile, since N2 becomes N3, then

we just have this simple term right there,

and these are the differential equations

which we want to solve.

We knew from last time that the solution to this equation

is pretty simple.

I'm not going to re-go through the derivation

there since I think that's kind of an easy one.

And N3, we know is pretty simple.

We used the conservation equation

to say that the total amount of all atoms in the system

has to be equal to N10 or N10.

So we know we have N10 equals and N1 plus N2

plus N3 for all time.

So we don't really have to solve for N3 because we can just

deal with it later.

The last thing that we need to derive

is what is the solution to N2.

And I want to correct a mistake that I

made because I'm going to chalk that up to exhaustion

assuming that integrating factor was zero.

It's not zero, so I want to show you why it's not now.

So how do we go about solving this?

What method did we use?

We chose the integration factor method

because it's a nice clean one.

So we rewrite this equation in terms of let's just

say N2 prime--

I'm sorry-- plus lambda 2 N2 minus lambda 1 N1.

And we don't necessarily want to have an N1 in there

because we want to have one variable only.

So instead of N1, we can substitute this whole thing

in there.

So N10 e to the minus lambda 1t equals zero.

And let's just draw a little thing around here

to help visually separate.

We know how to solve this type of differential equation

because we can define some integrating factor mu

equals e to the minus whatever is in front of the N2.

That's not too hard.

I'm sorry, just e to the integral, not

minus, of lambda 2 dt.

We're just equal to just e lambda 2t.

And we multiply every term in this equation

by mu because we're going to make sure that the stuff here--

after we multiply by mu and mu and nu

and mu for completeness--

that stuff in here should be something that looks

like the end of a product rule.

So if we multiply that through, we

get N2 prime e to the lambda 2t plus,

let's see, mu times lambda 2 e to the lambda

2t times n2 minus e to the lambda 2t lambda 1 n10

e to the minus lambda 1t equals zero.

And indeed, we've got right here what

looks like the end result of the product rule

where we have something, let's say, one function times

the derivative of another plus the derivative

of that function times the original other function.

So to compact that up, we can call that, let's say, N2

e to the lambda t--

sorry, lambda 2t prime minus--

and I'm going to combine these two exponents right here.

So we'll have minus lambda 1 and 10e to the--

let's see, is it lambda 2 minus lambda 1t equals zero.

Just going to take this term to the other side

of the equals sign, so I'll just do that, integrate both sides.

And we get N2 e to the lambda 2t equals--

let's see, that'll be lambda 1 N10 over lambda 2 minus lambda

1 times all that stuff.

I'm going to divide each side of the equation by--

I'll use a different color for that intermediate step--

e to the lambda 2t.

And that cancels these out.

That cancels these out.

And I forgot that integrating factor again, didn't I?

Yeah, so there's going to be a plus C somewhere here.

And we're just going to absorb this e to the lambda 2t

into this integrating constant because it's

an integrating constant.

We haven't defined it yet.

Did someone have a question I thought I saw?

OK, and so now this is where I went wrong last time

because I think I was exhausted and commuted in from Columbus.

I just assumed right away that C equals zero,

but it's not the case.

So if we plug-in the condition at t equals 0--

and two should equal 0--

let's see what we get.

That would become a zero.

That t would be a zero, which means that we just

end up with the equation lambda 1 N10

over lambda 2 minus lambda 1 plus c equals

zero so obviously the integration constant

is not like we thought it was.

So then C equals negative that stuff.

That make more sense.

So you guys see why the integrating constants not zero.

So in the end--

I'm going t5o skip ahead a little of the math because I

want to get into nuclear activation analysis--

we end up with N2 should equal, let's see,

lambda 1 N10 over lambda 1 minus lambda 1 times--

did I write that twice?

I think I did--

times e to the--

let's see, e to the minus lambda 1t minus e

to the minus lambda 2t.

And so since we know N1, we've found N10.

We know N3 from this conservation equation.

We've now fully determined what is

the concentration of every isotope

in this system for all time.

And because the solution to this is not that intuitive--

like I can't picture what the function looks like in my head.

I don't know about you guys.

Anyone?

No?

OK, I can't either.

I coded them up in this handy graphing calculator where

you can play around with the eye of the concentration N10, which

is just a multiplier for everything,

and the relative half lives lambda 1 and lambda 2.

And I'll share this with you guys,

so you can actually see generally how this works.

So let's start looking at a couple of cases--

move this a little over so we can see the axes.

Let's say don't worry about anything before T equals zero.

That's kind of an invalid part of the solution.

So I'll just shrink us over there.

And so I've coded up all three of these equations.

There is the solution to N1 highlighted right there.

That's as you'd expect simple exponential decay.

All N1 knows is that it's decaying

according to its own half life or exponential decay equation.

And two, here in the blue, which expands, of course,

looks a little more complicated.

So what we notice here is that N2 is tied directly

to the slope of N1.

That should follow pretty intuitively

from the differential equations because if you

look at the slope of N2, well, it depends directly

on the value of N1.

For very, very short times, this is

the sort of limiting behavior in the and the graphical guidance.

I want to give you two solve questions like,

what's on the exam or how to do a nuclear activation analysis.