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SAL: The notion of a half-life
is useful, if we're dealing

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with increments of time that are
multiples of a half-life.

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For example, where time
equals zero, we

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have 100% of our substance.

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Then after time equals one
half-life, we'd have 50% of

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our substance.

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At time is equal to two
half-lives, we'd have 25% of

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our substance, and so
on and so forth.

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So if I say that three
half-lives have gone by-- in

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the case of carbon that would
be, what, roughly 15,000

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years-- I can tell you roughly,
or almost exactly,

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what percentage of my original
element I still have. In the

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case of carbon-14, I'll tell
you what percentage of my

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original carbon-14 has not
decayed into nitrogen, as yet,

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nitrogen-14.

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And that's useful, but what if I
care about how much carbon I

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have after 1/2 a year, or after
1/2 a half life, or

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after three billion years,
or after 10 minutes?

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What if I want a general
function.

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A general function, as a
function of time, that tells

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me the number, or the amount,
of my decaying substance I

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have. So that's what we're going
to do in this video.

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And it's going to be a little
mathy, but I think the math is

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pretty straightforward,
especially if you've taken a

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first-year course in calculus.

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And this is actually a pretty
neat application of it.

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So let's just think a little bit
about the rate of change,

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or the probability, or the
number particles that are

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changing at any given time.

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So if we say, the difference
or change in our number of

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particles, or the amount of
particles, in any very small

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period of time, what's this
going to be dependent on?

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This is the number particles
we have in a

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given period time.

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This is our rate of change.

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So one thing, we know that our
rate of change is going down.

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We know it's a negative
number.

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We know that, in the case of
radioactive decay, I could do

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the same exercise with
compounding growth, where I

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would say, oh no, it's not a
negative number, that our

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growth is dependent on how much
we have. In this case the

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amount we're decaying is
proportional, but it's going

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to be the negative of how much
of the actual compound we

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already have.

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Let me explain that.

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So what I'm saying is, look,
our amount of decay is

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proportional to the amount of
the substance that we already

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are dealing with.

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And just to maybe make that a
little bit more intuitive,

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imagine a situation here
where you have 1

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times 10 to the 9th.

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You have a billion carbon atoms.
And let's say over here

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you have 1 times 10 to the 6th
carbon atoms. And if you look

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at it at over some small period
of time, let's say, if

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you look at it over one second,
let's say our dt.

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dt as an infinitesimally small
time, but let's say it's a

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change in time.

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It's a delta t.

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And let's say over one second,
you observe that this sample

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had, I don't know, let's say you
saw 1000 carbon particles.

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You really wouldn't see that
with carbon-14, but this is

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just for the sake of
our intuition.

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Let's say over one second you
saw 1000 carbon particles per

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second here.

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Well here you have 1000th of the
number particles in this

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sample as this one.

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So, for every thousand particles
you saw decaying

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here, you'd really expect to
see one carbon particle per

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second here.

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Just because you have
a smaller amount.

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Now I don't know what the
actual constant is.

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But we know that no matter what
substance we're talking

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about, this constant is
dependent on the substance.

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Carbon's going to be different
from uranium, is going to be

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different from, you know,
we looked at radon.

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They're all going to
have different

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quantities right here.

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And we can see that.

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We'll actually do it in the next
video, you can actually

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calculate this from
the half-life.

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But the rate of change is always
going to be dependent

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on the number of particles
you have, right?

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I mean, we saw that here
with half-life.

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When you have 1/2 the
number of particles,

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you lose 1/2 as much.

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Here, if we start with 100
particles here, we went to 50

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particles, then we went to 25.

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When you start with 50, in a
period of time you lose 25.

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When you start with
100, you lose 50.

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So clearly the amount you lose
is dependent on the amount you

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started with, right?

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Over any fraction of
time, and here it's

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a very small fraction.

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So what I set up here is really
fairly simple, but it

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doesn't sound so simple to a lot
of people if you say it's

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a differential equation.

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We can actually solve this using
pretty straightforward

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techniques.

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This is actually a separation
of variables problem.

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And so, what can we do?

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Let's divide both sides by N.

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We want to get all the N's on
this side and all the t stuff

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on the other side.

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So if we have 1 over
N, dN over dt is

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equal to minus lambda.

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I just divided both sides
of this by N.

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And then I can multiply both
sides of this by dt, and I get

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1 over N dN is equal
to minus lambda dt.

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Now I can take the integral of
both sides of this equation.

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And what do I get?

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What's the antiderivative?

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I'm taking the indefinite
integral or the

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antiderivative.

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What's the antiderivative
of 1 over N?

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Well that's the natural log of
N plus some constant-- I'll

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just do that in blue--
plus some constant.

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And then that equals-- What's
the antiderivative of just

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some constant?

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Well it's just that
constant times the

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derivative, the variable.

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We're taking the antiderivative

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with respect to.

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So minus lambda, times t,
plus some constant.

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These are different constants,
but they're arbitrary.

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So if we want, we can just
subtract that constant from

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that constant, and put them all
on one side and then we

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just get another constant.

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So this boils down to our
solution to our differential

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equation is the natural log of
N is equal to minus lambda-t,

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plus some other constant, I call
it c3, it doesn't matter.

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And now if we want to just make
this a function of N in

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terms of t, let's take both of
these, or both take e to the

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power of both sides of this.

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You can view that as kind of
the inverse natural log.

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So e to the power of ln of N,
ln of N is just saying what

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power do you raise
e to to get to N?

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So if you raise e to that
power, you get N.

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So I'm just raising both
sides of this equation.

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I'm raising e to both sides
of this equation.

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e to the ln of N is just N.

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And that is equal to e to the
minus lambda-t, plus c3.

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And now this can be rewritten
as, N is equal to e to the

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minus lambda-t, times
e to the c3.

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And now once again this is an
arbitrary constant, so we can

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just really rename that as,
I don't know, let me

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rename it as c4.

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So, our solution to our
differential equation, N, as a

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function of t, is equal to
our c4 constant, c4e

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to the minus lambda-t.

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Now let's say, even better,
let's say is N equals 0.

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Let's say that N equals 0.

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We have N sub 0 of our sample.

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That's how much we're
starting off with.

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So let's see if we can
substitute that into our

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equation to solve for c4.

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So we said N sub-0 is equal to,
let's put 0 in here, so

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let's see, that's equal
to N sub naught.

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And that's equal to c4 times e
to the minus lambda, times 0.

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Well, minus anything
times 0 is 0.

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So it's e to the 0.

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So that's just 1.

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So c4 is equal to N naught,
our starting

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amount for the sample.

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So we've actually got
an expression.

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We have the number of particles,
or the amount as a

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function of t, is equal to the
amount that we start off with,

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at time is equal to 0,
times e to the minus

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lambda, times time.

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And we just have to be careful
that we're always using the

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time constant when we solve
for the different

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coefficients.

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So this seems all abstract.

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How does this relate
to half-life?

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Well let's try to figure out
this equation for carbon.

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This'll be true for
anything where we

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have radioactive decay.

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If we actually had a plus sign
here it'd be exponential

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growth as well.

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We know that carbon, c-14, has
a 5,700-year half-life.

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So the way you could think about
it, is if at time equals

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0 you start off with t-- So time
equals 0. t equals-- let

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me write that down.

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If at N of 0 is equal to--
and we could write

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100 there if we want.

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Actually why don't we do that?

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If N of 0 we start
off with 100.

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And then at N of 5,700 years--
so we're going to take t to be

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in years, you just have to be
consistent with your units--

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how much will we have left?

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We'll have 50 left.

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We could have written x and x
over two here, and it would

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have all have worked
out in the end.

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So let's see, let's apply that
to this equation and try to

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solve this for lambda.

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So we know N of 0
is equal to 100.

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So we immediately know that we
can write this equation as N

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of t is equal to 100e, to the
minus lambda-t, at least in

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this exact circumstance.

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And we also know that N of
5,700-- so that means, N of

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5,700-- that is equal to,
we just said, that's one

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half-life away.

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So we have 1/2 as much
of our compound left.

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00:10:19,020 --> 00:10:25,560
That's equal to 50, which is
equal to the 5,700th power

208
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times lambda.

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00:10:26,830 --> 00:10:30,500
So it's equal to 100 times
e, to the minus

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00:10:30,500 --> 00:10:34,770
lambda, times 5,700.

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00:10:34,770 --> 00:10:35,940
And now we just solve
for lambda.

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00:10:35,940 --> 00:10:38,240
Then we'll have a general
equation for how much carbon

213
00:10:38,240 --> 00:10:40,470
we have at any given
moment in time.

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00:10:40,470 --> 00:10:44,850
So if you divide both sides
of this by 100.

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00:10:44,850 --> 00:10:46,160
What do we get?

216
00:10:46,160 --> 00:10:52,230
We get 0.5, we have 1/2, is
equal to e to the-- let me

217
00:10:52,230 --> 00:10:56,200
just write minus 5,700 lambda,
and then we could take the

218
00:10:56,200 --> 00:10:58,010
natural log of both sides.

219
00:10:58,010 --> 00:11:02,350
So then we get-- scroll down a
bit-- the natural log of 1/2

220
00:11:02,350 --> 00:11:05,140
is equal to the-- the natural
log of this is just minus

221
00:11:05,140 --> 00:11:07,890
5,700 lambda.

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00:11:07,890 --> 00:11:10,510
To solve for lambda, you get
lambda is equal to the natural

223
00:11:10,510 --> 00:11:16,210
log of 1/2, over minus 5,700.

224
00:11:16,210 --> 00:11:19,990
So let me see what that is.

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00:11:19,990 --> 00:11:20,980
Let's see what that is.

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00:11:20,980 --> 00:11:30,980
So 0.5 natural log is that,
divided by minus 5,700.

227
00:11:30,980 --> 00:11:38,456
5,700 negative is equal to 1.2
times 10 to the negative 4.

228
00:11:38,456 --> 00:11:43,410
Is equal to 1.21 times
10 to the minus 4.

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00:11:43,410 --> 00:11:45,370
So there you have it, we
figured out our lambda.

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00:11:45,370 --> 00:11:49,440
So the general equation for
how much carbon-14 we can

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00:11:49,440 --> 00:11:55,340
expect at any moment in time,
t, where t is in years, is N

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00:11:55,340 --> 00:11:58,340
of t is equal to the amount of
carbon we start off with,

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times e to the minus lambda.

234
00:12:01,280 --> 00:12:04,780
The minus lambda is 1.21
times 10 to the minus

235
00:12:04,780 --> 00:12:08,350
4, times t in years.

236
00:12:08,350 --> 00:12:11,500
So now if you say after 1/2 a
year, you just plug it in and,

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00:12:11,500 --> 00:12:13,630
you have to tell me how much you
started off with, and then

238
00:12:13,630 --> 00:12:15,870
I can tell you how much you
have after 1/2 a year, or

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00:12:15,870 --> 00:12:18,060
after a billion years, or
after a gazillion years.

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00:12:18,060 --> 00:12:19,360
And we'll do a lot
more of these

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00:12:19,360 --> 00:12:21,350
problems in the next video.

242
00:12:21,350 --> 00:00:00,000