WEBVTT

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SAL: Two videos ago we learned
about half-lives.

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And we saw that they're good if
we are trying to figure out

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how much of a compound we have
left after one half-life, or

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two half-lives, or
three half-lives.

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We can just take 1/2 of the
compound at every period.

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But it's not as useful if we're
trying to figure out how

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much of a compound we have after
1/2 of a half-life, or

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after one day, or 10 seconds,
or 10 billion years.

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And to address that issue in the
last video, I proved that

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it involved a little bit
of sophisticated math.

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And if you haven't taken
calculus, you can really just

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skip that video.

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You don't have to watch it
for an intro math class.

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But if you're curious, that's
where we proved

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the following formula.

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That at any given point of
time, if you have some

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decaying atom, some element,
it can be described as the

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amount of element you have at
any period of time is equal to

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the amount you started off
with, times e to some

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constant-- in the last
video I use lambda.

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I could use k this time--
minus k times t.

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And then for a particular
element with a particular

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half-life you can just solve for
the k, and then apply it

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to your problem.

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So let's do that in this video,
just so that all of

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these variables can become a
little bit more concrete.

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So let's figure out the general
formula for carbon.

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Carbon-14, that's the one that
we addressed in the half-life.

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We saw that carbon-14 has a
half-life of 5,730 years.

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So let's see if we can somehow
take this information and

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apply it to this equation.

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So this tells us that after
one half-life-- so

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t is equal to 5,730.

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N of 5,730 is equal to the
amount we start off with.

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So we're starting off with,
well, we're starting off with

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N sub 0 times e to the minus--
wherever you see the t you put

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the minus 5,730-- so minus
k, times 5,730.

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That's how many years
have gone by.

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And half-life tells us that
after 5,730 years we'll have

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1/2 of our initial
sample left.

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So we'll have 1/2 of our
initial sample left.

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So if we try to solve
this equation for

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k, what do we get?

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Divide both sides by N naught.

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Get rid of that variable, and
then we're left with e to the

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minus 5,730k-- I'm just
switching these two around--

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is equal to 1/2.

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If we take the natural log of
both sides, what do we get?

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The natural log of e to
anything, the natural log of e

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to the a is just a.

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So the natural log of this is
minus 5,730k is equal to the

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natural log of 1/2.

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I just took the natural
log of both sides.

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The natural log and natural
log of both sides of that.

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And so to solve for k, we could
just say, k is equal to

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the natural log of 1/2 over
minus 5,730, which we did in

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the previous video.

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But let's see if we can do that
again here, to avoid--

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for those who might
have skipped it.

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So if you have 1/2, 0.5, take
the natural log, and then you

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divide it by 5,730, it's a
negative 5,730, you get 1.2

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times 10 to the negative 4.

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So it equals 1.2 times
10 to the minus 4.

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So now we have the general
formula for carbon-14, given

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its half-life.

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At any given point in time,
after our starting point-- so

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this is for, let's call this for
carbon-14, for c-14-- the

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amount of carbon-14 we're going
to have left is going to

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be the amount that we started
with times e to the minus k. k

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we just solved for.

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1.2 times 10 to the minus 4,
times the amount of time that

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has passed by.

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This is our formula for
carbon, for carbon-14.

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If we were doing this for some
other element, we would use

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that element's half-life to
figure out how much we're

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going to have at any given
period of time to figure out

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the k value.

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So let's use this to
solve a problem.

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Let's say that I start off
with, I don't know, say I

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start off with 300 grams
of carbon, carbon-14.

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And I want to know, how much do
I have after, I don't know,

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after 2000 years?

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How much do I have?

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Well I just plug into
the formula.

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N of 2000 is equal to the amount
that I started off

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with, 300 grams, times e to the
minus 1.2 times 10 to the

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minus 4, times t, is times
2000, times 2000.

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So what is that?

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So I already have that 1.2 times
10 the minus 4 there.

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So let me say, times 2000
equals-- and of course, this

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throws a negative out there,
so let me put the negative

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number out there.

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So there's a negative.

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And I have to raise
e to this power.

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So it's 0.241.

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So this is equal to N of 2000.

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The amount of the substance I
can expect after 2000 years is

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equal to 300 times e to
the minus 0.2419.

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And let's see, my calculator
doesn't have an e to the

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power, so Let me just take e.

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I need to get a better
calculator.

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I should get my scientific
calculator back.

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But e is, let's say 2.71-- I
can keep adding digits but

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I'll just do 2.71-- to the 0.24
negative, which is equal

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to 0.78 times the amount that I
started off with, times 300,

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which is equal to 236 grams. So
this is equal to 236 grams.

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So just like that, using this
exponential decay formula, I

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was able to figure out how
much of the carbon I have

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after kind of an unusual
period of time, a

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non-half-life period of time.

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Let's do another
one like this.

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Let's go the other way around.

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Let's say, I'm trying
to figure out.

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Let's say I start off with
400 grams of c-14.

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And I want to know how long--
so I want to know a certain

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amount of time-- does it
take for me to get to

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350 grams of c-14?

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So, you just say that 350
grams is how much

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I'm ending up with.

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It's equal to the amount that I
started off with, 400 grams,

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times e to the minus k.

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That's minus 1.2 times 10 to
the minus 4, times time.

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And now we solve for time.

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How do we do that?

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Well we could divide
both sides by 400.

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What's 350 divided by 400?

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350 by 400.

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It's 7/8.

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So 0.875.

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So you get 0.875 is equal to e
to the minus 1.2 times 10 to

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the minus 4t.

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You take the natural
log of both sides.

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You get the natural log of
0.875 is equal to-- the

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natural log of e to anything is
just the anything-- so it's

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equal to minus 1.2 times
10 to the minus 4t.

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And so t is equal to this
divided by 1.2 times 10

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to the minus 4.

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So the natural log, 0.875
divided by minus 1.2 times 10

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to the minus 4, is equal to the
amount of time it would

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take us to get from
400 grams to 350.

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[PHONE RINGS]

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My cell phone is ringing,
let me turn that off.

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To 350.

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So let me do the math.

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So if you have 0.875, and we
want to take the natural log

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of it, and divide it by minus
1.-- So divided by 1.2e 4

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negative, 10 to the
negative 4.

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This is all a negative number.

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Oh, I'll just divide it by this,
and then just take the

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negative of that.

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Equals that and then I have
to take a negative.

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So this is equal to 1,112 years
to get from 400 to 350

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grams of my substance.

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This might seem a little
complicated, but if there's

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one thing you just have to do,
is you just have to remember

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this formula.

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And if you want to know where
it came from, watch the

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previous video.

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For any particular element you
solve for this k value.

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And then you just substitute
what you know, and then solve

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for what you don't know.

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I'll do a couple more of these
in the next video.