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# K-Ar dating calculation

Video transcript

In the last video, we
give a bit of an overview of potassium-argon dating. In this video, I want to go
through a concrete example. And it'll get a
little bit mathy, usually involving a
little bit of algebra or a little bit of
exponential decay, but to really show you how
you can actually figure out the age of some volcanic
rock using this technique, using a little bit
of mathematics. So we know that anything that is
experiencing radioactive decay, it's experiencing
exponential decay. And we know that there's
a generalized way to describe that. And we go into
more depth and kind of prove it in other
Khan Academy videos. But we know that the amount
as a function of time-- so if we say N is the amount
of a radioactive sample we have at some time-- we know that's
equal to the initial amount we have. We'll call that N sub 0, times
e to the negative kt-- where this constant is particular
to that thing's half-life. In order to do this for the
example of potassium-40, we know that when time
is 1.25 billion years, that the amount we have left
is half of our initial amount. So let's write it that way. So let's say we start with
N0, whatever that might be. It might be 1 gram,
kilogram, 5 grams-- whatever it might be-- whatever
we start with, we take e to the negative
k times 1.25 billion years. That's the half-life
of potassium-40. So 1.25 billion years. We know, after that long,
that half of the sample will be left. So we will have 1/2 N0 left. Whatever we started
with, we're going to have half left after
1.25 billion years. Divide both sides by N0. And then to solve
for k, we can take the natural log of both sides. So you get the
natural log of 1/2-- we don't have that
N0 there anymore-- is equal to the natural
log of this thing. The natural log is
just saying-- to what power do I have to
raise e to get e to the negative k
times 1.25 billion? So the natural log of this--
the power they'd have to raise e to to get to e to the negative
k times 1.25 billion-- is just negative k
times 1.25 billion. Or I could write it as negative
1.25-- let me write times-- 10 to the ninth k. That's the same thing
as 1.25 billion. We have our negative
sign, and we have our k. And then, to solve for k,
we can divide both sides by negative 1.25 billion. And so we get k. And I'll just flip
the sides here. k is equal to the natural log
of 1/2 times negative 1.25 times 10 to the ninth power. And what we can do
is we can multiply the negative times the top. Or you could view
it as multiplying the numerator and the
denominator by a negative so that a negative
shows up at the top. And so we could make this
as over 1.25 times 10 to the ninth. It's just 1.25 billion. Let me write it over here
in a different color. The negative natural log-- well,
I could just write it this way. If I have a natural log of
b-- we know from our logarithm properties, this
is the same thing as the natural log
of b to the a power. So the negative
natural log of 1/2 is the same thing as
the natural log of 1/2 to the negative 1 power. And so this is the same thing. Anything to the
negative power is just its multiplicative inverse. So this is just the
natural log of 2. So negative natural
log of 1 half is just the natural
log of 2 over here. So we were able to
figure out our k. It's essentially
the natural log of 2 over the half-life
of the substance. So we could actually
generalize this if we were talking about some
other radioactive substance. And now let's think
about a situation-- now that we've figured out a k--
let's think about a situation where we find in some
sample-- so let's say the potassium that we
find is 1 milligram. I'm just going to
make up these numbers. And usually, these
aren't measured directly, and you really care about
the relative amounts. But let's say you were
able to figure out the potassium is 1 milligram. And let's say that the
argon-- actually, I'm going to say the
potassium-40 found, and let's say the
argon-40 found-- let's say it is 0.01 milligram. So how can we use this
information-- in what we just figured out here, which is
derived from the half-life-- to figure out how old this
sample right over here? How do we figure out
how old this sample is right over there? Well, what we need
to figure out-- we know that n, the
amount we were left with, is this thing right over here. So we know that we're
left with 1 milligram. And that's going to be equal
to some initial amount-- when we use both
of this information to figure that initial
amount out-- times e to the negative kt. And we know what k is. And we'll figure it out later. So k is this thing
right over here. So we need to figure out
what our initial amount is. We know what k is, and
then we can solve for t. How old is this sample? And to figure out
our initial amount, we just have to remember that
for every argon-40 we see, that must have decayed from--
when you have potassium-40, when it decays, 11% decays into
argon-40 and the rest-- 89%-- decays into calcium-40. We saw that in the last video. So however much argon-40, that
is 11% of the decay product. So if you want to think
about the total number of potassium-40s
that have decayed since this was kind
of stuck in the lava. And we learned that anything
that was there before, any argon-40 that
was there before would have been able to
get out of the liquid lava before it froze or
before it hardened. So to figure out how
much potassium-40 this is derived from, we
just divide it by 11%. So maybe I could say k initial--
the potassium-40 initial-- is going to be equal to the amount
of potassium 40 we have today-- 1 milligram-- plus the
amount of potassium-40 we needed to get this
amount of argon-40. We have this amount of
argon-40-- 0.01 milligrams. And that number of milligrams
there, it's really just 11% of the original potassium-40
that it had to come from. The rest of it turned
into calcium-40. So we divide it by 11% or 0.11. And this isn't the exact number,
but it'll get the general idea. And so our initial--
which is really this thing right over here. I could call this N0. This is going to
be equal to-- and I won't do any of the
math-- so we have 1 milligram we have left is
equal to 1 milligram-- which is what we found-- plus
0.01 milligram over 0.11. And then, all of that
times e to the negative kt. And what you see
here is, when we want to solve for t--
assuming we know k, and we do know k now-- that
really, the absolute amount doesn't matter. What actually
matters is the ratio. Because if we're
solving for t, you want to divide both
sides of this equation by this quantity
right over here. So you get this side--
the left-hand side-- divide both sides. You get 1 milligram over
this quantity-- I'll write it in blue-- over this
quantity is going to be 1 plus-- I'm just
going to assume, actually, that the units here
are milligrams. So you get 1 over
this quantity, which is 1 plus 0.01 over the 11%. That is equal to e
to the negative kt. And then, if you
want to solve for t, you want to take the
natural log of both sides. This is equal right over here. You want to take the
natural log of both sides. So you get the natural
log of 1 over 1 plus 0.01 over 0.11 or 11%
is equal to negative kt. And then, to solve for
t, you divide both sides by negative k. So I'll write it over here. And you can see, this a little
bit cumbersome mathematically, but we're getting to the answer. So we got the natural log of
1 over 1 plus 0.01 over 0.11 over negative k. Well, what is negative k? We're just dividing both sides
of this equation by negative k. Negative k is the
negative of this over the negative natural
log of 2 over 1.25 times 10 to the ninth. And now, we can get our
calculator out and just solve for what this time is. And it's going to be
in years because that's how we figured
out this constant. So let's get my handy TI-85. First, I'll do this part. So this is 1 divided by 1
plus 0.01 divided by 0.11. So that's this part
right over here. That gives us that number. And then, we want to take
the natural log of that. So let's take the natural
log of our previous answer. So it's the natural
log of 0.9166667. It gives us negative 0.087. So that's this
numerator over here. And we're going to divide that. So this number is our
numerator right over here. We're going to divide
that by the negative-- I'll use parentheses carefully--
the negative natural log of 2-- that's that there-- divided
by 1.25 times 10 to the ninth. So it's negative natural
log of 2 divided by 1.25. E9 means times 10 to the ninth. And I closed both parentheses. And now, we need our drum roll. So this should give
us our t in years. Let's see how many--
this is thousands, so it's 3,000-- so we
get 156 million or 156.9 million years if we round. So this is approximately a
157-million-year-old sample. So the whole point of this-- I
know the math was a little bit involved, but it's something
that you would actually see in a pre-calculus
class or an algebra 2 class when you're studying
exponential growth and decay. But the whole point
I wanted to do this is to show you that it's
not some crazy voodoo here. And, you know, Sal, gave this
very high-level explanation, and then, you say,
oh, well, there must be some super difficult
mathematics after that. The mathematics
really is something that you would see
in high school. If you saw a sample that
had this ratio of argon-40 to potassium-40,
you would actually be able to do that high
school mathematics. You would be able to do
that to figure out this is a 157-million-year-old
sample of volcanic rock.