Back

Introduction to Exponential Decay

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Introduction to Exponential Decay

相關課程

0 / 750

SAL: Two videos ago we learned about half-lives.
And we saw that they're good if we are trying to figure out
how much of a compound we have left after one half-life, or
two half-lives, or three half-lives.
We can just take 1/2 of the compound at every period.
But it's not as useful if we're trying to figure out how
much of a compound we have after 1/2 of a half-life, or
after one day, or 10 seconds, or 10 billion years.
And to address that issue in the last video, I proved that
it involved a little bit of sophisticated math.
And if you haven't taken calculus, you can really just
skip that video.
You don't have to watch it for an intro math class.
But if you're curious, that's where we proved
the following formula.
That at any given point of time, if you have some
decaying atom, some element, it can be described as the
amount of element you have at any period of time is equal to
the amount you started off with, times e to some
constant-- in the last video I use lambda.
I could use k this time-- minus k times t.
And then for a particular element with a particular
half-life you can just solve for the k, and then apply it
to your problem.
So let's do that in this video, just so that all of
these variables can become a little bit more concrete.
So let's figure out the general formula for carbon.
Carbon-14, that's the one that we addressed in the half-life.
We saw that carbon-14 has a half-life of 5,730 years.
So let's see if we can somehow take this information and
apply it to this equation.
So this tells us that after one half-life-- so
t is equal to 5,730.
N of 5,730 is equal to the amount we start off with.
So we're starting off with, well, we're starting off with
N sub 0 times e to the minus-- wherever you see the t you put
the minus 5,730-- so minus k, times 5,730.
That's how many years have gone by.
And half-life tells us that after 5,730 years we'll have
1/2 of our initial sample left.
So we'll have 1/2 of our initial sample left.
So if we try to solve this equation for
k, what do we get?
Divide both sides by N naught.
Get rid of that variable, and then we're left with e to the
minus 5,730k-- I'm just switching these two around--
is equal to 1/2.
If we take the natural log of both sides, what do we get?
The natural log of e to anything, the natural log of e
to the a is just a.
So the natural log of this is minus 5,730k is equal to the
natural log of 1/2.
I just took the natural log of both sides.
The natural log and natural log of both sides of that.
And so to solve for k, we could just say, k is equal to
the natural log of 1/2 over minus 5,730, which we did in
the previous video.
But let's see if we can do that again here, to avoid--
for those who might have skipped it.
So if you have 1/2, 0.5, take the natural log, and then you
divide it by 5,730, it's a negative 5,730, you get 1.2
times 10 to the negative 4.
So it equals 1.2 times 10 to the minus 4.
So now we have the general formula for carbon-14, given
its half-life.
At any given point in time, after our starting point-- so
this is for, let's call this for carbon-14, for c-14-- the
amount of carbon-14 we're going to have left is going to
be the amount that we started with times e to the minus k. k
we just solved for.
1.2 times 10 to the minus 4, times the amount of time that
has passed by.
This is our formula for carbon, for carbon-14.
If we were doing this for some other element, we would use
that element's half-life to figure out how much we're
going to have at any given period of time to figure out
the k value.
So let's use this to solve a problem.
Let's say that I start off with, I don't know, say I
start off with 300 grams of carbon, carbon-14.
And I want to know, how much do I have after, I don't know,
after 2000 years?
How much do I have?
Well I just plug into the formula.
N of 2000 is equal to the amount that I started off
with, 300 grams, times e to the minus 1.2 times 10 to the
minus 4, times t, is times 2000, times 2000.
So what is that?
So I already have that 1.2 times 10 the minus 4 there.
So let me say, times 2000 equals-- and of course, this
throws a negative out there, so let me put the negative
number out there.
So there's a negative.
And I have to raise e to this power.
So it's 0.241.
So this is equal to N of 2000.
The amount of the substance I can expect after 2000 years is
equal to 300 times e to the minus 0.2419.
And let's see, my calculator doesn't have an e to the
power, so Let me just take e.
I need to get a better calculator.
I should get my scientific calculator back.
But e is, let's say 2.71-- I can keep adding digits but
I'll just do 2.71-- to the 0.24 negative, which is equal
to 0.78 times the amount that I started off with, times 300,
which is equal to 236 grams. So this is equal to 236 grams.
So just like that, using this exponential decay formula, I
was able to figure out how much of the carbon I have
after kind of an unusual period of time, a
non-half-life period of time.
Let's do another one like this.
Let's go the other way around.
Let's say, I'm trying to figure out.
Let's say I start off with 400 grams of c-14.
And I want to know how long-- so I want to know a certain
amount of time-- does it take for me to get to
350 grams of c-14?
So, you just say that 350 grams is how much
I'm ending up with.
It's equal to the amount that I started off with, 400 grams,
times e to the minus k.
That's minus 1.2 times 10 to the minus 4, times time.
And now we solve for time.
How do we do that?
Well we could divide both sides by 400.
What's 350 divided by 400?
350 by 400.
It's 7/8.
So 0.875.
So you get 0.875 is equal to e to the minus 1.2 times 10 to
the minus 4t.
You take the natural log of both sides.
You get the natural log of 0.875 is equal to-- the
natural log of e to anything is just the anything-- so it's
equal to minus 1.2 times 10 to the minus 4t.
And so t is equal to this divided by 1.2 times 10
to the minus 4.
So the natural log, 0.875 divided by minus 1.2 times 10
to the minus 4, is equal to the amount of time it would
take us to get from 400 grams to 350.
[PHONE RINGS]
My cell phone is ringing, let me turn that off.
To 350.
So let me do the math.
So if you have 0.875, and we want to take the natural log
of it, and divide it by minus 1.-- So divided by 1.2e 4
negative, 10 to the negative 4.
This is all a negative number.
Oh, I'll just divide it by this, and then just take the
negative of that.
Equals that and then I have to take a negative.
So this is equal to 1,112 years to get from 400 to 350
grams of my substance.
This might seem a little complicated, but if there's
one thing you just have to do, is you just have to remember
this formula.
And if you want to know where it came from, watch the
previous video.
For any particular element you solve for this k value.
And then you just substitute what you know, and then solve
for what you don't know.
I'll do a couple more of these in the next video.

載入中...

Introduction to Exponential Decay

Introduction to Exponential Decay