SAL: Let's do a couple more of these exponential decay
problems, because a lot of this really is just practice
and being very comfortable with the general formula, and
I'll write it again.
Where the amount of the element that's decaying, that
we have at any period in time, is equal to the amount that we
started with, times e to the minus kt.
Where the k value is specific to any certain element with a
certain half-life, and sometimes they don't even give
you the half-life.
So let's try this interesting situation.
Let's say that I have an element.
Let me just give you a formula.
Let's say that I have some magic element here, where its
formula is, its k value I give to you, k is equal to minus,
let me think of a-- [coughs]
Excuse me, I just had a lot of walnuts and my throat is dry.
Let's say that k is equal to, well k, we're putting a minus
in front of it, so I'll say the k value
is a positive 0.05.
So its exponential decay formula would be the amount
that you start off with, times e to the minus 0.05t.
My question to you is, given this, what is the half-life of
the compound that we're talking about?
What is the half-life?
So to figure that out, we need to figure out what t value can
we put here, so that if we start off with whatever value
here, we end up with 1/2 of that value there.
So let's do that.
So we're starting off with N sub 0 This is just some value,
our initial starting point.
We could put 100 there.
Actually, let's do that, just to keep things less abstract.
So let's say we start with 100.
I'm just picking 100 out of air.
I could have left it abstract with N.
Let's say I'm starting with 100.
And I take the 100 times e to the minus 0.05, times t.
t is whatever our half-life.
So after our half-life we're going to have 1/2 of this
stuff left.
So this should be equal to 50.
We just solved for t.
Divide both sides by 100.
You get e to the minus 0.05t, is equal to 1/2.
You take the natural log of both sides of this.
The natural log of this, the natural log of that.
And then you get-- the natural log of e to anything, I've
said it before, is just the anything.
So it is minus 0.05t is equal to the natural log of 1/2.
And then you get t is equal to the natural log of 1/2,
divided by minus 0.05.
So let's figure out what that is.
Actually, someone just made a comment, and I
might as well do that.
I could just put this minus up here.
I could make this a plus, and this a minus, if I just
multiply the numerator and the denominator by negative 1.
And if I want to, just to make the calculator math a little
easier, if you put a minus in front of a natural log, or any
logarithm, that's the same thing as the log of the
inverse of 2 over 0.05.
It makes the calculator math a little bit easier.
The same thing.
So if I do 2 natural log, divided by 0.05,
it is equal to 13.86.
So when t is equal to 13.86.
And I'm assuming that we're dealing with time in years.
That tends to be the convention, although sometimes
it could be something else and you'd always have
to convert to years.
But assuming that this original formula, where they
gave this k value 0.05, that was with the assumption that t
is in years, and I've just solved its half-life.
I just solved that after 13.86 years, you can expect to have
1/2 of the substance left.
We started with 100, we ended up with 50.
I could have started with x and ended up with x over 2.
Let's do one more of these problems, just so that we're
really comfortable with the formula.
Let's say that I have something with a half-life of,
I don't know, let's say I have it as one month.
Half-life of one month.
And after, well let's say that I-- well let me just for the
sake of time, let me make it a little bit simpler.
Let's say I just have my k value is equal to-- I mean you
can go from half-life to a k value, we did that in the
previous video.
Let's say my k value is equal to 0.001.
So my general formula is the amount of product I have, is
equal to the amount that I started with times e to the
minus 0.001 times t.
And I gave you this, if you have to figure it out from
half-life, I did that in the previous video with carbon-14.
But let's say this is the formula.
And let's say that after, I don't know, let's say after
1000 years I have 500 grams of whatever element is described.
The decay formula for whatever element is
described by this formula.
How much did I start off with?
So essentially I need to figure out N sub 0, right?
I'm saying that after 1000 years, so N of 1000, which is
equal to N sub naught times e to the minus
0.001, times 1000.
Right?
That's the N of 1000.
And I'm saying that that's equal to 500 grams. That
equals 500 grams. So I just have to
solve for N sub naught.
So what's the e value?
So if I have 0.0001 times 1000, so this is N sub naught.
This is 1/1000 of a 1000-- so times e to the minus 1 is
equal to 500 grams. Or I could multiply both sides by e, and
I have N sub naught is equal to 500e, which is about 2.71.
So it's 500 times 2.71.
I don't actually have e on this calculator or at least I
don't see it.
So we'll have 1,355 grams. So it's equal to 1,355 grams, or
1.355 kilograms. That's what I started with.
So hopefully you see now.
I mean, I think we've approached this pretty much at
almost any direction that a chemistry test or teacher
could throw the problem at you.
But you really just need to remember this formula.
And this applies to a lot of things.
Later you'll learn, you know, when you do compound interest
in finance, the k will just be a positive value, but it's
essentially the same formula.
And there's a lot of things that this formula actually
describes well beyond just radioactive decay.
But the simple idea is, use information they give you to
solve for as many of these constants as you can.
And then whatever they're asking for, solve for
whatever's left over.
And hopefully I've given you enough examples of that.
But let me know, I'm happy to do more.