Exponential Decay Formula Proof (can skip, involves Calculus) Showing that N(t)=Ne^(-kt) describes the amount of a radioactive substance we have at time T. For students with background in Calculus. Not necessary for intro chemistry class.
Exponential Decay Formula Proof (can skip, involves Calculus)
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- SAL: The notion of a half-life is useful, if we're dealing
- with increments of time that are multiples of a half-life.
- For example, where time equals zero, we
- have 100% of our substance.
- Then after time equals one half-life, we'd have 50% of
- our substance.
- At time is equal to two half-lives, we'd have 25% of
- our substance, and so on and so forth.
- So if I say that three half-lives have gone by-- in
- the case of carbon that would be, what, roughly 15,000
- years-- I can tell you roughly, or almost exactly,
- what percentage of my original element I still have. In the
- case of carbon-14, I'll tell you what percentage of my
- original carbon-14 has not decayed into nitrogen, as yet,
- And that's useful, but what if I care about how much carbon I
- have after 1/2 a year, or after 1/2 a half life, or
- after three billion years, or after 10 minutes?
- What if I want a general function.
- A general function, as a function of time, that tells
- me the number, or the amount, of my decaying substance I
- have. So that's what we're going to do in this video.
- And it's going to be a little mathy, but I think the math is
- pretty straightforward, especially if you've taken a
- first-year course in calculus.
- And this is actually a pretty neat application of it.
- So let's just think a little bit about the rate of change,
- or the probability, or the number particles that are
- changing at any given time.
- So if we say, the difference or change in our number of
- particles, or the amount of particles, in any very small
- period of time, what's this going to be dependent on?
- This is the number particles we have in a
- given period time.
- This is our rate of change.
- So one thing, we know that our rate of change is going down.
- We know it's a negative number.
- We know that, in the case of radioactive decay, I could do
- the same exercise with compounding growth, where I
- would say, oh no, it's not a negative number, that our
- growth is dependent on how much we have. In this case the
- amount we're decaying is proportional, but it's going
- to be the negative of how much of the actual compound we
- already have.
- Let me explain that.
- So what I'm saying is, look, our amount of decay is
- proportional to the amount of the substance that we already
- are dealing with.
- And just to maybe make that a little bit more intuitive,
- imagine a situation here where you have 1
- times 10 to the 9th.
- You have a billion carbon atoms. And let's say over here
- you have 1 times 10 to the 6th carbon atoms. And if you look
- at it at over some small period of time, let's say, if
- you look at it over one second, let's say our dt.
- dt as an infinitesimally small time, but let's say it's a
- change in time.
- It's a delta t.
- And let's say over one second, you observe that this sample
- had, I don't know, let's say you saw 1000 carbon particles.
- You really wouldn't see that with carbon-14, but this is
- just for the sake of our intuition.
- Let's say over one second you saw 1000 carbon particles per
- second here.
- Well here you have 1000th of the number particles in this
- sample as this one.
- So, for every thousand particles you saw decaying
- here, you'd really expect to see one carbon particle per
- second here.
- Just because you have a smaller amount.
- Now I don't know what the actual constant is.
- But we know that no matter what substance we're talking
- about, this constant is dependent on the substance.
- Carbon's going to be different from uranium, is going to be
- different from, you know, we looked at radon.
- They're all going to have different
- quantities right here.
- And we can see that.
- We'll actually do it in the next video, you can actually
- calculate this from the half-life.
- But the rate of change is always going to be dependent
- on the number of particles you have, right?
- I mean, we saw that here with half-life.
- When you have 1/2 the number of particles,
- you lose 1/2 as much.
- Here, if we start with 100 particles here, we went to 50
- particles, then we went to 25.
- When you start with 50, in a period of time you lose 25.
- When you start with 100, you lose 50.
- So clearly the amount you lose is dependent on the amount you
- started with, right?
- Over any fraction of time, and here it's
- a very small fraction.
- So what I set up here is really fairly simple, but it
- doesn't sound so simple to a lot of people if you say it's
- a differential equation.
- We can actually solve this using pretty straightforward
- This is actually a separation of variables problem.
- And so, what can we do?
- Let's divide both sides by N.
- We want to get all the N's on this side and all the t stuff
- on the other side.
- So if we have 1 over N, dN over dt is
- equal to minus lambda.
- I just divided both sides of this by N.
- And then I can multiply both sides of this by dt, and I get
- 1 over N dN is equal to minus lambda dt.
- Now I can take the integral of both sides of this equation.
- And what do I get?
- What's the antiderivative?
- I'm taking the indefinite integral or the
- What's the antiderivative of 1 over N?
- Well that's the natural log of N plus some constant-- I'll
- just do that in blue-- plus some constant.
- And then that equals-- What's the antiderivative of just
- some constant?
- Well it's just that constant times the
- derivative, the variable.
- We're taking the antiderivative
- with respect to.
- So minus lambda, times t, plus some constant.
- These are different constants, but they're arbitrary.
- So if we want, we can just subtract that constant from
- that constant, and put them all on one side and then we
- just get another constant.
- So this boils down to our solution to our differential
- equation is the natural log of N is equal to minus lambda-t,
- plus some other constant, I call it c3, it doesn't matter.
- And now if we want to just make this a function of N in
- terms of t, let's take both of these, or both take e to the
- power of both sides of this.
- You can view that as kind of the inverse natural log.
- So e to the power of ln of N, ln of N is just saying what
- power do you raise e to to get to N?
- So if you raise e to that power, you get N.
- So I'm just raising both sides of this equation.
- I'm raising e to both sides of this equation.
- e to the ln of N is just N.
- And that is equal to e to the minus lambda-t, plus c3.
- And now this can be rewritten as, N is equal to e to the
- minus lambda-t, times e to the c3.
- And now once again this is an arbitrary constant, so we can
- just really rename that as, I don't know, let me
- rename it as c4.
- So, our solution to our differential equation, N, as a
- function of t, is equal to our c4 constant, c4e
- to the minus lambda-t.
- Now let's say, even better, let's say is N equals 0.
- Let's say that N equals 0.
- We have N sub 0 of our sample.
- That's how much we're starting off with.
- So let's see if we can substitute that into our
- equation to solve for c4.
- So we said N sub-0 is equal to, let's put 0 in here, so
- let's see, that's equal to N sub naught.
- And that's equal to c4 times e to the minus lambda, times 0.
- Well, minus anything times 0 is 0.
- So it's e to the 0.
- So that's just 1.
- So c4 is equal to N naught, our starting
- amount for the sample.
- So we've actually got an expression.
- We have the number of particles, or the amount as a
- function of t, is equal to the amount that we start off with,
- at time is equal to 0, times e to the minus
- lambda, times time.
- And we just have to be careful that we're always using the
- time constant when we solve for the different
- So this seems all abstract.
- How does this relate to half-life?
- Well let's try to figure out this equation for carbon.
- This'll be true for anything where we
- have radioactive decay.
- If we actually had a plus sign here it'd be exponential
- growth as well.
- We know that carbon, c-14, has a 5,700-year half-life.
- So the way you could think about it, is if at time equals
- 0 you start off with t-- So time equals 0. t equals-- let
- me write that down.
- If at N of 0 is equal to-- and we could write
- 100 there if we want.
- Actually why don't we do that?
- If N of 0 we start off with 100.
- And then at N of 5,700 years-- so we're going to take t to be
- in years, you just have to be consistent with your units--
- how much will we have left?
- We'll have 50 left.
- We could have written x and x over two here, and it would
- have all have worked out in the end.
- So let's see, let's apply that to this equation and try to
- solve this for lambda.
- So we know N of 0 is equal to 100.
- So we immediately know that we can write this equation as N
- of t is equal to 100e, to the minus lambda-t, at least in
- this exact circumstance.
- And we also know that N of 5,700-- so that means, N of
- 5,700-- that is equal to, we just said, that's one
- half-life away.
- So we have 1/2 as much of our compound left.
- That's equal to 50, which is equal to the 5,700th power
- times lambda.
- So it's equal to 100 times e, to the minus
- lambda, times 5,700.
- And now we just solve for lambda.
- Then we'll have a general equation for how much carbon
- we have at any given moment in time.
- So if you divide both sides of this by 100.
- What do we get?
- We get 0.5, we have 1/2, is equal to e to the-- let me
- just write minus 5,700 lambda, and then we could take the
- natural log of both sides.
- So then we get-- scroll down a bit-- the natural log of 1/2
- is equal to the-- the natural log of this is just minus
- 5,700 lambda.
- To solve for lambda, you get lambda is equal to the natural
- log of 1/2, over minus 5,700.
- So let me see what that is.
- Let's see what that is.
- So 0.5 natural log is that, divided by minus 5,700.
- 5,700 negative is equal to 1.2 times 10 to the negative 4.
- Is equal to 1.21 times 10 to the minus 4.
- So there you have it, we figured out our lambda.
- So the general equation for how much carbon-14 we can
- expect at any moment in time, t, where t is in years, is N
- of t is equal to the amount of carbon we start off with,
- times e to the minus lambda.
- The minus lambda is 1.21 times 10 to the minus
- 4, times t in years.
- So now if you say after 1/2 a year, you just plug it in and,
- you have to tell me how much you started off with, and then
- I can tell you how much you have after 1/2 a year, or
- after a billion years, or after a gazillion years.
- And we'll do a lot more of these
- problems in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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