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ემზადებით გამოცდისთვის? მოემზადეთ ამ 4 გაკვეთილის დახმარებით შემდეგ თემაზე: Quantum Physics

იხილეთ 4 გაკვეთილი

# 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, nitrogen-14. 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 techniques. 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 antiderivative. 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 coefficients. 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.