Introduction to exponential decay Introduction to Exponential Decay
Introduction to exponential decay
- 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.
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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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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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