There's some unstable atoms that have a very short half-life (in terms of seconds),how could we study them or even know that they exist if they are changing in such a rate?

Usually we simply observe the things that they have decayed into. We know what the 'rules' of nuclear decay are, so we can work out what unstable atom was there. However, while this does tell us about the physical presence of such atoms, it doesn't tell us anything about their chemical properties, so for very unstable elements -- particularly the very heavy ones with atomic numbers over 100 -- we can only estimate the pure elements melting point, colour, density, etc, based on what we know about other, similar elements.

at 2:47 sal says something 'natralog' ,what does it means?

lt is the natural logarithm. You need to have mastered logarithms and exponential function before attempting to study chemistry, as they are used quite extensively. This is an Algebra II topic -- and you must know this material before being able to study General Chemistry: https://www.khanacademy.org/math/algebra2/logarithms-tutorial

What is the symbol for a natural log?

ln is the way we denote natural log. ln(e) = 1.

How come for some questions you make the constant k positive and for radioactive decay, the constant k becomes negative? Is it supposed to be like this for these two types of examples?

If you want decay, you need to raise e to a negative power. If you want growth, the power needs to be positive.

An ion is an atom or molecule that has become positively charged (known as a "cation") or negatively charged (known as an "anion"). The charge is the result of the loss or gain of a valence electron. Losing an electron will cause a positive charge, gaining an electron will cause a negative charge.

Hi, at 6:25 when you substituted the initial amount of C-14 into the equation, shouldn't it be converted into SI units (kg)? In addition to this, does time have an SI units of seconds in physics, meaning the values must be converted to the appropriate SI units before being substituted in? Thanks!

Strictly speaking, kg ought to be used. If you have an instructor who is insistent upon strict adherence to SI units, then you must use kg. However, kg is typically too big of a unit for practical use in laboratory chemistry. Thus, we often use g, mg, or µg even though they are not SI. For that matter, we use liter and mL when neither of those is SI because m³ is just to much volume for practical lab use. In fact, SI has officially listed the liter as a "non-SI unit acceptable within SI" because very few people are willing to try to use m³ instead of a liter. So, yes, strictly speaking the only mass SI accepts is the kg. But, unless your instructor insists otherwise, in chemistry grams are much more likely to be used.

Why is the formula `Ne^(-kt)`? Wouldn't it make more intuitive sense for the formula to be `N(1/2)^(t/h)` where "h" is the half life?

The half-life formula is derived using 1st order kinetics since radioactive decay is a first order reaction. A first order reaction has the general form of: A -> products. For radioactive decay problems you can imagine the reactant decaying into new nuclides where the rate of the reaction only depends on the original radioactive nuclide. The rate law is written as: Rate=k[A], where 'k' is the rate constant and [A] is the concentration of the reactant in molarity. This can be rewritten as: -Δ[A]/Δt = k[A], where the rate is being expressed as the disappearance of reactant A per unit of time. If you use some calculus to figure out the integrated rate law (it's a separable differential equation) you arrive at: ln[A] = -kt + ln[A]o, where [A]o is the initial concentration of A. This can be rearranged into: ln([A]/[A]o) = -kt Which can be rearranged into: [A] = [A]o*e^(-kt) which is the form Sal is using which ultimately originated from the rate law of a first order reaction. The other equation is derived from ln([A]/[A]o) = -kt. At the time of half life (h), half of the original sample has decayed which can be written as: ln((1/2*[A]o)/[A]o) = -kh. Which simplifies into ln(1/2) = -kh. And if we solve for half life we get: h = -ln(1/2)/k, which is where Sal got his equation. So all the equations result from chemical kinetics of a first order reaction. Hope that clears things up.

If radioactive atoms are constantly decaying, how do they exist in the first place?

They get created in stars, and then for some of them it takes a long time to decay. If they decay fast then they are really rare!

I understand that e is a mathematical number but whats its exact number or rounded number

It's like pi, you can't write all the digits of it. You can easily use the internet to find it to however many decimal places you like.

hi, i would like to know the unit of lambda. Thanks

Its unit is 1/time. If time is in seconds lambda unit is 1/s. This makes e^(t/lambda) work.

Main content

Course: AP®︎/College Physics 2 > Unit 8

Lesson 3: Nuclear physics

Exponential decay problem solving

Name: Exponential decay problem solving
Uploaded: 2011-02-20T16:35:31Z
Description: Exponential decay refers to a process in which a quantity decreases over time, with the rate of decrease becoming proportionally smaller as the quantity gets smaller. Use the exponential decay formula to calculate k, calculating the mass of carbon-14 remaining after a given time, and calculating the time it takes to have a specific mass remaining.

Google Classroom

Exponential decay refers to a process in which a quantity decreases over time, with the rate of decrease becoming proportionally smaller as the quantity gets smaller. Use the exponential decay formula to calculate k, calculating the mass of carbon-14 remaining after a given time, and calculating the time it takes to have a specific mass remaining. Created by Sal Khan.

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Anas Lahrichi
Posted 11 years ago. Direct link to Anas Lahrichi's post “There's some unstable ato...”
There's some unstable atoms that have a very short half-life (in terms of seconds),how could we study them or even know that they exist if they are changing in such a rate?
Button navigates to signup pageComment on Anas Lahrichi's post “There's some unstable ato...”
(18 votes)
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- Ben Willetts
  Posted 11 years ago. Direct link to Ben Willetts's post “Usually we simply observe...”
  Usually we simply observe the things that they have decayed into. We know what the 'rules' of nuclear decay are, so we can work out what unstable atom was there. However, while this does tell us about the physical presence of such atoms, it doesn't tell us anything about their chemical properties, so for very unstable elements -- particularly the very heavy ones with atomic numbers over 100 -- we can only estimate the pure elements melting point, colour, density, etc, based on what we know about other, similar elements.
  Button navigates to signup page
  (8 votes)
fickus42
Posted 11 years ago. Direct link to fickus42's post “what's an ion?”
what's an ion?
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(2 votes)
Answer
- TSlackluster
  Posted 11 years ago. Direct link to TSlackluster's post “An ion is an atom or mole...”
  An ion is an atom or molecule that has become positively charged (known as a "cation") or negatively charged (known as an "anion"). The charge is the result of the loss or gain of a valence electron. Losing an electron will cause a positive charge, gaining an electron will cause a negative charge.
  Button navigates to signup page
  (23 votes)
Antreev
Posted 10 years ago. Direct link to Antreev's post “at 2:47 sal says somethin...”
at
2:47
sal says something 'natralog' ,what does it means?
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(5 votes)
Answer
- Just Keith
  Posted 10 years ago. Direct link to Just Keith's post “lt is the natural logarit...”
  lt is the natural logarithm. You need to have mastered logarithms and exponential function before attempting to study chemistry, as they are used quite extensively.
  This is an Algebra II topic -- and you must know this material before being able to study General Chemistry:
  https://www.khanacademy.org/math/algebra2/logarithms-tutorial
  Button navigates to signup page
  (9 votes)
Steven
Posted 11 years ago. Direct link to Steven's post “What is the symbol for a ...”
What is the symbol for a natural log?
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(4 votes)
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- Andrew M
  Posted 11 years ago. Direct link to Andrew M's post “ln is the way we denote n...”
  ln is the way we denote natural log. ln(e) = 1.
  Button navigates to signup page
  (4 votes)
Henry K. Tran
Posted 10 years ago. Direct link to Henry K. Tran's post “How come for some questio...”
How come for some questions you make the constant k positive and for radioactive decay, the constant k becomes negative? Is it supposed to be like this for these two types of examples?
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(4 votes)
Answer
- Andrew M
  Posted 10 years ago. Direct link to Andrew M's post “If you want decay, you ne...”
  If you want decay, you need to raise e to a negative power. If you want growth, the power needs to be positive.
  Comment on Andrew M's post “If you want decay, you ne...”
  (2 votes)
Matthew Chen
Posted 5 years ago. Direct link to Matthew Chen's post “If radioactive atoms are ...”
If radioactive atoms are constantly decaying, how do they exist in the first place?
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(1 vote)
Answer
- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “They get created in stars...”
  They get created in stars, and then for some of them it takes a long time to decay. If they decay fast then they are really rare!
  Button navigates to signup page
  (5 votes)
kclay501
Posted 10 years ago. Direct link to kclay501's post “Hi, at 6:25 when you subs...”
Hi, at
6:25
when you substituted the initial amount of C-14 into the equation, shouldn't it be converted into SI units (kg)? In addition to this, does time have an SI units of seconds in physics, meaning the values must be converted to the appropriate SI units before being substituted in? Thanks!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Just Keith
  Posted 10 years ago. Direct link to Just Keith's post “Strictly speaking, kg oug...”
  Strictly speaking, kg ought to be used. If you have an instructor who is insistent upon strict adherence to SI units, then you must use kg.
  
  However, kg is typically too big of a unit for practical use in laboratory chemistry. Thus, we often use g, mg, or µg even though they are not SI. For that matter, we use liter and mL when neither of those is SI because m³ is just to much volume for practical lab use. In fact, SI has officially listed the liter as a "non-SI unit acceptable within SI" because very few people are willing to try to use m³ instead of a liter.
  
  So, yes, strictly speaking the only mass SI accepts is the kg. But, unless your instructor insists otherwise, in chemistry grams are much more likely to be used.
  Comment on Just Keith's post “Strictly speaking, kg oug...”
  (2 votes)
Cobra Coder
Posted 3 years ago. Direct link to Cobra Coder's post “Why is the formula `Ne^(-...”
Why is the formula Ne^(-kt)? Wouldn't it make more intuitive sense for the formula to be N(1/2)^(t/h) where "h" is the half life?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Richard
  Posted 3 years ago. Direct link to Richard's post “The half-life formula is ...”
  The half-life formula is derived using 1st order kinetics since radioactive decay is a first order reaction.
  
  A first order reaction has the general form of: A -> products. For radioactive decay problems you can imagine the reactant decaying into new nuclides where the rate of the reaction only depends on the original radioactive nuclide.
  
  The rate law is written as: Rate=k[A], where 'k' is the rate constant and [A] is the concentration of the reactant in molarity. This can be rewritten as: -Δ[A]/Δt = k[A], where the rate is being expressed as the disappearance of reactant A per unit of time. If you use some calculus to figure out the integrated rate law (it's a separable differential equation) you arrive at: ln[A] = -kt + ln[A]o, where [A]o is the initial concentration of A.
  This can be rearranged into: ln([A]/[A]o) = -kt
  Which can be rearranged into: [A] = [A]o*e^(-kt) which is the form Sal is using which ultimately originated from the rate law of a first order reaction.
  
  The other equation is derived from ln([A]/[A]o) = -kt. At the time of half life (h), half of the original sample has decayed which can be written as: ln((1/2*[A]o)/[A]o) = -kh.
  Which simplifies into ln(1/2) = -kh. And if we solve for half life we get: h = -ln(1/2)/k, which is where Sal got his equation. So all the equations result from chemical kinetics of a first order reaction. Hope that clears things up.
  Comment on Richard's post “The half-life formula is ...”
  (2 votes)
brandon.kaiser
Posted 8 years ago. Direct link to brandon.kaiser's post “I understand that e is a ...”
I understand that e is a mathematical number but whats its exact number or rounded number
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(1 vote)
Answer
- Andrew M
  Posted 8 years ago. Direct link to Andrew M's post “It's like pi, you can't w...”
  It's like pi, you can't write all the digits of it.
  You can easily use the internet to find it to however many decimal places you like.
  Button navigates to signup page
  (3 votes)
Badr Laabouri
Posted 6 years ago. Direct link to Badr Laabouri's post “hi, i would like to know ...”
hi, i would like to know the unit of lambda. Thanks
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(1 vote)
Answer
- Andrew M
  Posted 6 years ago. Direct link to Andrew M's post “Its unit is 1/time. If t...”
  Its unit is 1/time. If time is in seconds lambda unit is 1/s.
  This makes e^(t/lambda) work.
  Button navigates to signup page
  (4 votes)

Video transcript

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.