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## Class 12 Physics (India)

### Unit 13: Lesson 3

Half life and decay rate- Half-life and carbon dating
- Half-life (qualitative)
- Half life (intermediate)
- Exponential decay formula proof (can skip, involves calculus)
- Activity and Mean life
- Exponential decay problem solving
- More exponential decay examples
- Half-life plot
- Worked example: Fraction of undecayed nuclei
- Exponential decay and semi-log plots

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# Exponential decay problem solving

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.

## Want to join the conversation?

- 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?(18 votes)
- 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.(7 votes)

- 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.(23 votes)

- at2:47sal says something 'natralog' ,what does it means?(5 votes)
- 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(9 votes)

- Sal said the half-life of carbon-14 is 5,730 years, however in the half-life video he said that it was 5740 years. What is the correct half-life of carbon-14 then?

I've seen on Wikipedia that it is 5,730 ± 40.(5 votes)- 5730 +/- 40 includes 5740 so there really is no point in trying to pin it down to a single number of 5730 or 5740. In the other video, a correction appears to say 5730 not 5740, but since the range of uncertainty goes from 5690 to 5770 then both 5730 and 5740 can be considered the same.(3 votes)

- What is the symbol for a natural log?(4 votes)
- ln is the way we denote natural log. ln(e) = 1.(4 votes)

- 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?(4 votes)
- If you want decay, you need to raise e to a negative power. If you want growth, the power needs to be positive.(2 votes)

- If radioactive atoms are constantly decaying, how do they exist in the first place?(1 vote)
- 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!(5 votes)

- When I calculated the Carbon-14's half-life, my solution was this.

1/2=0.5

log(0.5)/5,730 = 5.2535775857588341224038201522599e-5

However, the video showed me that the solution was 1.2096809433855939082325167913755e-4.

Where am I wrong?(1 vote)- I think you are trying to calculate the decay constant.

In that case the formula is ln2/(half life) and not ln0.5/(half life) as done by you.(4 votes)

- 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?(2 votes)- 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.(2 votes)

- Hi, at6:25when 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!(2 votes)
- 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.(1 vote)

## 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.