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## AP®︎/College Physics 2

### Course: AP®︎/College Physics 2 > Unit 8

Lesson 3: Nuclear physics- Mass defect and binding energy
- Nuclear stability and nuclear equations
- Types of decay
- Writing nuclear equations for alpha, beta, and gamma decay
- Half-life and carbon dating
- Half-life plot
- Exponential decay formula proof (can skip, involves calculus)
- Exponential decay problem solving
- More exponential decay examples
- Exponential decay and semi-log plots

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# More exponential decay examples

A few more examples of exponential decay. Practice calculating k from half-life, and calculating initial mass. Created by Sal Khan.

## Want to join the conversation?

- Sal says that after one half life, you could expect 50% of the substance to decay. Since the particles have only a 50% chance of decaying, Is it possible for the substance to not decay at all after one half life?

Thanks in advance :)(1 vote)- possible: yes

likely: no

If you have lets say 131 grams of Iodine(131) which has a half-life of 8 days and you let it sit for 8 days, the probability for "no atom of Iodine(131) has decayed" would be the same as if you throw 6*10^23 coins and**all of them**would show "heads".(22 votes)

- at3:03what's the log inverse.(5 votes)
- 10 to the power of X

i.e log (100) = 2

Therefore 10 ^ 2 = 100 which is true :)

As a more general rule Log a ( b ) = Log ( a ^ b )

In chemistry log base 10 is used almost all the time so: a will always = 10

edit: Just saw that in this video Log base e is used and not log base 10.

i.e the inverse of Log base 10 ( b ) = 10 ^ b

and the inverse of Log base e ( b ) = e ^ b

Hope that helps :)(5 votes)

- Can you show me how to calculate a 1/2 life using disentegrations per minute. For example, I start with a compound that is 9.7 x 10^5 dpm, then several days later I end up with 2.2dpm, how do I figure out the half life?(3 votes)
- Disintegrations per minute are handled the same way as number of atoms. DPM are considered an activity A(t), so instead of the equation N(t)=No*exp(-kt) you have A(t)=Ao*exp(-kt). From the talk, he mentions that half-life = ln(2)/k, so use the above equation for A(t) above, solve for k, and then convert to half-life.(4 votes)

- Does anyone know of somewhere I can practice problems like these?(2 votes)
- You can Google practice problems and PDFs usually pop up with the answers at the bottom for practice, that's how I practiced chemistry problems. Good luck!(5 votes)

- What doe kt stand for?0:18(3 votes)
- I'm having a hard time understanding what K is. Help!(3 votes)
- The K is the decay constant. If you're confused about where Sal got that, I'd refer you to the "Exponential decay formula proof" video earlier in this playlist. He uses the Greek letter Lambda instead of K in that video, but both letters just mean the decay constant.

Here's the link:

https://www.khanacademy.org/science/chemistry/radioactive-decay/v/exponential-decay-formula-proof-can-skip-involves-calculus(2 votes)

- this question might seem a little off topic but if atoms can decay is there something opposite to that?(3 votes)
- No. Nuclei don't spontaneously add protons or neutrons.

You might call electron capture the opposite of beta decay.

In beta decay, a nucleus spontaneously emits an electron. In electron capture, a nucleus captures an inner electron.

However, the same nucleus doesn't undergo both processes.(2 votes)

- Can I use the formula N = N_0 (0.5) ^ (t / n), where N is the final amount, N_0 is the starting amount, t is the time and n is the half life to solve these problems as well?(3 votes)
- I have a question about the time units. Do I always have to convert any unit time given to years? Or can I use another unit as well? Thank you.(1 vote)
- The time units will depend on the process being modeled, but must be used consistently.

For example, if you determined**k**using**t**in seconds, then the half-life would also be in seconds.

Similarly, if you are given a**k**then you need to know what time units were used to derive that value.

Does that help?(4 votes)

- At3:30, Sal say that he is assuming that we are dealing with time in years. He then adds that if it were something else, we would have to convert to years.

Doesn't the formula work with any time unit as long as we are consistent, and don't switch from one to another?(2 votes)

## Video transcript

SAL: Let's do a couple more
of these exponential decay problems, because a lot of this
really is just practice and being very comfortable with
the general formula, and I'll write it again. Where the amount of the element
that's decaying, that we have at any period in time,
is equal to the amount that we started with, times
e to the minus kt. Where the k value is specific to
any certain element with a certain half-life, and sometimes
they don't even give you the half-life. So let's try this interesting
situation. Let's say that I have
an element. Let me just give
you a formula. Let's say that I have some magic
element here, where its formula is, its k value I give
to you, k is equal to minus, let me think of a-- [coughs] Excuse me, I just had a lot of
walnuts and my throat is dry. Let's say that k is equal to,
well k, we're putting a minus in front of it, so I'll
say the k value is a positive 0.05. So its exponential decay formula
would be the amount that you start off with, times
e to the minus 0.05t. My question to you is, given
this, what is the half-life of the compound that we're
talking about? What is the half-life? So to figure that out, we need
to figure out what t value can we put here, so that if we start
off with whatever value here, we end up with 1/2
of that value there. So let's do that. So we're starting off with N sub
0 This is just some value, our initial starting point. We could put 100 there. Actually, let's do that, just to
keep things less abstract. So let's say we start
with 100. I'm just picking
100 out of air. I could have left it
abstract with N. Let's say I'm starting
with 100. And I take the 100 times e to
the minus 0.05, times t. t is whatever our half-life. So after our half-life we're
going to have 1/2 of this stuff left. So this should be equal to 50. We just solved for t. Divide both sides by 100. You get e to the minus 0.05t,
is equal to 1/2. You take the natural log
of both sides of this. The natural log of this, the
natural log of that. And then you get-- the natural
log of e to anything, I've said it before, is just
the anything. So it is minus 0.05t is equal
to the natural log of 1/2. And then you get t is equal
to the natural log of 1/2, divided by minus 0.05. So let's figure out
what that is. Actually, someone just
made a comment, and I might as well do that. I could just put this
minus up here. I could make this a plus, and
this a minus, if I just multiply the numerator and the
denominator by negative 1. And if I want to, just to make
the calculator math a little easier, if you put a minus in
front of a natural log, or any logarithm, that's the same
thing as the log of the inverse of 2 over 0.05. It makes the calculator math
a little bit easier. The same thing. So if I do 2 natural log,
divided by 0.05, it is equal to 13.86. So when t is equal to 13.86. And I'm assuming that we're
dealing with time in years. That tends to be the convention,
although sometimes it could be something else
and you'd always have to convert to years. But assuming that this original
formula, where they gave this k value 0.05, that was
with the assumption that t is in years, and I've just
solved its half-life. I just solved that after 13.86
years, you can expect to have 1/2 of the substance left. We started with 100, we
ended up with 50. I could have started with x and
ended up with x over 2. Let's do one more of these
problems, just so that we're really comfortable
with the formula. Let's say that I have something
with a half-life of, I don't know, let's say I
have it as one month. Half-life of one month. And after, well let's say that
I-- well let me just for the sake of time, let me make
it a little bit simpler. Let's say I just have my k value
is equal to-- I mean you can go from half-life to a k
value, we did that in the previous video. Let's say my k value
is equal to 0.001. So my general formula is the
amount of product I have, is equal to the amount that I
started with times e to the minus 0.001 times t. And I gave you this, if you
have to figure it out from half-life, I did that in the
previous video with carbon-14. But let's say this
is the formula. And let's say that after, I
don't know, let's say after 1000 years I have 500 grams of
whatever element is described. The decay formula for
whatever element is described by this formula. How much did I start off with? So essentially I need to figure
out N sub 0, right? I'm saying that after 1000
years, so N of 1000, which is equal to N sub naught
times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's
equal to 500 grams. That equals 500 grams. So
I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000,
so this is N sub naught. This is 1/1000 of a 1000-- so
times e to the minus 1 is equal to 500 grams. Or I could
multiply both sides by e, and I have N sub naught is equal to
500e, which is about 2.71. So it's 500 times 2.71. I don't actually have e on this
calculator or at least I don't see it. So we'll have 1,355 grams. So
it's equal to 1,355 grams, or 1.355 kilograms. That's
what I started with. So hopefully you see now. I mean, I think we've approached
this pretty much at almost any direction that a
chemistry test or teacher could throw the problem
at you. But you really just need to
remember this formula. And this applies to
a lot of things. Later you'll learn, you know,
when you do compound interest in finance, the k will just be
a positive value, but it's essentially the same formula. And there's a lot of things
that this formula actually describes well beyond just
radioactive decay. But the simple idea is, use
information they give you to solve for as many of these
constants as you can. And then whatever they're
asking for, solve for whatever's left over. And hopefully I've given you
enough examples of that. But let me know, I'm
happy to do more.