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## MCAT

### Course: MCAT > Unit 8

Lesson 22: Atomic nucleus- Atomic nucleus questions
- Radioactive decay types article
- Decay graphs and half lives article
- Atomic number, mass number, and isotopes
- Atomic mass
- Mass defect and binding energy
- Nuclear stability and nuclear equations
- Writing nuclear equations for alpha, beta, and gamma decay
- Types of decay
- Half-life and carbon dating
- Half-life plot
- Exponential decay formula proof (can skip, involves calculus)
- Introduction to exponential decay
- Exponential decay and semi-log plots
- More exponential decay examples
- Mass spectrometer

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# Introduction to exponential decay

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video. Created by Sal Khan.

## Want to join the conversation?

- If this is an MCAT channel, I'm not sure why problems that cannot solved without using a calculator are examples?(33 votes)
- What is the mathematical process of using Carbon-14 to calculate the age of very old objects?(3 votes)
- Good question, @Wudaifu! In order to use 14C dating, the sample/object in question must have at one point been alive, which is why this method is often used to date fossils. There is a known background level of 14C in the biosphere, which is caused by the conversion of nitrogen to 14C in the atmosphere as a result of cosmic radiation and which remains practically constant through the ages. While organisms are alive, they are continuously exchanging carbon with their environment, and therefore the concentration of 14C in their biomass is equal to the background concentration in the environment. After the organism dies, the 14C decays but is no longer replaced because there ceases any interaction between the organism and its surroundings.

Thus, if we were to date a sample, we would utilize the equation in the video, substituting the background 14C concentration (which would be equal to the original concentration in the sample) for Nₒ and the sample's concentration for N; lambda would be equal to the value given in the video. We would then solve the equation for "t" like Sal did starting at5:02, which would give us the number of years that had elapsed since the organism died.

I hope my answer was helpful, however if that is not the case, please let me know in the comments section! :)(1 vote)

- How do do we solve natural log problems without a calculator?(4 votes)
- How would you solve this without a calculator?(2 votes)
- Why doesn't t appear to the right AND left side of the equation in the last example? It appears twice in the first example.(1 vote)

## Video transcript

- [Voiceover] 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-life or three half-lives. We can just take half 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 one half of a half-life or after one day or 10 seconds or 10 billion years. And to solve 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 used lambda. I could use "k" this time, minus "k" times "t". And then for a particular element with a particular half-life you could 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 "N" sub zero times "e" to the minus wherever you see the "t", you put the 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 half of our initial sample left. So we'll have half 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 we're left with "e" to the minus 5,730 "k", I'm just switching these two around, is equal to one half. Take the natural log of both sides. What do we get? We get the natural log of "e" to anything, the natural log of "e" to the "a" is just "a". The natural log of this is minus 5,730 "k" is equal to the natural log of one half. I just took the natural log of both sides. 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 one half over minus 5,730 so it equals 1.2 times 10 to the minus four. 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 or 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 four, times the amount of time that has passed by. This is our formula for carbon, if we were doing it 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... let's say that I start off with 300 grams of carbon-14, and I want to know how much do I have after 2,000 years. How much do I have? Well, I just plug into the formula. "N" of 2,000 is equal to the amount that I started off with, 300 grams, times "e" to the minus 1.2 times 10 to the minus four times "t", times 2,000. So what is that? 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 say 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 wanna know how long so I wanna 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 four, times "time" and now we solve for "time". So you get .875 is equal to "e" to the minus 1.2 times 10 to the minus four "t". Take the natural log of both sides. You get the natural of .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 four "t". So "t" is equal to this divided by 1.2 times 10 to the minus four. So the natural log of .875 divided by minus 1.2 times 10 to the minus four is equal to the amount of time it would take us to get from 400 grams to 350. My cell phone is ringing let me turn that off. To 350 so let me do the math. 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 you just have to. If there's one thing you just have to do, is you just have to remember this formula.