If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:9:19

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 ten seconds or ten billion years and to solve to address that issue in the last video I proved and 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 prove the following formula at any given point of time if you have some decaying 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 start it 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 5730 years so let's see what that 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 5730 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 5,700 you put the 5730 so minus K times 5730 that's how many years have gone by and half-life tells us that after five thousand seven hundred and thirty years we'll have half of our initial sample left so we'll have half of our initial sample yet left so if we try to solve this equation 4k what do we get divide both sides by and not get rid of a t variable and then we're left with E to the minus 5730 K 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 we get 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 5730 K is equal to the natural log of 1/2 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 K is equal to the natural log of 1/2 / - 5730 which we did in the previous video but let's see if we can let's do that again here to avoid for those of you who might have skipped it so if you have 1/2 0.5 take the natural log and then you divide it by 5732 negative 5730 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 if we were do it for carbon-14 if we were doing this for some other element we would use that elements 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 let's say I start off with 300 grams of carbon carbon 14 carbon 14 and I want to know how much do I have after after I don't know after 2000 years 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 mm so what is that that's so I already have that 1.2 times 10 to the minus 4 there so let me say times 2000 equals and of course this there's a negative out there so let me put the negative number out there so this is a negative and I have to raise e to this power so it's 0.2 for 1 so this is equal to so n of 2000 the amount of the substance I can expect after 2000 years is equal to 300 times e to the minus 0.2 for 1/9 and let's see does my does my 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 2.7 one I could keep adding digits but I'll just do 2.7 1 to the point 2 4 negative which is equal to 0.78 times the amount that I started off with times 300 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 say let's go the other way around let's say I'm trying to figure out let's say I start off with let's say I start off with 400 grams of c14 and I want to know how long so I want to I don't I want to know I want to a certain amount of time does it take for me to get to 350 grams of c14 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 300 350 divided by 400 350 by 400 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 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 4 T and so T is equal to this divided by 1.2 times 10 to the minus 4 so the natural log point eight seven five 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 if you have 0.875 we want to take the natural log of it natural log and divide it by minus one point so divided by one point to e for negative 10 to the negative 4 and 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 have to take a negative so this is equal to one thousand one hundred and twelve years to get from 400 to three 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 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 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