![]() ![]() And let's see, my calculatorĭoesn't have an e to the power, so Let me just take e. The amount of the substance IĬan expect after 2000 years is equal to 300 times e to So let me put the negative number out there. So let me say, times 2000Įquals- and of course, this throws a negative out there, So what is that? So I already have that 1.2 timesġ0 the minus 4 there. Minus 1.2 times 10 to the minus 4, times t, is timesĢ000, times 2000. That I started off with, 300 grams, times e to the I have after, I don't know, after 2000 years? How much do I have? Well I just plug into With, I don't know, say I start off with 300 grams Period of time to figure out the k value. Other element, we would use that element's half-life toįigure out how much we're going to have at any given This is our formula forĬarbon, for carbon-14. ![]() Times the amount of time that has passed by. To have left is going to be the amount that we started At any given point in time,Īfter our starting point- so this is for, let's call this forĬarbon-14, for c-14- the amount of carbon-14 we're going So now we have the generalįormula for carbon-14, given its half-life. Negative 5,730, you get 1.2 times 10 to the negative 4. The natural log, and then you divide it by 5,730, it's a But let's see if we can do thatĪgain here, to avoid- for those who might Minus 5,730, which we did in the previous video. Just say, k is equal to the natural log of 1/2 over Minus 5,730k is equal to the natural log of 1/2. If we take the natural log ofīoth sides, what do we get? The natural log of e toĪnything, the natural log of e to the a is just a. Switching these two around- is equal to 1/2. Then we're left with e to the minus 5,730k- I'm just This equation for k, what do we get? Divide both sides by N naught. And half-life tells us thatĪfter 5,730 years we'll have 1/2 of our initial Wherever you see the t you put the minus 5,730- so minus ![]() Well, we're starting off with N sub 0 times e to the minus. N of 5,730 is equal to theĪmount we start off with. Take this information and apply it to this equation. So let's figure out the generalįormula for carbon. Just so that all of these variables can become a The k, and then apply it to your problem. And then for a particularĮlement with a particular half-life you can just solve for With, times e to some constant- in the last It can be described as the amount of element you have atĪny period of time is equal to the amount you started off Time, if you have some decaying atom, some element, You don't have to watch itįor an intro math class. And if you haven't takenĬalculus, you can really just skip that video. Last video, I proved that it involved a little bit Trying to figure out how much of a compound we have afterġ/2 of a half-life, or after one day, or 10 seconds, We can just take 1/2 of theĬompound at every period. Left after one half-life, or two half-lives, or ![]() We are trying to figure out how much of a compound we have Hope that clears things up.Ībout half-lives. So all the equations result from chemical kinetics of a first order reaction. And if we solve for half life we get: h = -ln(1/2)/k, which is where Sal got his equation. At the time of half life (h), half of the original sample has decayed which can be written as: ln((1/2*o)/o) = -kh. The other equation is derived from ln(/o) = -kt. Which can be rearranged into: = o*e^(-kt) which is the form Sal is using which ultimately originated from the rate law of a first order reaction. This can be rearranged into: ln(/o) = -kt If you use some calculus to figure out the integrated rate law (it's a separable differential equation) you arrive at: ln = -kt + lno, where o is the initial concentration of A. This can be rewritten as: -Δ/Δt = k, where the rate is being expressed as the disappearance of reactant A per unit of time. The rate law is written as: Rate=k, where 'k' is the rate constant and is the concentration of the reactant in molarity. 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. Rewrite the function in terms of half-life.The half-life formula is derived using 1st order kinetics since radioactive decay is a first order reaction.Ī first order reaction has the general form of: A -> products. ![]()
0 Comments
Leave a Reply. |
Details
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |