Solving Radioactive Decay Problems

Solving Radioactive Decay Problems-18
We can actually solve this using pretty straightforward techniques. We want to get all the N's on this side and all the t stuff on the other side.This is actually a separation of variables problem. So if we have 1 over N, d N over dt is equal to minus lambda. And then I can multiply both sides of this by dt, and I get 1 over N d N is equal to minus lambda dt. I'm taking the indefinite integral or the antiderivative. Well that's the natural log of N plus some constant-- I'll just do that in blue-- plus some constant.

Tags: America Education EssayThematic Essay On Belief SystemsResume Cover Letter For Insurance UnderwriterWho Changed My Life EssayWhat To Write A College Essay On3000 Solved Problems In Linear AlgebraHow To Make A Conclusion In Research PaperThesis On Water Conservation

In the case of carbon-14, I'll tell you what percentage of my original carbon-14 has not decayed into nitrogen, as yet, nitrogen-14.

And that's useful, but what if I care about how much carbon I have after 1/2 a year, or after 1/2 a half life, or after three billion years, or after 10 minutes? A general function, as a function of time, that tells me the number, or the amount, of my decaying substance I have.

Here, if we start with 100 particles here, we went to 50 particles, then we went to 25.

When you start with 50, in a period of time you lose 25. So clearly the amount you lose is dependent on the amount you started with, right?

And now once again this is an arbitrary constant, so we can just really rename that as, I don't know, let me rename it as c4. So let's see if we can substitute that into our equation to solve for c4. So c4 is equal to N naught, our starting amount for the sample. We have the number of particles, or the amount as a function of t, is equal to the amount that we start off with, at time is equal to 0, times e to the minus lambda, times time. Well let's try to figure out this equation for carbon.

So, our solution to our differential equation, N, as a function of t, is equal to our c4 constant, c4e to the minus lambda-t. So we said N sub-0 is equal to, let's put 0 in here, so let's see, that's equal to N sub naught. And we just have to be careful that we're always using the time constant when we solve for the different coefficients. This'll be true for anything where we have radioactive decay.You really wouldn't see that with carbon-14, but this is just for the sake of our intuition.Let's say over one second you saw 1000 carbon particles per second here.Over any fraction of time, and here it's a very small fraction.So what I set up here is really fairly simple, but it doesn't sound so simple to a lot of people if you say it's a differential equation.For example, where time equals zero, we have 100% of our substance.Then after time equals one half-life, we'd have 50% of our substance.We're taking the antiderivative with respect to. These are different constants, but they're arbitrary.So if we want, we can just subtract that constant from that constant, and put them all on one side and then we just get another constant.Well here you have 1000th of the number particles in this sample as this one. But we know that no matter what substance we're talking about, this constant is dependent on the substance.So, for every thousand particles you saw decaying here, you'd really expect to see one carbon particle per second here. Carbon's going to be different from uranium, is going to be different from, you know, we looked at radon.


Comments Solving Radioactive Decay Problems

The Latest from ©