Solving Radioactive Decay Problems

Solving Radioactive Decay Problems-32
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.

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.

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You can view that as kind of the inverse natural log. And that is equal to e to the minus lambda-t, plus c3.

So e to the power of ln of N, ln of N is just saying what power do you raise e to to get to N? So I'm just raising both sides of this equation. And now this can be rewritten as, N is equal to e to the minus lambda-t, times e to the c3.

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.

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.

And if you look at it at over some small period of time, let's say, if you look at it over one second, let's say our dt.

dt as an infinitesimally small time, but let's say it's a change in time. And let's say over one second, you observe that this sample had, I don't know, let's say you saw 1000 carbon particles.

They're all going to have different quantities right here. We'll actually do it in the next video, you can actually calculate this from the half-life.

But the rate of change is always going to be dependent on the number of particles you have, right? When you have 1/2 the number of particles, you lose 1/2 as much.

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