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The half-life of a substance is 100 years_ Find the exact k_rate of decaySimplify in terms of x sec (sin-1 [use triangle to find your answer}...

Question

The half-life of a substance is 100 years_ Find the exact k_rate of decaySimplify in terms of x sec (sin-1 [use triangle to find your answer}

The half-life of a substance is 100 years_ Find the exact k_rate of decay Simplify in terms of x sec (sin-1 [use triangle to find your answer}



Answers

A strontium isotope has a half-life of 90 years. What is the continuous compound rate of decay?

This question is walking us through continuous compound rates of decay for radioactive substances. And we're given this formula up here and that this substance has a half life of 90 years. We want to determine obviously that rate of case. Let's recall real quick that this half life is equal to t. In our formula. Because we're working with half lives whatever we started at, suppose it's too we need to have ended with half of it, right? Because in 90 years we're going to have half of what we started with. So let's just use two and one for simplicity, plugging in these values that we have, one is equal to two times E. To the power of Our times are half life, which is 90 going ahead. And solving for this, dividing both sides by two, we get 0.5 is equal to E. To the power of our times 90. Go ahead and take the natural log of both sides so that we're able to drop that E. On the right side and also bring down our exponents. We don't. So we can actually solve for R. Now it's pretty obvious that we can just divide both sides then by 90, and we'll have our continuous compound rate of decay. Plugging this into our calculators that are is then equal to negative 0.0077-016353.

But the rate of decay will be for three separate new Clyde's, each having varying half lives of 12,000 years, 12 hours and finally 12 seconds. To figure this out, we will have to utilize a certain equation. This equation has been derived from a larger one, but let's simplify it here will identify it as K is equal to the negative natural log of To this to represents the fact that rear losing half of every sample and the ratio would therefore be 2 to 1 from beginning to end will divide that by the time it takes for the sample to decay. And we will take that entire thing and multiply it by our starting material. And in this case, it's by Adam. Now, in this case, every one of the samples has been identified as being one mole. And as we hopefully remember, for every one mole, there is a God Rose number of Adam's that is 6.22 times 10 to the 23rd. So and we're gonna identify K as the decay rate, which is what is being asked for. All these decay rates are being asked for in seconds, so we're going to look at the 1st 1 and solve for it as such will take the negative log negative natural log of two. And in this case, we need our time to be in seconds to be consistent with what the question is asking. So if we were given 12,000 years, we will first need to convert that into seconds. So first on, multiplied by 365 days per year and in every day there's 24 hours and in 24 hours there is 3600 seconds, which I derived from 60 minutes multiplied by 60 seconds per minutes. So when I multiply this out, I will get and amount of 3.78 times 10 to the 11th seconds. So when I take the negative natural log of two divided by the time and then multiply it by a va God rose number, I will find that the decay rate ends up being negative 1.1 times 10 to the 12th decays per seconds and consequently that will work out as the answer. So now if we change this up so that we are instead going to be looking at 1/2 life of 12 hours, we'll take the same procedure. So in 12 hours there is 3600 seconds per hour and almost play that out and find it to be 43,200 seconds. So if I multiply that through with avocados number just like I did last time. So negative natural log of two divided by 43,200 seconds, multiplied by avocados number. And that will results in a value of nine points seven times 10 to the 18th decays per seconds. And then, finally, the last part's asked for What will it be? The decay rate, that is, if it only takes 12 seconds to take place for a single half life. So no further calculation necessary on this one. So negative natural log of two divided by 12 times of a God rose number, and it should end up being just about 3.5 times 10 to the 20 seconds decays for seconds

Hearing this problem Laid initial number of atoms. Initial number of atoms is equal to and not and final number of atoms. Final number of atoms is equal to end. So I can write the question at an is equal to and not into the power minus lame that we have simplifying it further. I can devaluate. One by two is equal to a to the power minus named dirty hub. On going forward, I can write the Value Edge two is equal to it to the power Linda Behalf taking log rhythm on both sides Natural logarithms so long to is equal to Linda the Uh huh. So I can write the value H t r is equal to learn to buy Lambda. So finally I gave the value the hub at learn to buy 8.9 into 10 to the power minus three purse again on solving I get the half physical to 70 eights again. So this is our answer note long to each 0.693 So dividing this value I get 70 eights again as the help like period of the given radioactive such tons

In this problem were asked to find the final amount if 10,000 DK's a rate of 12% over the course of 15 years. And to do this we're going to use the exponential decay formula, which says, See, Time's one minus are to the power of X C. Is your initial amounts in this case 10,000 one minus R R as your d K factor? Uh, 12% is the decay factor, but we're gonna write that as a decimal 0.12 and then X is the amount of years or your time frame, just 15 and then we'll just simplify this rate it a little easier as one minus 10.12 is 20 a to the power of 15. When she evaluates 2 1000 469 0.738 it goes on from there, but I rounded and the promise you around to the nearest whole, so we're gonna rate 1470


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