The half-life of cesium-137 is 30 years. Suppose we have a
100 -mg sample.
(a) Find the mass that remains after $t$ years.
(b) Find the mass that remains after $t$ years.
(b) How much of the sample remains after 100 years?
(c) After how long will only 1 mg remain?

Answer

a) $$
100 e^{-\frac{\ln 2}{30} t}
$$
b) $$
9.92 \mathrm{mg}
$$
c) $$
199.3
$$

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INVERSE FUNCTIONS

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Video Transcript

Hi, guys. Welcome back This problem. We know that the initial value is 100 milligrams at the half. Life is 30 years, quite a long half life, actually the 1st 1 to find the equation. So we know that will follow the form change. And why, with respect to t two k y pure to generate both sides. That comes out to why it's equal to the initial value. Y subzero times e raised the Katie. Now we could plug some dollars into this equation. We know. And since we have 1/2 life of 30 are half life will occur when there is 0.5 of the sample left. About 30 years, it will be equal 2.5 times. 100. It will be half of what the initial value us. And that comes out to 50. We have. Why 30 deal of 50. So you plug in 50 for a Y value. Our initial Y value we know is 100. If e race the Katie Que is unknown, R t value is 30 years now. We want to solve for K. We can divide both sides by 100. We're left with 0.5 this evil to e 30 k We can take the natural log of both sides. Upon doing that, we get the natural log of 0.5 is equal to 30 k and this K is equal to the natural log of 0.5 zero point five all over 30. Sorry about that. All over 30. Now we can plug that back in tow our equation. So we know that any time why we have why is equal to y subzero? Just 100 e race to the Ln 0.5 divided by 30 Cumpsty. It'll be our equation for the second part of the problem. The rest to figure out how much of the sample remains. After 100 years, part B t equals 100 t equals 100. What is? Why equal to between this plug this right indoor equation that we just created of why is equal to 100 he raised to the Ellen 0.5 over 30 times? 100? This is all erased all that. And when you do all that math, just quite a bunch of algebra, you're left with nine 0.92 We know that Why? Of 100 nine point nine to no units on that are milligram part C, or as to figure out what happens or how much time it takes with sampled to go down to only one milligram. How much time to sample was only one milligram? So again, we want to use our answer from part A wise he will do 100 e erase the l one point 5/30 times t we're gonna figure out what happens when. Why? Siegel toe one. So we have one is equal to 100. You raised to the Ln 0.5 over 30 t and now we just want to solve for T so we can divide both sides by 100. So we have 1000.0 one x equal to E Ellen 0.5 over 30 times. T we take the natural longer both sides are left with Ellen. 0.1 is able to Ellen 0.5 over 30 times t that would solve for t. You go to 30 times Ln We know one over Ellen 0.5. On doing that, you got an answer of 199 0.3. It will take Sorry. 199.3. It will take 100 in 99 0.3 years. The sample to reduce down to only one milligram or other words. If you were to graft this, this point will be on the graph you have. Why 199 0.3 Xarelto one. And that is the answer to part C. And that concludes all three parts of the problem. Thanks for watching.

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