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The half-life of bismuth-210, $ ^{210} Bi $, is 5 days.

(a) If a sample has a mass of 200 mg, find the amount remaining after 15 days.

(b) Find the amount remaining after $ t $ days.

(c) Estimate the amount remaining after 3 weeks.

(d) Use a graph to estimate the time required for the mass to be reduced to 1 mg.

a) 25 $\mathrm{mg}$

b) $y=200 \cdot 2^{-t / 5} \mathrm{mg}$

c) $y=200 \cdot 2^{-21 / 5} \approx 10.9 \mathrm{mg}$

d) 38.2 days.

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We're working on 1/2 life problem and we know the half life of the substance is five days. So if we start with 200 milligrams five days later, there's half of it. So there's 100 milligrams. Five days later, there's half of that 50 milligrams and five days later, half of that. So when 15 days have elapsed, there are 25 milligrams left. So that's the answer for part A. For part B. We want to generalize. So let's look at what's happening so that we can figure out how many milligrams will be left when T days have gone by. So we started with 200 then we multiply 200 by 1/2 so that could be considered multiplied by 1/2 to the first. And then we multiplied it by 1/2 again. So we multiplied it by 1/2 to the second power and then again, so we multiplied it by 1/2 to the third power. So what we can see is the exponents. Here is the number of days divided by five the number of days divided by the half life. So if we generalize that we could say that if T days have gone by, we have 200 times 1/2 times the number of days divided by the half life, and that gives us our answer for Part B. Remember that T is in days now. We're going to use that as our formula or as our model, and we're going to find the estimated amount of bismuth to 10 left after three weeks, so three weeks would be 21 days. We need to convert today's since this whole problem has been done in days, and then we're going to substitute 21 into our equation for tea, and then we'll put it in the calculator and we get approximately 10.88 milligrams remaining. Finally, we'll use a graphing calculator, and we'll graph the function that we found 200 times, 1/2 to the X, divided by five. And we'll also graph the line y equals one because we're interested in finding the time it takes to get down toe one milligram. So the line Y equals one will represent one milligram, then for a window. I've chosen to go from zero to our from negative 1 to 80 on the X axis and from negative 1 to 50 on my Y axis, and you just choose some numbers and fiddle around with them until you find what you like. Now we'll graph that. So here receive the exponential decay curve that represents the half life. And then we see this red line at the bottom. That is the line at a height of one that's one milligram and were interested in knowing that point of intersection so we can go into the calculate menu. Choose number five Intersect. Put the cursor on the first curve, press enter, put the cursor on the second curve, press enter and then move the cursor over to the intersection. Point press enter, and what we get you can see at the bottom of the screen is about 38.2. So that tells us that it takes about 38.2 days for the substance to be down to just one milligram remaining