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Question 1187659: The half-life of two products of a chemical disaster are shown in the chart below:
Substance Half-life
Iodine-131 8.1 days
Cesium-144 282 days
Draw a graph showing the percent remaining during the first 5 half-lives.
b. Express the percent remaining as a function of the number of half-lives
elapsed n.
c. What percent of the substance remains after one week?
d. How long is it until the percent remaining of each substance is 10%?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! formula for growth is f = p * (1 + r) ^ n
in this problem, the time period is in days.
to find the rate for the half life, make f = 1/2 and p = 1.
formula becomes 1/2 = 1 * (1 + r) ^ n
since 1 * anything is anything, the formula becomes:
1/2 = (1 + r) ^ n
for iodine, the half life is represented by 1/2 = (1 + r) ^ 8.l
for cesium, the half life is represented by 1/2 = (1 + r) ^ 282
when you solve for r, you get the rate of growth per day.
a rate of growth less than 1 is a decay.
to find the rate of growth for iodine, raise both sides of the equation to the (1/8.1) power to get:
(1/2) ^ (1/8.1) = (1 + r)
solve for (1 + r) to get:
(1 + r) = .9179854612.
to find the rate of growth for cesium raise both sides of the equation to the (1/282) power to get:
(1/2) ^ (1/282) = (1 + r)
solve for (1 + r) to get:
(1 + r) = .9975450496.
iodine will have half its life remaining every 8.1 days.
cesium will have half its life remaining every 282 days.
the growth formula for iodine will be f = .9179854612^(8.1*x).
the growth formula for cesium will be f = .9975450496^(282*x).
these formulas will give you the percent remaining for each half life up to 5 half lives.
the graph of these equations are shown below:
the coordinate points on these graphs tell you the percent remaining based on the number of half lives that have elapsed.
for example, when x = 1, .5 = 50 percent is remaining and when x = 5, .0312 or 3.12% is remaining.
these results are rounded in the graph.
to get a more detailed answer, make a table as shown below.
x = 1, 50% remaining.
x = 2, 25% remaining.
x = 3, 12.5% remaining.
x = 4, 6.25% remaining.
x = 5, 3.125% remaining.
the graph was accurate except for the last one, where it looks like it truncated the 5 from 3.125 to get 3.12, or .0312 as shown on the graph.
the graph showed the rate.
the percent is 100 * the rate.
x represented the number of half lives.
it was the same in both graphs, except the exponent in the first graph was 8.1 * x and the exponent in the second graph was 282 * x.
that's because the half life of the iodine was every 8.1 days and the half life of the cesium was every 282 days.
to find the percent remaining after 7 days, the formulas to use are:
for iodine, f = .9179854612^x
for cesium, f = .9975450496^x
the x represents the number of days.
the growth rate shown is the growth per day.
for iodine, the half life will be achieve for every multiple of x = 8.1.
for cesium, the half life will be achieved for every multiple of x = 282.
here are two graphs that show that, the first for iodine, the second for cesium.
the following graphs show the percent remaining after 7 days.
the first graph is for iodine, where the percent remaining will be f = .9179854612^7 = .5493518875 * 100 = 54.93518875%.
the second graph is for cesium, where the percent remaining will be f = .9975450496^7 = .9829413931 * 100 = 98.29413931%.
the results on the graph are the rate, not the percent, and they are rounded to 3 decimal places; .549 for iodine and .983 for cesium.
to find how long it takes to have 10% of the original product remaining, the formulas will be:
for iodine, f = .9179854612^x becomes .10 = .9179854612^x.
for cesium, f = .9975450496^x becomes .10 = .9975450496^x.
to find the value of x, you use logs.
take the log of both sides of the equation to get:
for iodine:
log(.10) = log(.9179854612^x).
this becomes:
log(.10) = x * log(.9179854612).
solve for x to get:
x = log(.10) / log(.9179854612) = 26.90761757.
that's the number of days to get to 10% of the original amount.
for cesium:
log(.10) = log(.9975450496^x).
this becomes:
log(.10) = x * log(.9975450496).
solve for x to get:
x = log(.10) / log(.9975450496) = 936.7837228.
that's the number of days to get to 10% of the original amount.
the following graphs show those results.
the first is for iodine; the second is for cesium.
let me know if you have any questions.
theo
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