SOLUTION: 7.) Cesium-137 is a particularly dangerous by-product of nuclear reactors. It has a half-life of 30 years. It can be readily absorbed into the food chain and is one of the material

Algebra ->  Linear-equations -> SOLUTION: 7.) Cesium-137 is a particularly dangerous by-product of nuclear reactors. It has a half-life of 30 years. It can be readily absorbed into the food chain and is one of the material      Log On


   

Question 1000708: 7.) Cesium-137 is a particularly dangerous by-product of nuclear reactors. It has a half-life of 30 years. It can be readily absorbed into the food chain and is one of the materials that would be stored in the proposed waste
site at Yucca Mountain. Suppose we place 3000 grams of cesium-137 in a nuclear waste site.
a. How much cesium-137 will be present after 30 years, or one half-life?
b. How much cesium-137 will be present after 60 years, or two half-lives?
c. Find an exponential formula that gives the amount G in grams of cesium-137 remaining in the site after h half-lives.
d. How many half-lives of cesium-137 is 200 years? Round your answer to two decimal places.
e. Use your answer to part (c) to determine how much cesium-137 will be present after 200 years. Round your answer to the nearest whole number.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Cesium-137 is a particularly dangerous by-product of nuclear reactors. It has a half-life of 30 years. It can be readily absorbed into the food chain and is one of the materials that would be stored in the proposed waste site at Yucca Mountain.
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A(t) = Ao(1/2)^(x/30)

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Suppose we place 3000 grams of cesium-137 in a nuclear waste site.
a. How much cesium-137 will be present after 30 years, or one half-life?
Ans:: 1500 grams
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b. How much cesium-137 will be present after 60 years, or two half-lives?
Ans: 750 grams
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c. Find an exponential formula that gives the amount G in grams of cesium-137 remaining in the site after h half-lives.
Ans: 3000*(1/2)^h
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d. How many half-lives of cesium-137 is 200 years? Round your answer to two decimal places.
Ans: 200/30 = 20/3 = 6 2/3 half-lives
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e. Use your answer to part (c) to determine how much cesium-137 will be present after 200 years. Round your answer to the nearest whole number.
A(200) = 3000(1/2)^(200/30)
A(200) = 3000*(1/2)^6.666 = 29.53 grams
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Cheers,
Stan H.
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