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Question 134870This question is from textbook
: Iodine-131 is a radioactive material used to treat a certain types of cancer. Iodine-131 has half-life of 8 days, which means that it tkes 8 days for half of an initial sample to decay. The percent of radioactive material that remains after t days can be detemined from the expression 100(1/2)^t/h, where h is the half-life in days.
a.what percent of a sample of iodine-131 will remain after 20 days?
b.Another form of raioactive iodine used in cancer treatment is iodine-125. Iodine-125 has a half-life of 59 days. A hospital has 20 g of iodine-125 and 20 g of Iodine-131 left over from treating patients. How much more iodine-125 than iodine-131 will remain after a period of 30 days?
This question is from textbook
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Iodine-131 is a radioactive material used to treat a certain types of cancer. Iodine-131 has half-life of 8 days, which means that it takes 8 days for half of an initial sample to decay. The percent of radioactive material that remains after t days can be determined from the expression 100(1/2)^t/h, where h is the half-life in days.
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a.what percent of a sample of iodine-131 will remain after 20 days?
A% = 100(1/2)^(20/8)
A% = 100(1/2)^2.5
A% = 100 * .1767767
A% = 17.68 %
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b. Another form of radioactive iodine used in cancer treatment is iodine-125. Iodine-125 has a half-life of 59 days. A hospital has 20 g of iodine-125 and 20 g of Iodine-131 left over from treating patients. How much more iodine-125 than iodine-131 will remain after a period of 30 days?
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Find the amt of Iodine-125 will be left after 30 days
I-125 = 20(1/2)^(30/59)
I-125 = 20 * .2554
I-125 = 5.08 g after 30 days
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Find the amt of Iodine-131 will be left after 30 days
I-131 = 20(1/2)^(30/8)
I=131 = 20 * .0743
I-131 = 1.49 g after 30 days
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5.08 - 1.49 = 3.61 g more of the Iodine-125
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