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Half-Life Notes
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Headsium Decay # of ½ lives IndividualLab Island ClassCourse 0 1 2 3 4 30
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½ Life Activity (Bring back Calculators) 1.Go back to the lab with your partner and grab a ½ Life activity kit for EACH of you 2.You all have 30 pennies (radioactive isotopes) to begin with. Place them all HEADS UP. Shake the box (each trial = ½ life). 3.Take out the tails (these are the decayed isotopes). Record how many “headsium” isotopes remain 4.Repeat 3 more times and record results
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The Half-Life The half-life T 1/2 of an isotope is the time in which one- half of its unstable nuclei will decay. NoNo Number of Half-lives Number Undecayed Nuclei 1432 Where n is number of half-lives Number of Half-lives:
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Example 1: A sample of iodine-131 has a half life of 8 days. If you start out with 15.0 grams of I-131, how much is left of the sample 32 days later? First we determine the number of half-lives: n = 32 days/8days = 4 half-lives N = (15.0g) (1/2) 4 = 0.938 grams remaining 15.0 g 7.50 3.75 1.88 0.938 g
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Example 2: Cobalt-60 is a radioisotope used as a source of ionizing radiation in cancer treatment; the radiation it emits is effective in killing rapidly dividing cancer cells. Co-60 has a ½ life of five years. a.If the hospital starts with a 1000. mg supply, how many mg of Co-60 would they have to purchase after 10 yrs. to replenish their original supply? b.How many ½ lives would it take for the supply to Co-60 to dwindle to 1.Less than 10.0 % of the original? 2.Less ant 1.00 % of the original? 3.Less than.100 % of the original?
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a. If the hospital starts with a 1000. mg supply, how many mg of Co-60 would they have to purchase after 10 yrs. to replenish their original supply? First we determine the number of half-lives: N 0 = 1000. mg = 10 yrs/5 yrs = 2 half-lives =(1000. mg) (1/2) 2 = 250.0 mg remain 1000. mg – 250.0 mg = 750. mg needed to replenish 1000. -> 500.0 -> 250.0 mg remain
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b.How many ½ lives would it take for the supply to Co-60 to dwindle to 1.Less than 10.0 % of the original? 2.Less ant 1.00 % of the original? 3.Less than.100 % of the original? 1. “4” - 50% left after 1 half-life, 25 % left after 2 half-lives, 12.5 % left after 3 half- lives, and 6.25 % after 4 half-lives 2. “7” -.781 % left after 7 half-lives 3. “10” -.0977 % left after 10 half-lives
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Section 24.2 16. Calculate how much of a 10.0 g sample of americium- 241 remains after four half-lives. Americium-241is a radioisotope commonly used in smoke detectors and has a half-life of 430 y 0.625 g 10.0 → 5.00 → 2.50 → 1.25 → =(10.0 g) (1/2) 4 = 0.625 mg remain
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17. After 2.00 y, 1.986 g of a radioisotope remains from a sample that had an original mass of 2.000 g A. Calculate the half-lifeN = N 0 (1/2) t/T 1.986 g = 2.000 g x (1/2) (2.00 y)/T 1.986 g = (1/2) (2.00 y)/T 0.9930 = 0.5 (2.00 y)/T 2.000 gln 0.9930 = ln (0.5 (2.00 y)/T ) ln 0.9930 = ln (0.5) x (2.00 y)/T ln 0.9930 = ln (0.5) x (2.00 y)/T T = ln (0.5) x (2.00 y) = 197 years ln 0.9930 ln 0.9930 Section 24.2
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17. After 2.00 y, 1.986 g of a radioisotope remains from a sample that had an original mass of 2.000 g B. How much of the radioisotope remains after 10.00 y? N = 2.000 g x (1/2) 10.00 y/197 y = 1.93 g
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18. A sample of polonium- 214 originally has a mass of 1.0 g. Express the mass remaining as a percent of the original sample after a period of one, two, and three half-lives. Graph the percent remaining versus the number of half-lives. Approximately how much time has elapsed when 20% of the original sample remains? Section 24.2 NoNo Number of Half-lives Number Undecayed Nuclei 1432 50% 25% 12.5% 12.5% ~ 2.3 half lives From Table 24.5 pg 871: T 1/2 = 163.7 μs 163.7 x 2.3 ~ 380 μs
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