Time 5- What is the slope of the line in number 4- What is the value o.docx
Time 5. What is the slope of the line in number 4? What is the value of the rate constant, k, iIt km? The Slope is -o.oo3? . The rat e constant. \'s o.?3??. 6. Using the relationship between the half-live and the rate constant (In 2 -kts), determine the half-life of your radioactive substance,tw 7. Is your value of ty consistent with the accepted value of approximately 2.552 minutes for the halif-life of barium-137? What is the percent error? No, our hak lifevalue was 3.44 minutes, whicis o.sG2 minutes above the accepted value. 3.lit-2.ssaIx1oo,22.02% 8. What fraction of the initial activity of your barium sample would remain after 25 minu barium 137 9. Was a good assumption that the counts in the last five minutes would be due entirey to non-barium sources? Solution 8) given : n=t(given)/t(half) . here, n=25/3.114. . n=8. now,again: ms/ml=(1/2)^n. therefore ms/ml=(1/2)^8. . ms/ml=0.39%. 9) we see that amount of barium left is almost negigible.So yes,we can say that counts in the last 5 minutes would be entirely due to non barium sources. .
Time 5- What is the slope of the line in number 4- What is the value o.docx
Time 5. What is the slope of the line in number 4? What is the value of the rate constant, k, iIt km? The Slope is -o.oo3? . The rat e constant. \'s o.?3??. 6. Using the relationship between the half-live and the rate constant (In 2 -kts), determine the half-life of your radioactive substance,tw 7. Is your value of ty consistent with the accepted value of approximately 2.552 minutes for the halif-life of barium-137? What is the percent error? No, our hak lifevalue was 3.44 minutes, whicis o.sG2 minutes above the accepted value. 3.lit-2.ssaIx1oo,22.02% 8. What fraction of the initial activity of your barium sample would remain after 25 minu barium 137 9. Was a good assumption that the counts in the last five minutes would be due entirey to non-barium sources? Solution 8) given : n=t(given)/t(half) . here, n=25/3.114. . n=8. now,again: ms/ml=(1/2)^n. therefore ms/ml=(1/2)^8. . ms/ml=0.39%. 9) we see that amount of barium left is almost negigible.So yes,we can say that counts in the last 5 minutes would be entirely due to non barium sources. .
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Time 5- What is the slope of the line in number 4- What is the value o.docx
- 1. Time 5. What is the slope of the line in number 4? What is the value of the rate constant, k, iIt km? The Slope is -o.oo3? . The rat e constant. 's o.?3??. 6. Using the relationship between the half-live and the rate constant (In 2 -kts), determine the half-life of your radioactive substance,tw 7. Is your value of ty consistent with the accepted value of approximately 2.552 minutes for the halif-life of barium-137? What is the percent error? No, our hak lifevalue was 3.44 minutes, whicis o.sG2 minutes above the accepted value. 3.lit-2.ssaIx1oo,22.02% 8. What fraction of the initial activity of your barium sample would remain after 25 minu barium 137 9. Was a good assumption that the counts in the last five minutes would be due entirey to non-barium sources? Solution 8) given : n=t(given)/t(half) . here, n=25/3.114. . n=8. now,again: ms/ml=(1/2)^n. therefore ms/ml=(1/2)^8. . ms/ml=0.39%. 9) we see that amount of barium left is almost negigible.So yes,we can say that counts in the last 5 minutes would be entirely due to non barium sources.