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Calculate the activity and write your solution in a separate sheet of paper. Calculate the activity due to ¹2C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

Calculate the activity and write your solution in a separate sheet of paper. Calculate the activity due to ¹2C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter29: Nuclear Physics
Section: Chapter Questions
Problem 19P
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Example Calculate the activity due to ¹4C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. To find the activity R using the equation 0.693N R t₁ 2 we must know N and t₁. The half-life of 14C was stated as 5730 y. To find N, we first find the number of 12C nuclei in 2 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by 1.3 x 10-12 (the abundance of 14C in a carbon sample from a living organism) to get the number of 14C nuclei in a living organism. Solution One mole of carbon has a mass of 12.0 g, since it is nearly pure ¹2C. (A mole has a mass in grams equal in magnitude to A found in the periodic table.) Thus, the number of carbon nuclei in a kilogram is N(¹2C) = 6.02x1023 mol-¹ 12.0g/mol -x (1000g) 5.02x10²5 So, the number of ¹4C nuclie in 1 kg of carbon is N(¹4C) = (5.02x1025) (1.3x10-12) = 6.52x10¹3 Now the activity R is found using the equation 0.693N 0.693 (6.52x10¹3) 5730 y R = Entering known values gives R = = 7.89x10²y-¹ t1 2 R = (7.89x10'y-¹) 1.00 y 3.16 x 10's or 7.89x10⁹ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus, = 250 Bq R = 250 Bq 3.7 x 10¹0 Bq/Ci or 250 decays per second. To express R in curies, we use the definition of a curie, = 6.67 x 10 Ci Thus, 6.67 nCi Activity Calculate the activity and write your solution in a separate sheet of paper. Calculate the activity due to ¹2C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. =
Transcribed Image Text:Example Calculate the activity due to ¹4C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. To find the activity R using the equation 0.693N R t₁ 2 we must know N and t₁. The half-life of 14C was stated as 5730 y. To find N, we first find the number of 12C nuclei in 2 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by 1.3 x 10-12 (the abundance of 14C in a carbon sample from a living organism) to get the number of 14C nuclei in a living organism. Solution One mole of carbon has a mass of 12.0 g, since it is nearly pure ¹2C. (A mole has a mass in grams equal in magnitude to A found in the periodic table.) Thus, the number of carbon nuclei in a kilogram is N(¹2C) = 6.02x1023 mol-¹ 12.0g/mol -x (1000g) 5.02x10²5 So, the number of ¹4C nuclie in 1 kg of carbon is N(¹4C) = (5.02x1025) (1.3x10-12) = 6.52x10¹3 Now the activity R is found using the equation 0.693N 0.693 (6.52x10¹3) 5730 y R = Entering known values gives R = = 7.89x10²y-¹ t1 2 R = (7.89x10'y-¹) 1.00 y 3.16 x 10's or 7.89x10⁹ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus, = 250 Bq R = 250 Bq 3.7 x 10¹0 Bq/Ci or 250 decays per second. To express R in curies, we use the definition of a curie, = 6.67 x 10 Ci Thus, 6.67 nCi Activity Calculate the activity and write your solution in a separate sheet of paper. Calculate the activity due to ¹2C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. =
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