general physics 2

Example Calculate the activity due to \"C in 1.00 kg of carbon found in a living organism. Express the activity in units of Hg and Ct. _ 0.693N _ t1 2' we must know N and 1. The half-life at 1\"if; was stated as 5730 y. To find N. we rst find the number of 120 nuclei in 2 1.00 kg at carbon using the concept at a mole. As indicated. we then multiply by 1.3 x 1012 {the abundance or 14t: in a carbon sample from a living organism} to getthe number of 1\"('5 nuclei in a living organism. To find the activity R using the equation Solution One mole of carbon has a mass of 12.0 9, since it is nearly pure 120. (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 6.02x1023m0r1 \"(IZC) = W 1(10009) = 5321:1025 So. the number of \"C nuclie in 1 kg of carbon is \"(145) =(5.02x1025}(1.3x10'13)= 6.52x1013 Now the activity R is found using the equation (1,5935! _ 0.693(652x1013) _ Entering known values gives R - = 7.89x109y\" I1 5730 y '1 1.00 y or 7.3925109 deca er ar. To convert this to the unit B , we R = 7_ 9 m9 -1 , = 2 0 B \"p ye q (8x y )3.16x10s 5 q stmplv convert years to seconds. Thus, or 250 decays per second. To express R in curies, we R = $32\"! = 667 x 10'9Ci Thus, 6.67 nCi use the definition of a curie, ' " Activity Calculate the activity and write your solution in a separate sheet of paper. Calculate the activity due to 1"'C in 1.00 kg at carbon found in a living organism. Express the activity in units of Be and Cr\