This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 57 10. How many atoms are there in a 1.20 MBq sourceof (a) 24Naand (b) 238U? Solution:
BecauseA ==>'N we have N(atoms) = A(Bq)j>.(sl). (a) For 24Na we find from Table A.4 that Tl/2 = 14.96 h = 5.385 X 104s. Then>.. = In2/TI/2 = 1.287x 105 s1 Thus the number of atoms of 24Na for a 1.20 MBq sourceis A N=:x=
1.200 X 106 1.287 X 105 = 9.32 X lWo atoms. (b) For 238U e find from Table A.4 that T 1/2 = 4.468X 109 Y = 1.410X 1017s. w Then>. = In 2/Tl/2 = 4.916 X 1018 S1 Thus the number of atoms of 238U for a 1.20 MBq sourceis N = ~ 1.200
). 4.916 X 1018 = 2.44 X 1023 atoms. X 106 11. A very old specimenof wood contained 1012atoms of 14C in 1986. (a) How many 14c atoms did it contain in 9474B.C.? (b) How many 14c atoms did it contain in 1986B.C.? Solution:
For 14C the halflife is 5730 y so that its decay constant is >. = In 2/T1/2 = 1.210 X 104 y1. From the radioactive decay law N(t) = N(O)exp[>.t]. In this problem we identify the year 1986as t = 0 with earlier times being negative. (a) For t = 1986 9474 = 11,460 y, the number of 14C atoms in the specimen at 9474 BC is y) = 1012 xp[+1.210 X 104x 11,460] = 4.00 X lW2 atoms. e (b) For t = 19861985 = 3972 y, the number of 14Catoms in the specimen at 1986BC is N(3972 y) = 1012exp[+1.210x 104 X 3972] = 1.62 X lW2 atoms. July 24, 2002 N(11,460 ...
View
Full
Document
This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell.
 Spring '06
 CADY

Click to edit the document details