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Unformatted text preview: 5-7 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>.(s-l). (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 10-5 s-1 Thus the number of atoms of 24Na for a 1.20 MBq sourceis A N=-:x=
1.200 X 106 1.287 X 10-5 = 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 10-18 S-1 Thus the number of atoms of 238U for a 1.20 MBq sourceis N = ~ -1.200
). -4.916 X 10-18 = 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 half-life is 5730 y so that its decay constant is >. = In 2/T1/2 = 1.210 X 10-4 y-1. 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 10-4x 11,460] = 4.00 X lW2 atoms. e (b) For t = -1986-1985 = -3972 y, the number of 14Catoms in the specimen at 1986BC is N(-3972 y) = 1012exp[+1.210x 10-4 X 3972] = 1.62 X lW2 atoms. July 24, 2002 N(-11,460 ...
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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