HW1-13b - 1-5 (b) The volumeV of the uraniumis V = Mu/pu =...

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Unformatted text preview: 1-5 (b) The volumeV of the uraniumis V = Mu/pu = (4000 )/(19.2 g/cm3) = g 208.3 m3. Hence he atomdensities re c t a Ns = ~ AsV MaNa = (791.9 )(6.022x 1023 toms/mol) = 9.740 X 1021 cm-3 g a (235 g/mol)(208.3 cm3) (3208g)(6.022 x 1023atoms/mol) = 3.896 X 1022 cm-3 AaV = (238g/mol)(208.3cm3) 14. A sampleof uranium is enrichedto 3.2 atom-percentin 235U ith the remainder w being 238U.What is the enrichment of 235Uin weight-percent? Solution: Let the subscripts 5, 8 and U refer to 235U, 238U,and uranium, respectively. For the given atom-% enrichment, The number of atoms in a sample of the uranium are Ns = O.O320Nu and Ns = O.9680Nu. The mass Ms and Ms of 23SU nd 23SUin the sample is a Ms = O.O320Nums and M8 = O.9680Num8, where m5 and ms is the massof an atom of 235Uand 23SU, espectively. r The enrichment in weight-% is thus e(wt-%) = 100 x Ms M5 + Ms100 X 0.0320 O.O320m5 , = 100 x 1 ---';;' D.0320ms + 0.9680mB 100 x 0.0320 = 3.16 wt-%. 15. A crystal of NaI has a density of 2.17 gjcm3. What is the atom density of sodium in the crystal? Solution: Atomic weights for Na and Cl are obtained from Table A.3, so that A(NaCl) = A(Na) + A(Cl) = 22.990+ 35.453= 58.443gjmol. Thus the atom density of Na is N(Na) = N(NaCI) = PNac.Na = 2.17 x 6.022 x 1023 = 2.24 X 1022 cm-3. A(NaCl) July 24, 2002 58.443 Ns= ...
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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 University (Engineering School).

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