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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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= ...
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 University (Engineering School).

Ask a homework question - tutors are online