# hw6sol - H omework #6 Solutions From the text: 3.091 Fall...

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3.091 Fall Term 2004 Homework #6 Solutions From the text: 9-50. 3 , Av atomic molar molar N n V aV M ρ == + solve for n , the number of atoms/unit cell. 23 38 3 6.02 10 0.856 (5.247 10 ) 1.90 2 39.0893 Av atomic N na M ×× = × ×=≅ thus, K must be BCC with 2 atoms/unit cell 9-51. V unit cell = 2 51.996 g Cr 1 mol × 1 mol 6.022 × 10 23 7.20 g cm 3 V unit cell = a 3 = 2.40 × 10 –23 cm 3 a = 2.40 × 10 23 cm 3 3 = 2.88 × 10 -8 cm d 2 body = a 2 + a 2 + a 2 = 3a 2 d body = 3a = 3 × (2.88 × 10 -8 cm) = 5.00 × 10 -8 cm d body = 4 r Cr r Cr = 5.00 × 10 8 cm 4 = 1.25 × 10 -8 cm 9-59. From the geometry of the body-centered cubic structure, we find the shortest distance between barium atoms will occur down the body diagonal where d body = 3 a = 4r Ba r Ba = 3 4 × 0.5025 nm × 1m 10 9 nm × 100 cm = 2.178 × 10 –8 cm The shortest distance between barium atoms will be 2 r Ba . The Ba–Ba distance = 4.356 × 10 –8 cm.

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3.091 Fall Term 2004 Homework #6 Solutions page 2 Additional questions: 0. in BCC there are 2 atoms / unit cell, so 3 2 Av molar N aV = , where V molar = A/ ρ (A is atomic mass of iron) 3 1 3 8 2 24 3 1.24 10 Av Av N aA A ar N rc m ρ = ⎛⎞ ∴= = ⎜⎟ ⎝⎠ ∴= × if we assume that change of phase does not change the radius of the iron atom, then we repeat the calculation in the context of an FCC crystal structure, i.e., 4 atoms per unit cell and 22 = 3 3 4 8.60 (2 2 ) Av A gcm Nr ==
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## This note was uploaded on 11/02/2009 for the course CHEMISTRY 3.091 taught by Professor Donsadoway during the Fall '04 term at MIT.

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hw6sol - H omework #6 Solutions From the text: 3.091 Fall...

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