ENGINEERING MATERIAL - ANSWER 12.3 : (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0

ENGINEERING MATERIAL - EXAMPLE 12.3 : Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.

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Answer Posted / kangchuentat

ENGINEERING MATERIAL - ANSWER 12.3 : (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0376 cubic nanometers. (b) Mass per unit cell = 2 x 92.9 (g / mol) [ 1 mol / (6.02 x 10 ^ 23 atoms) ] = 3.086 x 10 ^ (-22) g = m. V = 0.0376 x [ 10 ^ (-7) ] ^ 3 = 0.0376 x 10 ^ (-21) cubic cm. Then density is m / V = 8.207 g / cubic cm.

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