Gold and silver both have the FCC crystal structure, and Ag forms a substitution

Question

Gold and silver both have the FCC crystal structure, and Ag forms a substitutional solid solution for all concentrations at room temperature. Compute the unit cell edge length for a 55 wt% Au - 45 wt% Ag alloy. The room-temperature density and atomic weight of Au are 19.32 g/cm^3 and 196.97 g/mol, the room-temperature density and atomic weight of Ag are 10.49 g/cm^3 and 107.87 g/mol, respectively. In this problem, you are asked to find the unit cell edge length for a substitutional solid solution. What is the average density of this solid solution? What is the average atomic weight? What is the volume of the unit cell in cm^3? Be sure to review Section 5.6. What equation should you use to find the required unit cell dimension? Once you have the volume of the unit cell, recall the relationship between the volume and the unit cell dimension. Review Sections 3.3, 3.4, and 3.5. Check your mathematics. If you are having difficulty with this problem, you may wish to review: Solving Equations Review of 2D and 3D Geometry What is the unit cell dimension in cm?

Explanation / Answer

unit cell edge (a) for a cell = n{[(100/(CAu/AAu + CAg/AAg)]/[(100/CAu/pAu + CAg/pAg)](NA)}^1/3

with,

C is for metal percentage in alloy

A is atomic mass of metals

p is for densitiy of metals

NA is avogadro's number

n = 4 for FCC

we get,

a = 4{[(100/(55/196.97 + 45/107.87)]/[(100/(55/19.32 + 45/10.49)](6.023 x 10^23)}^1/3

   = 4(1.70 x 10^-23)^1/3

   = 4 x 2.57 x 10^-8

   = 1.03 x 10^-7 cm

Thus, the unit cell dimension is 1.03 x 10^-7 cm

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