Cobalt has an HCP crystal structure, an atomic radius of \(0.1253 \mathrm{nm},\) and a \(c / a\) ratio of 1.623 Compute the volume of the unit cell for Co.

Short Answer

Expert verified
Answer: The volume of the unit cell for cobalt is approximately 0.0661 nm³.

Step by step solution

01

Find the lattice parameter a from the atomic radius

In an HCP structure, the relationship between the lattice parameter a and the atomic radius r can be expressed as: \(a = 2\sqrt{2} \cdot r\) Given, r = 0.1253 nm Now substituting the value of r in the equation, \(a = 2\sqrt{2} \cdot 0.1253 = 0.3544 \mathrm{nm}\)
02

Find the lattice parameter c from the c/a ratio

Given, c/a ratio = 1.623 We found the value of a in step 1, so now we can compute the value of c as: \(c = (c/a) \cdot a\) \(c = 1.623 \cdot 0.3544 = 0.5753 \mathrm{nm}\)
03

Calculate the volume of the unit cell

Using the formula \(V_{cell} = (3\sqrt{3}/2) \cdot a^2 \cdot c\), and substituting the values of a and c found in steps 1 and 2, we get: \(V_{cell} = (3\sqrt{3}/2) \cdot (0.3544)^2 \cdot 0.5753\) \(V_{cell} \approx 0.0661 \mathrm{nm^3}\) The volume of the unit cell for cobalt is approximately 0.0661 nm³.

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