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Chapter 3: Problem 7

Molybdenum (Mo) has a BCC crystal structure, an atomic radius of \(0.1363 \mathrm{~nm}\), and an atomic weight of \(95.94 \mathrm{~g} / \mathrm{mol}\). Compute and compare its theoretical density with the experimental value found inside the front cover of the book.

Short Answer

Expert verified
Question: Calculate the theoretical density of Molybdenum (Mo) with a given atomic radius of 0.1363 nm, atomic weight of 95.94 g/mol, and a Body Centered Cubic (BCC) crystal structure. Compare the theoretical density with the experimental density value. Answer: The theoretical density of Molybdenum (Mo) with the given atomic radius, weight, and BCC structure is calculated to be 10.27 g/cm³.

Step by step solution

01

Understand the BCC structure

In a BCC structure, there are two atoms present within each unit cell. One atom is at the center of the unit cell, while the other atom is at the corners (1/8 of an atom at each corner, total 1).
02

Calculate the lattice constant (a)

For a BCC structure, lattice constant 'a' can be calculated using the relationship between atomic radius 'r' and lattice constant 'a': \(a = 4r/\sqrt{3}\). Given atomic radius \(r = 0.1363\,\mathrm{nm}\), we compute the lattice constant 'a': \(a = \frac{4(0.1363\,\mathrm{nm})}{\sqrt{3}} = 0.2963\,\mathrm{nm}\).
03

Find the volume of the unit cell

The volume of the unit cell can be calculated as: \(V = a^3\). With the computed lattice constant 'a', we find the volume of Mo's unit cell: \(V = (0.2963\,\mathrm{nm})^3 = 2.599 \times 10^{-2}\,\mathrm{nm^3}\).
04

Find the number of Mo atoms in the unit cell

As mentioned earlier, there are 2 Mo atoms in each unit cell of BCC structure.
05

Calculate Avogadro's number

Avogadro's number is a well-known constant which represents the number of atoms in one mole of substance, and is approximately equal to \(6.022 \times 10^{23}\,\mathrm{atoms/mol}\).
06

Calculate the theoretical density

Now, we can calculate the theoretical density (\(\rho\)) of Mo using the following formula: \(\rho = \frac{n (atoms/unit\,cell) \times M (g/mol)}{V(unit\,cell\,in\,cm^3) \times N_A (atoms/mol)}\). Here, n = 2, M = 95.94 g/mol, V = 2.599 x 10^{-2} nm^3 (convert to cm^3), and N_A = 6.022 x 10^{23} atoms/mol. \(\rho = \frac{2 \times 95.94\,\mathrm{g/mol}}{(2.599 \times 10^{-23}\,\mathrm{cm^3}) \times (6.022 \times 10^{23}\,\mathrm{atoms/mol})} = 10.27\,\mathrm{g/cm^3}\).
07

Compare theoretical and experimental density

Now, compare the calculated theoretical density of 10.27 g/cm³ with the experimental density value found inside the front cover of the book. The difference between the two values indicates the accuracy of our calculation.

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Most popular questions from this chapter

(a) What are the direction indices for a vector that passes from point \(\frac{1}{3} \frac{1}{2} 0\) to point \(\frac{2}{3} \frac{3}{4} \frac{1}{2}\) in a tetragonal unit cell? (b) Repeat part (a) for a rhombohedral unit cell.

Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters \(c / a\) is \(1.568\). If the radius of the Be atom is \(0.1143 \mathrm{~nm}\), (a) determine the unit cell volume, and (b) calculate the theoretical density of Be and compare it with the literature value.

Show that the atomic packing factor for HCP is \(0.74\).

If the atomic radius of lead is \(0.175 \mathrm{~nm}\), calculate the volume of its unit cell in cubic meters.

Magnesium (Mg) has an HCP crystal structure, a \(c / a\) ratio of \(1.624\), and a density of \(1.74 \mathrm{~g} / \mathrm{cm}^{3}\). Compute the atomic radius for \(\mathrm{Mg}\).
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