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Chapter 9: Problem 57

Titanium metal has a body-centered cubic unit cell. The density of titanium is \(4.50 \mathrm{g} / \mathrm{cm}^{3} .\) Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

Short Answer

Expert verified
Using the given density of titanium, we first find the mass of one titanium atom, the volume of the unit cell, and then the edge length of the unit cell by calculating the cube root of the volume. With the edge length, we calculate the body diagonal using the Pythagorean theorem and finally, we determine the atomic radius of titanium as one-fourth of the body diagonal.

Step by step solution

01

Calculate the volume of the unit cell

To start, we will first need to get the total volume of the unit cell using the formula: \(Density (\rho) = \frac{Mass}{Volume}\) \(Volume = \frac{Mass}{\rho}\) In a body-centered cubic unit cell, there are 2 atoms per unit cell. The molar mass of titanium is 47.87 g/mol and we know that the density is given as 4.50 g/cm³.
02

Determine the mass of one titanium atom

To get the mass of 1 titanium atom present in the unit cell, we'll use Avogadro's number (6.022 x 10²³ particles/mol): \(Mass_{Ti-atom} = \frac{Molar\,Mass_{Ti}}{Avogadro's\,number}\) \(Mass_{Ti-atom} = \frac{47.87\,g/mol}{6.022 \times 10^{23}\,particles/mol}\)
03

Find the volume of the unit cell

Now, we can find the volume of the unit cell by plugging the mass of 1 titanium atom into the density formula: \(Volume_{Unit\,Cell} = \frac{2 \times Mass_{Ti-atom}}{Density}\)
04

Calculate the edge length of the unit cell

We know the volume of a cube is equal to \(edge^3\). Therefore, by solving for the edge length of the unit cell we get: \(Edge^3 = Volume_{Unit\,Cell}\) \(Edge = \sqrt[3]{Volume_{Unit\,Cell}}\)
05

Calculate the body diagonal of the unit cell

Let d be the body diagonal of the unit cell and a be the edge length. Using the Pythagorean theorem, we have: \(d^2 = a^2 + a^2 + a^2\) \(d^2 = 3a^2\)
06

Calculate the atomic radius of titanium

In a body-centered cubic unit cell, the atoms touch across the body diagonal. Therefore, the body diagonal is equal to the sum of four atomic radii (2 on each end and 2 in the center): \(d = 4 \times atomic\,radius\) Now, we can calculate the atomic radius: \(atomic\,radius = \frac{d}{4}\) By following these step-by-step calculations, we will find the edge length of the unit cell and the atomic radius of titanium.

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Most popular questions from this chapter

The structure of manganese fluoride can be described as a simple cubic array of manganese ions with fluoride ions at the center of each edge of the cubic unit cell. What is the charge of the manganese ions in this compound?

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) has a vapor pressure of 213 torr at \(40 .^{\circ} \mathrm{C}\) and 836 torr at \(80 .^{\circ} \mathrm{C} .\) What is the normal boiling point of \(\mathrm{CCl}_{4} ?\)

Some ionic compounds contain a mixture of different charged cations. For example, some titanium oxides contain a mixture of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions. Consider a certain oxide of titanium that is \(28.31 \%\) oxygen by mass and contains a mixture of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions. Determine the formula of the compound and the relative numbers of \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ti}^{3+}\) ions.

Compare and contrast the structures of the following solids. a. diamond versus graphite b. silica versus silicates versus glass

Explain how a p-n junction makes an excellent rectifier.
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