🤓 Based on our data, we think this question is relevant for Professor N/A's class at Ryerson University.

A **body-centered cubic (BCC) unit cell** is composed of a cube with one atom at each of its corners and one atom at the center of the cube.

Recall that the **edge length (a)** of** **a BCC unit cell can be calculated using the equation:

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{\mathbf{4}\mathbf{r}}{\sqrt{\mathbf{3}}}}$

**Step 1: Calculate the volume of 1 unit cell using density**

- molar mass Ba = 137.33 g/mol
- 1 mole = 6.022x10
^{23}entities (Avogadro' number)*entities = atoms, ions, molecules, formula units*

*recall the # of atoms present **per 1 BCC unit cell*: *corner atoms contribute 1/8 and the center atom contribute 1*

$\mathbf{\#}\mathbf{}\mathbf{of}\mathbf{}\mathbf{atoms}\mathbf{=}\left(\mathbf{8}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{8}}\right)\mathbf{+}\mathbf{1}$

**# of atoms = 2 per 1 unit cell**

• Solving **volume of 1 unit cell:**

$\mathbf{volume}\mathbf{=}\frac{\mathbf{137}\mathbf{.}\mathbf{33}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ba}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ba}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ba}}}{(6.022\times {10}^{23})\mathbf{}\overline{)\mathbf{Ba}\mathbf{}\mathbf{atoms}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\overline{)\mathbf{Ba}\mathbf{}\mathbf{atoms}}}{\mathbf{1}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{cell}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}{\mathbf{cm}}^{\mathbf{3}}}{\mathbf{3}\mathbf{.}\mathbf{59}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ba}}}$

**volume =1.27x10 ^{-22} cm^{3}/1 unit cell**

Barium has a density of 3.59 g/cm^{3} and crystallizes with a body-centered cubic unit cell. Calculate the radius of a barium atom.

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What scientific concept do you need to know in order to solve this problem?

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Based on our data, we think this problem is relevant for Professor N/A's class at Ryerson University.