If a face-centered cubic unit cell has a volume of 1.45 * 10⁸pm³ and the atoms at the corners touch the atom on the face, what must be the atom's radius?

Question

Quizplus · Accepted Answer

In a face-centered cubic (FCC) unit cell, the diagonal across the face of the cell, which goes through the center of two atoms on opposite faces and touches the edges of the cell, is equal to $4r$, where $r$ is the radius of the atoms. The length of each edge of the unit cell ($a$) can be related to the radius of the atoms by the equation $a = 2\sqrt{2}r$. Given the volume of the unit cell is $1.45 \times 10^{3} pm^{3}$, we can find the edge length $a$ by taking the cube root of the volume, $a = \sqrt[3]{1.45 \times 10^{3}} pm$. To find $r$, rearrange the equation to $r = \frac{a}{2\sqrt{2}}$. Plugging in the values and solving for $r$ gives a radius close to 186 pm, making option A correct.

If a Face-Centered Cubic Unit Cell Has a Volume of 1.45