Question

question 5 (1 point) a fictional metallic element (atomic mass m = 76.94) crystallizes with a face - centred cubic structure and a density of 2.70 g cm^(-3). determine the atomic radius of one atom in a crystal of this bulk metal, in units of pm. your answer: answer units

Explanation:

Step1: Recall the formula for density of a face - centred cubic (FCC) unit cell

The density formula for a unit cell is $
ho=\frac{nM}{N_{A}V}$, where $
ho$ is density, $n$ is the number of atoms per unit cell, $M$ is the molar mass, $N_{A}=6.022\times 10^{23}\text{ mol}^{-1}$ is Avogadro's number and $V$ is the volume of the unit cell. For an FCC unit cell, $n = 4$.

Step2: Rearrange the density formula to find the volume of the unit cell

We can rewrite the density formula for $V$ as $V=\frac{nM}{N_{A}
ho}$. Substituting $n = 4$, $M=76.94\text{ g/mol}$, $N_{A}=6.022\times 10^{23}\text{ mol}^{-1}$ and $
ho = 2.70\text{ g/cm}^3$:
\[V=\frac{4\times76.94}{6.022\times 10^{23}\times2.70}\text{ cm}^3\]
\[V=\frac{307.76}{1.62594\times 10^{24}}\text{ cm}^3\approx1.893\times 10^{-22}\text{ cm}^3\]

Step3: Find the edge - length ($a$) of the unit cell

Since $V = a^{3}$ for a cubic unit cell, then $a=\sqrt[3]{V}$. So, $a=\sqrt[3]{1.893\times 10^{-22}\text{ cm}^3}\approx5.74\times 10^{-8}\text{ cm}$

Step4: Relate the edge - length to the atomic radius ($r$) in an FCC unit cell

In an FCC unit cell, the relationship between the edge - length $a$ and the atomic radius $r$ is $a = 2\sqrt{2}r$. So, $r=\frac{a}{2\sqrt{2}}$.
Substituting $a = 5.74\times 10^{-8}\text{ cm}$ into the formula for $r$:
\[r=\frac{5.74\times 10^{-8}\text{ cm}}{2\sqrt{2}}\approx2.03\times 10^{-8}\text{ cm}\]

Step5: Convert the radius to picometers

Since $1\text{ cm}=10^{10}\text{ pm}$, then $r = 2.03\times 10^{-8}\text{ cm}\times10^{10}\text{ pm/cm}=203\text{ pm}$

Answer:

203