The atomic radius of a hypothetical face-centered cubic (FCC) metal is (√2 / 10) nm. The atomic weight of the metal is 24.092 g/mol. Taking Avogadro’s number to be 6.023×10²³ atoms/mol, the density of the metal is ____________ kg/m³ . (Answer in integer)

The correct answer is 2500.

Based on the provided question and the data shown in the reference diagram, we can calculate the density of the face-centered cubic (FCC) metal step-by-step:

Step 1: Find the edge length (a) of the unit cell
For a face-centered cubic (FCC) crystal structure, the relation between the edge length $a$ and the atomic radius $r$ is given by:
$$a=2\sqrt{2}r$$
Given the atomic radius from the image and text:
$$r=\frac{\sqrt{2}}{10}\text{nm}$$
Substitute the value of $r$ into the relation:
$$a=2\sqrt{2}\times \frac{\sqrt{2}}{10}=\frac{4}{10}\text{nm}=0.4\text{nm}=4\times {10}^{-10}\text{m}$$

Step 2: Calculate the volume (V) of the unit cell
The volume $V$ of a cubic unit cell is:
$$V=a^3={\left(4\times {10}^{-10}\text{m}\right)}^3=64\times {10}^{-30}{\text{m}}^3$$

Step 3: Calculate the density (ρ) of the metal
For an FCC unit cell, the number of atoms per unit cell ($N_e$) is 4.
The atomic weight of the metal is:
$$M=24.092\text{g/mol}=24.092\times {10}^{-3}\text{kg/mol}$$
Avogadro's number is given as:
$$N_A=6.023\times {10}^{23}\text{atoms/mol}$$
Using the density formula:
$$\rho =\frac{N_e\times M}{N_A\times V}$$
Substitute the values into the equation:
$$\rho =\frac{4\times 24.092\times {10}^{-3}}{6.023\times {10}^{23}\times 64\times {10}^{-30}}$$
Simplify the calculation:
$$\rho =\frac{96.368}{385.472\times {10}^{-7}}=0.25\times {10}^4=2500{\text{kg/m}}^3$$