Polonium with an Atomic Radius of 0.168 nm Crystallizes in a Simple Cubic Structure: What is its Density (g/cm³)?

Polonium, an element with atomic number 84 and the symbol Po, exhibits unique properties when it crystallizes in a simple cubic structure, characterized by its atomic radius of 0.168 nm. To determine its density, we must consider the relationship between atomic radius, crystal structure, and atomic mass.

For a simple cubic structure, the edge length (a) of the unit cell is directly related to the atomic radius (r) by the equation: a = 2 * r

Given the atomic radius of 0.168 nm, we can calculate the edge length: a = 2 * 0.168 nm = 0.336 nm

The density (ρ) of a crystal can be determined using the following formula: ρ = (M * N_A) / (V * N)

where:

  • M is the atomic mass of polonium (209 g/mol)
  • N_A is Avogadro's number (6.022 × 10²³ mol⁻¹)
  • V is the volume of the unit cell
  • N is the number of atoms in the unit cell

For a simple cubic structure, the volume of the unit cell is calculated as: V = a³ = (0.336 nm)³ = 0.037 nm³

Since there is one atom per unit cell in a simple cubic structure, N = 1. Plugging these values into the density formula: ρ = (209 g/mol * 6.022 × 10²³ mol⁻¹) / (0.037 nm³ * 1)

Converting nanometers to centimeters: 1 nm = 10⁻⁹ cm

We get: ρ = (209 g/mol * 6.022 × 10²³ mol⁻¹) / (3.7 × 10⁻²⁴ cm³)

ρ ≈ 9.22 g/cm³

Therefore, the density of polonium in a simple cubic structure with an atomic radius of 0.168 nm is approximately 9.22 g/cm³.

  1. What type of crystal structure does polonium form with an atomic radius of 0.168 nm?
  2. How is the density of a crystal related to its atomic mass, atomic radius, and crystal structure?
  3. What is the volume of the unit cell for a simple cubic structure with an edge length (a) of 0.336 nm?
  4. How many atoms are present in a simple cubic unit cell?
  5. What is the density of polonium in a hexagonal close-packed structure, given its atomic radius of 0.168 nm?
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