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Platinum (atomic radius = 1.38 Å) crysta...

Platinum (atomic radius `= 1.38 Å`) crystallises in a cubic closed packed structure. Calculate the edge length of the face-centred cubic unit cell and the density of the platinum (Pt = 195)

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To solve the problem, we need to calculate the edge length of the face-centered cubic (FCC) unit cell and the density of platinum (Pt). ### Step 1: Calculate the Edge Length of the FCC Unit Cell 1. **Understanding the FCC Structure**: In a face-centered cubic (FCC) structure, atoms are located at each corner of the cube and at the center of each face. Each corner atom contributes 1/8 of its volume to the unit cell, and each face-centered atom contributes 1/2. 2. **Relationship Between Atomic Radius and Edge Length**: In an FCC structure, the face diagonal can be expressed in terms of the atomic radius (r). The face diagonal (d) can be calculated as: \[ ...
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In face -centered cubic unit cell, edge length is

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Knowledge Check

  • In face -centered cubic unit cell, edge length is

    A
    `4/sqrt3` r
    B
    `4/sqrt2` r
    C
    2r
    D
    `sqrt3/2` r
  • Aluminium crystallises in a cubic close-packed structure. Radius of the atom in the metal is 125 pm. What is the length of the side of the unit cell?

    A
    354 pm, 2.254 x `10^22` unit cells
    B
    254 pm, 3.254 x `10^22` unit cells
    C
    454 pm, 4.254 x `10^22 ` unit cells
    D
    654 pm, 5.254 x `10^22` unit cells
  • Aluminium crystallises in a cubic close-packed structure. Radius of the atom in the metal is 125 pm. How many unit cells are there in 1 cm of aluminium?

    A
    354 pm, 2.254 x `10^22` unit cells
    B
    254 pm, 3.254 x `10^22` unit cells
    C
    454 pm, 4.254 x `10^22 ` unit cells
    D
    654 pm, 5.254 x `10^22` unit cells
  • Similar Questions

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