हिंदी

In case of hcp structure, how are spheres in first, second and third layers arranged?

Advertisements
Advertisements

प्रश्न

In case of hcp structure, how are spheres in first, second and third layers arranged?

टिप्पणी लिखिए
Advertisements

उत्तर

  1. Hexagonal close-packed structure is obtained by stacking two dimensional hexagonal close-packed layers.
  2. So, the spheres in the first layer are arranged to form two dimensional hexagonal close packing.
  3. The spheres of the second layer are placed in the depressions of the first layer. If the first layer is labelled as ‘A’ layer, the second layer is labelled as ‘B’ layer because the two layers are aligned differently.
  4. The spheres of the third layer are aligned with the spheres of the first layer. The resulting pattern of the layers will be ‘ABAB....’. This arrangement results in a hexagonal close-packed (hcp) structure.
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Solid State - Short answer questions (Type- II)

APPEARS IN

संबंधित प्रश्न

Answer the following in one or two sentences.

Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use?


Answer the following in one or two sentences.

Mention two properties that are common to both hcp and ccp lattices.


Answer the following in brief.

Cesium chloride crystallizes in a cubic unit cell with Cl ions at the corners and a Cs+ ion in the center of the cube. How many CsCl molecules are there in the unit cell?


Cu crystallizes in fcc unit cell with edge length of 495 pm. What is the radius of Cu atom?


Calculate the packing efficiency for bcc lattice.


A substance crystallizes in fcc structure. The unit cell edge length is 367.8 pm. Calculate the molar mass of the substance if its density is 21.5 g/cm3.


The number of particles in 1 g of a metallic crystal is equal to ____________.


The vacant space in simple cubic lattice is ____________.


The percentage of vacant space of bcc unit cell is ____________.


Which among the following crystal structures the edge length of unit cell is equal to twice the radius of one atom?


What is the edge length of fcc type of unit cell having density and atomic mass 6.22 g cm−3 and 60 g respectively?


Atoms of elements A and B crystallize in hep lattice to form a molecule. Element A occupies 2/3 of tetrahedral voids, the formula of molecule is ______.


Copper crystallizes as face centered cubic lattice, with edge length of unit cell 361 pm. Calculate the radius of copper atom.


What is the percentage of void space in bcc type of in unit cell?


Silver crystallizes in face centred cubic structure, if radius of silver atom is 144.5 pm. What is the edge length of unit cell?


Gold crystallizes in face centred cubic structure. If atomic mass of gold is 197 g mol-1, the mass of unit cell of gold is ______.


The relation between the radius of the sphere and the edge length in the body-centred cubic lattice is given by the formula ______.


A compound made of elements C and D crystallizes in a fee structure. Atoms of C are present at the corners of the cube. Atoms of D are at the centres of the faces of the cube. What is the formula of the compound?


 The packing efficiency of bcc is ______.


A compound has a hep structure. Calculate the number of voids in 0.4 mol of it.


Find the edge length of bcc unit cell if radius of metal atom is 126 pm.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×