Advertisements
Advertisements
Question
If ‘a’ stands for the edge length of the cubic system; sc, bcc, and fcc. Then the ratio of radii of spheres in these systems will be respectively.
Options
`(1/2 "a" : sqrt3/2 "a" : sqrt2/2 "a")`
`(sqrt1 "a" : sqrt3 "a" : sqrt2 "a")`
`(1/2 "a" : sqrt3/4 "a" : 1/(2sqrt2) "a")`
`(1/2 "a" : sqrt3 "a" : 1/sqrt2 "a")`
Solution
`(1/2 "a" : sqrt3/4 "a" : 1/(2sqrt2) "a")`
APPEARS IN
RELATED QUESTIONS
The number of unit cells in 8 gm of an element X (atomic mass 40) which crystallizes in bcc pattern is (NA is the Avogadro number)
CsCl has bcc arrangement, its unit cell edge length is 400 pm, it's inter atomic distance is
The vacant space in bcc lattice unit cell is ____________.
The radius of an atom is 300 pm, if it crystallizes in a face centered cubic lattice, the length of the edge of the unit cell is ____________.
The fraction of total volume occupied by the atoms in a simple cubic is
If ‘a’ is the length of the side of the cube, the distance between the body centered atom and one corner atom in the cube will be
Potassium has a bcc structure with nearest neighbor distance 4.52 A0. its atomic weight is 39. its density will be
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. calculate the edge length of the unit cell.
An atom crystallizes in fcc crystal lattice and has a density of 10 g cm−3 with unit cell edge length of 100 pm. calculate the number of atoms present in 1 g of crystal.
Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?
