Advertisements
Advertisements
प्रश्न
KF crystallizes in fcc structure like sodium chloride. calculate the distance between K+ and F− in KF.
(Given: density of KF is 2.48 g cm−3)
उत्तर
Density of KF 2.48 g cm−3
ρ = `"nM"/("a"^3"N"_"A")`
n = 4 (For fcc)
ρ = `(4 xx 58)/("a"^3 xx 6.023 xx 10^23)`
2.48 = `(4 xx 58)/("a"^3 xx 6.023 xx 10^23)`
a3 = `(4 xx 58)/(2.48 xx 6.023 xx 10^23)`
a3 = `232/(14.93 xx 10^23)`
a3 = 15.54 × 10−23
a = 537.5 pm
d = `"a"/sqrt2` (For fcc)
d = `537.5/1.414`
d = 380.12 pm
APPEARS IN
संबंधित प्रश्न
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 ____________.
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.
Potassium has a bcc structure with nearest neighbor distance 4.52 A0. its atomic weight is 39. its density will be
Explain briefly seven types of unit cells.
Calculate the number of atoms in a fcc unit cell.
An element has bcc structure with a cell edge of 288 pm. the density of the element is 7.2 g cm−3. how many atoms are present in 208 g of the element.
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.
Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4.3 × 10−8 cm. calculate the radius of a sodium atom.
