A crystal belongs to a face cubic lattice with four atoms in the unit cell. The size of crystal is 1cm and its unit cell dimension is 1nm.Then the number of atoms in the crystal is in unit of 10^{21} is?
Crystal size = 1 cm
Size of unit cell = 1 nm
For fee structure, number of atoms per unit cell = 4
Number of atoms = 4x10^{21}
The correct answer is: 4
Metallic monovalent sodium crystallizes in bcc structures. If the length of unit cell is 4 x 10^{8} cm. The concentration of conduction electrons in metallic sodium is x * 10^{22 }cm^{3} . Find x?
Sodium has one conduction electron and bcc structure, for bcc structure, number of atoms in unit cell = 2.
Number of conduction electrons = 2x1 =2.
Concentration
The correct answer is: 3.125
Atomic radius of copper (fcc) is 1.28Â , then density of copper is (kg/m^{3} )
The correct answer is: 8260
Nearest neighbour distance in Na crystal is 1 . 8 3 Density of electrons in Na crystal in m^{3} is :
Na has bcc structure.
Nearest neighbour distance
The correct answer is: 0.212
Molecular weight of KCI (fcc) is 74.6amu and its density is 1.99 x 10^{3} kg/m^{3} . The lattice constant of KCI in units of is :
The correct answer is: 6.3
Smallest angle of diffraction for an Xray beam of wavelength 0.71 is (in degrees).
For [1 1 1] Plane, d=a/√(1^{2}+1^{2}+1^{2})=a/√3
2dsinθ=λ
For smallest angle, θ=sin^{1}(λ/2d)
θ=sin^{1}(λ/2d)
=sin^{1}(λ/(2a/√3)=sin^{1}(√3x0.71/2x6.3)
=> θ=5.6^{o}
So the correct answer is 5.6.
The value of lattice constant a for a fcc having in density 650 kg/m^{3} and atomic weight 60.2 in is :
FCC lattice (4R = 2 a) (n = 4) Atomic weight AW = 60.2 g/mol density r = 6250 kg/m3 = 6.25 g/cm3 N = 6.02 × 1026/kgmole = 6.02 × 1023/mole Volume of FCC = V = a3 Theoretical density r = A Volume n A N W = . . 6 02 10 a 4 60 2 23 3 = 6.25 a3 = 64 × 1024 cm3 a = 4 ×108 cm
If the angle between direction of Xray (incident) and diffracted one is 16°, the angle of incidence in degrees will be :
⇒ 2θ = 180°
⇒ θ = 82°
The correct answer is: 82
If 0.28nm is spacing between the nearest neighbouring ions in NaCI lattice, the unit cell parameter in is?
r  0.28 nm
⇒ 2r = 0.56 nm
= 5.6
The correct answer is: 5.6
If the miller indices for a plane is [1 1 1] and 2 be the lattice constant, then what is the area of the plane ?
Area of the plane=a√h^{2}+k^{2}+l^{2}
=>2√(1^{2}+1^{2}+1^{2})=3.464
The current answer: 3.464







