First slide

Crystal lattices and unit cells

Question

Density of a unit cell is represented as
$ρ = \frac{Effective no. of atom (s) \times Atomic mass / formula mass}{Volume of a unit cell} = \frac{Z \cdot M .}{N_{A} \cdot a^{3}}$
where, mass of unit cell = mass of effective no. of atom(s) or ion(s)
M = At. mass/formula mass
$N_{A} = Avogadro's no. \Rightarrow 6 .023 \times 10^{23}$
$α$ = edge length of unit cel

Moderate

Question

Silver crystallizes in a fcc lattice and has a density of 10.6 g/cm°. What is the length of an edge of the unit cell?

A

40.7 nm
B

0.2035 nm
C

0.101 nm
D

4.07 nm

Solution

$\begin{aligned} 10 .6 = \frac{4 \times 108}{a^{3} \times 6 .023 \times 10^{23}} \\ ∴ a = 40 .7 nm \end{aligned}$

Question

An element crystallizes in a structure having fcc unit cell of an edge 200 pm. Calculate the density, if 100 g of this element contains 12 × 10²³ atoms:

A

$41.66 g / {cm}^{3}$
B

$4.166 g / {cm}^{3}$
C

$1.025 g / {cm}^{3}$
D

$1.025 g / {cm}^{3}$

Solution

$Mass of 12 \times 10^{23} atoms = 100 gm$
$Mass of 6.022 \times 10^{23} atom$
$\begin{array}{l} = \frac{100}{12 \times 10^{23}} \times 6 .023 \times 10^{23} \\ = 50 .18 \\ ∴ ρ = \frac{4 \times 50 .18 \times 10}{6 .023 \times 10^{23} \times {(200 \times 10^{- 10})}^{3}} \\ = 41 .66 g / {cm}^{3} \end{array}$

Question

The density of KBr is 2.75 g/cm^-3. The length of the edge of the unit cell is 654 pm. To which type of cubic crystal, KBr belongs?

A

simple cubic
B

bcc
C

fcc
D

hcp

Solution

By hit and Trial : For fcc Z = 4 (4 pairs of KBr)
M = 39 + 80 = 119

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