The theoretical density of diamond given that the C-C distance and bond angle are 0.154 nm and 109.5° respectively is; 3.54 g/cm³
The first thing we will do is to get the unit cell edge length from the given C - C distance.
From the structure of diamond;
Φ = ¹/₂ * bond angle
Φ = ¹/₂ * 109.5°
Φ = 54.75°
Thus;
θ = 90° - 54.75°
θ = 35.25°
Let the height from a point to the base of the cube of the structure be x. Thus;
x = a/4 = y sin θ
where a is unit cell edge length and y is C -C length. Thus;
a = 4y sin θ
a = 4 * (0.154 * 10⁻⁹) * sin 35.25°
a = 3.56 * 10⁻⁸ cm
The unit cell volume is;
V_c = a³
Thus; V_c = (3.56 * 10⁻⁸)³
V_c = 4.51 * 10⁻²³ cm³
Now, there are 8 equivalent atoms per unit cell and as such density is calculated from;
ρ = (n * A_c)/(V_c * N_a)
A_c is atomic weight of diamond = 12.01 g/mol
N_a is avogadro's number = 6.022 * 10²² atoms/g
Thus;
ρ = (8 * 12.01)/(4.51 * 10⁻²³ * 6.022 * 10²²)
ρ = 3.54 g/cm³
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