For a body-centered orthorhombic unit cell of atoms of an element (Atomic mass = 20 amu), the unit cell dimensions, $a = \frac{1}{2}$ nm, $b = \frac{1}{5}$ nm, $c = \frac{1}{10}$ nm. Find the density in the unit of $\frac{a m u}{m^{3}}$ .

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Solution

For bcc unit cell, the number (Z) of atoms per unit cell is 2.
The atomic mass (M) is 20 amu.
Density $= \frac{Z \times M}{(a b c)}$
Here, a, b and c are the unit cell dimensions.
Density $= \frac{2 \times 20}{(0.5 \times 10^{- 9} \times 0.2 \times 10^{- 9} \times 0.1 \times 10^{- 9})}$
Density $= 4 \times 10^{30} a m u / m^{3}$
Note: nm are converted to m by using the relationship 1 nm $= 10^{- 9}$ m

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For a body-centered orthorhombic unit cell of atoms of an element (Atomic mass = 20 amu), the unit cell dimensions, a=12 nm, b=15 nm, c=110 nm. Find the density in the unit of amum3.

For bcc unit cell, the number (Z) of atoms per unit cell is 2.The atomic mass (M) is 20 amu.Density =Z×M(abc)Here, a, b and c are the unit cell dimensions.Density =2×20(0.5×10−9×0.2×10−9×0.1×10−9)Density =4×1030amu/m3Note: nm are converted to m by using the relationship 1 nm =10−9 m

For a body-centered orthorhombic unit cell of atoms of an element (Atomic mass = 20 amu), the unit cell dimensions, $a = \frac{1}{2}$ nm, $b = \frac{1}{5}$ nm, $c = \frac{1}{10}$ nm. Find the density in the unit of $\frac{a m u}{m^{3}}$ .