Question

If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density of the semiconductor in grams/cc ? Two significant digits, fixed point notation.

Answer 1

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are $6.023 \times 10^{26}$ atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = $\frac{38}{6.023 \times 10^{26}}$    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = $\frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}$

            = $37.06 \times 10^{-26}$

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = $a^{3}$

                   = $(0.503 \times 10^{-9})^{3}$

                   = $0.127 \times 10^{-27} m^{3}$

Formula to calculate density of diamond cell is as follows.

               Density = $\frac{mass}{volume}$

                             = $\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}$

                            = 2918.1 $g/m^{3}$

or,                         = 0.0029 g/cc       (as 1 $m^{3} = 10^{6} cm^{3}$)

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.