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Leto [7]
3 years ago
13

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.
Chemistry
1 answer:
adelina 88 [10]3 years ago
4 0

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.

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