Question

The number of ways in which a Schottky defect can occur in silicon is given by:

(N = total no. of atoms in a crystal of unit volume

n_s = number of Schottky defects per unit volume)

1. \(\dfrac{N!}{\left(N-n_s\right)!}\)
2. \(\dfrac{N!}{n_s!}\)
3. \(\dfrac{N!}{N-n_s!}\)
4. \(\dfrac{N!}{\left(N-n_s\right)!n_s!}\)

Answer 1

Correct Answer - Option 4 : \(\dfrac{N!}{\left(N-n_s\right)!n_s!}\)

Concept:

The Number of ways in which Schottky defect can occur in silicon depends on:

1) No. of atoms in a crystal of unit volume.

2) No. of Schottky atoms per unit volume.

If in given by \({}_{}^N{C_{{n_s}}}\)

Where N = No. of atoms

n_s = no. of schottky defects

\({}_{}^N{C_{{n_s}}} = \frac{{n!}}{{{n_s}!\left( {N - {n_s}} \right)!}}\)