Respuesta :
Answer:
(a) In the single crystal lattice,0.001 % of silicon atoms per unit volume that displaced.
(b)In the single crystal lattice,[tex]4\times 10^{-6} \%[/tex] of silicon atoms per unit volume that displaced.
Explanation:
Unit cell:
- 8 atoms at the corners at [tex]\frac18[/tex] each cell
- 6 atoms in the face at [tex]\frac12[/tex] each in cell.
- 4 atoms within cell.
Total number of atoms per unit cell of silica is [tex](8\times \frac18+6\times \frac12+4)[/tex]=8 .
Dimension of unit cell is [tex]5.43\times 10^{-8}\ cm[/tex].
The volume occupied by a single Si atom in the crystal structure is
[tex]V_{si}=\frac{a^3 \ cm^3/cell}{\textrm{Number of atom per cell}}[/tex]
[tex]=\frac{(5.43\times 10^{-8})^3\ cm^3 /cell}{8 \ atoms /cell}[/tex]
[tex]=2\times 10^{-23}[/tex] [tex]cm^3/atom[/tex]
The concentration of Si atom is
[tex]n_{si}=\frac{1}{V_{Si}}[/tex]
[tex]=\frac{1}{2\times 10^{-23}\ cm^3/atom}[/tex]
[tex]=5\times 10^{22}[/tex] [tex]atom/cm^3[/tex]
(a)
[tex]n_P=5\times 10^{17} \ atom/cm^3[/tex]
[tex]PCT=\frac{n_P}{n_{Si}}\times 100[/tex]
[tex]=\frac{5\times 10^{17} \ atom/cm^3}{5\times 10^{22}\ atom/cm^3}\times 100[/tex]
[tex]=10^{-3}\%[/tex]
=0.001 %
In the single crystal lattice,0.001 % of silicon atoms per unit volume that displaced.
(b)
[tex]n_b=2\times 10^{15}\ atom /cm^3[/tex]
[tex]PCT=\frac{n_b}{n_{Si}}\times 100[/tex]
[tex]=\frac{2\times 10^{15} \ atom/cm^3}{5\times 10^{22}\ atom/cm^3}\times 100[/tex]
[tex]=4\times 10^{-6}[/tex] %
In the single crystal lattice,[tex]4\times 10^{-6} \%[/tex] of silicon atoms per unit volume that displaced.
