Question

(a) If 5 × 10¹⁷ phosphorus atoms per cm³ are add to silicon as a substitutional impurity, determine the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b) Repeat part (a) for 2 × 10¹⁵ boron atoms per cm³ added to silicon.

Answer 1

(a) In the single crystal lattice,0.001 % of silicon atoms per unit volume that displaced.

(b)In the single crystal lattice,
`4* 10^(-6) \%` of silicon atoms per unit volume that displaced.

Step-by-step explanation:

Unit cell:

• 8 atoms at the corners at
  `\frac18` each cell
• 6 atoms in the face at
  `\frac12` each in cell.
• 4 atoms within cell.

Total number of atoms per unit cell of silica is
`(8* \frac18+6* \frac12+4)`=8 .

Dimension of unit cell is
`5.43* 10^(-8)\ cm`.

The volume occupied by a single Si atom in the crystal structure is

`V_(si)=\frac{a^3 \ cm^3/cell}{\textrm{Number of atom per cell}}`

`=((5.43* 10^(-8))^3\ cm^3 /cell)/(8 \ atoms /cell)`

`=2* 10^(-23)`
`cm^3/atom`

The concentration of Si atom is

`n_(si)=(1)/(V_(Si))`

`=(1)/(2* 10^(-23)\ cm^3/atom)`

`=5* 10^(22)`
`atom/cm^3`

(a)

`n_P=5* 10^(17) \ atom/cm^3`

`PCT=(n_P)/(n_(Si))* 100`

`=(5* 10^(17) \ atom/cm^3)/(5* 10^(22)\ atom/cm^3)* 100`

`=10^(-3)\%`

=0.001 %

In the single crystal lattice,0.001 % of silicon atoms per unit volume that displaced.

(b)

`n_b=2* 10^(15)\ atom /cm^3`

`PCT=(n_b)/(n_(Si))* 100`

`=(2* 10^(15) \ atom/cm^3)/(5* 10^(22)\ atom/cm^3)* 100`

`=4* 10^(-6)` %

In the single crystal lattice,
`4* 10^(-6) \%` of silicon atoms per unit volume that displaced.