Question

(a) If 5 x 10^17 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 x 10^15 boron atoms per cm3 added to silicon.

Answer 1

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 ×$10^{-5}$ %

Explanation:

No. of phosphorus atoms = 5 × $10^{17} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100$

PCT = $10^{-3} = 0.001$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × $10^{15} \ cm^{-3}$

The volume occupied by a single Si atom

$V_{si} = \frac{a^{3} }{8}$

$V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}$

$V_{si} =$ 2 × $10^{-23}$ $\frac{cm^{3} }{atom}$

$n_{si} = \frac{1}{V_{si} }$

$n_{si}$ = 5 × $10^{22}$ $\frac{atoms}{cm^{3} }$

$PCT = \frac{N_p}{N_{si}} 100$

Put the values in above equation we get

$PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100$

PCT = 0.4 ×$10^{-5}$ %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.