Respuesta :
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 ×[tex]10^{-5}[/tex] %
Explanation:
No. of phosphorus atoms = 5 × [tex]10^{17} \ cm^{-3}[/tex]
The volume occupied by a single Si atom
[tex]V_{si} = \frac{a^{3} }{8}[/tex]
[tex]V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}[/tex]
[tex]V_{si} =[/tex] 2 × [tex]10^{-23}[/tex] [tex]\frac{cm^{3} }{atom}[/tex]
[tex]n_{si} = \frac{1}{V_{si} }[/tex]
[tex]n_{si}[/tex] = 5 × [tex]10^{22}[/tex] [tex]\frac{atoms}{cm^{3} }[/tex]
[tex]PCT = \frac{N_p}{N_{si}} 100[/tex]
Put the values in above equation we get
[tex]PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100[/tex]
PCT = [tex]10^{-3} = 0.001[/tex] %
These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.
(b).
No. of boron atoms = 2 × [tex]10^{15} \ cm^{-3}[/tex]
The volume occupied by a single Si atom
[tex]V_{si} = \frac{a^{3} }{8}[/tex]
[tex]V_{si} = \frac{5.43^{3}(10^{-8} )^{3} }{8}[/tex]
[tex]V_{si} =[/tex] 2 × [tex]10^{-23}[/tex] [tex]\frac{cm^{3} }{atom}[/tex]
[tex]n_{si} = \frac{1}{V_{si} }[/tex]
[tex]n_{si}[/tex] = 5 × [tex]10^{22}[/tex] [tex]\frac{atoms}{cm^{3} }[/tex]
[tex]PCT = \frac{N_p}{N_{si}} 100[/tex]
Put the values in above equation we get
[tex]PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100[/tex]
PCT = 0.4 ×[tex]10^{-5}[/tex] %
These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.
