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The concentration of Schottky defects in a crystal with a density of 6.022 \times 10^{28} m^{-3} is ;

Option: 1

1.661 \times 10^{18} \mathrm{~m}^{-3}

Option: 2

6.022 \times 10^{22} \mathrm{~m}^{-3}

Option: 3

1.661 \times 10^{22} \mathrm{~m}^{-3}

Option: 4

6.022 \times 10^{18} m^{-3} \mid

Answers (1)


The concentration of Schottky defects  can be calculated using the formula:C_s=\frac{n}{N} \mathrm{C}_{-} \mathrm{s}=\mathrm{n} / \mathrm{N}where n is the number of Schottky defects and N  is the total number of lattice sites.

The density of the crystal (p) can be calculated using the formula:

p=N \times \frac{m}{V}, where m is the mass of one unit cell and V is the volume of one unit cell.

The number of lattice sites (N) can be calculated using the formula:

N=p \times \frac{V}{m}

Substituting the value of N in the formula for C_s, we get:

C_s=\frac{n}{\left(p \times \frac{V}{m}\right)}

Rearranging the above equation, we get:

n=C_s \times p \times \frac{V}{m}

Substituting the given values, we get:

n=\frac{1}{2} \times 6.022 \times 10^{28} \times 1.58 \times 10^{-29} \times \frac{\left(5.642 \times 10^{-10}\right)^3}{\left(6.023 \times 10^{23}\right)}

n=1.661 \times 10^6

Therefore, the concentration of Schottky defects is

1.661 \times 10^{18} m^{-3}

Posted by

manish painkra

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