Question

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm 3 of aluminium?

Answer 1

Hey there!

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Answer :

(i) r = 125 × $10^{-12}$ m, 4r = √2 a

⇒ a = $\frac{4r}{\sqrt{2}}$ = r × 2√2

= 125 × $10^{-12}$ × 2√2

= 250 × 1.414 ×$10^{-12}$ m

Edge length = a = 354 pm

⇒ $a^{3} = (354)^3$ × $(10^{-12})^3$ $m^{3}$

= $(3.54)^3$ × $10^{-30}$ $m^{3}$

= 44.36 × $10^{-30}$ $m^{3}$

(ii) Number of unit cells in 1 $cm^{3}$

= $\frac{Total Volume}{Volume of one unit cell}$

= $\frac{10^{-6}m^{3}}{44.36 × 10^{-30}m^{3}}$

= 2.254 × $10^{22}$ unit cells.

Answer 2

Hey !!

(a) r = 125 × 10⁻¹² m , 4r = √2a

=> a = 4r / √2 = r × 2√2

=> 125 × 10⁻¹² × 2√2

=> 250 × 1.414 × 10⁻¹² m

Now,

Edge Length = a = 354 pm

=> a³ = (354)³ × (10⁻¹²)³m³

=> (3.54)³ × 10⁻³⁰ m³

=> 44.36 × 10⁻³⁰ m³

which has 5 valence electrons. Out of those only 4 electrons are involved in bond formation 5th electron is not bonded and can be easily promoted to the conduction band. The conduction is thus mainly caused by the moment of electrons.

        So, due to presence of the negative charge such type of semi-conductor is called n-type semi conductor.

2) It is obtained by doping silicon with a group 13 element. Suppose Si doped with Ga which has 3 electrons, then these electrons form bonds with neighbouring Si atoms. A vacancy is left which can be filled by the transferd of valence electron from Si atom. This leaves behind hole which carries positive charge.

        The moment of this hole is responsible for the conduction of charge and such type are called p-type semiconductor.

(b) Number of unit cell is 1 cm³

            Total volume / Volume of one unit cell

     = 10⁻⁶ m³ / 44.36 × 10⁻³⁰ m³

     = 2.254 × 10²² unit cells.

GOOD LUCK !!