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The Solid State

Short Answer Type

141. How would you account for the following:
Schottky defects lower the density of related solids.

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142. How would you account for the following:
Impurity doped silicon is a semiconductor.

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143. Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 x 10^–8 cm and its density is 10.5 g cm^-3. Calculate on this basis the atomic mass of silver. (N_A= 6.02 x 10²³ mol^–1).

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144. Niobium crystallises in body centered cubic structure. If density is 8.55 g cm^–3, calculate atomic radius of niobium using its atomic mass 93 u.

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Long Answer Type

145. Copper crystallizes into a fcc lattice with edge length 3.61 x 10^–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^–3.

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Short Answer Type

146. Analysis shows that nickel oxide has formula Ni 0.98 O_1.00 What fractions of Nickel exist as Ni²⁺ and Ni³⁺ ions?

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147.

Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

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148. Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell?
(b) How many unit cells are there in 1.00 cm³ of aluminium?

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149. If NaCl is doped with 10^–3 mol % of SrCl_2'what is the concentration of cation vacancies?

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150. The ionic radius of CI^– ion is 181 pm. Consider the closest packed structure in which all anions are just touching:
(i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.

given:
Radius = 181 pm
thus,
Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
Cation having radius 74.934 pm will just fit into octahedral voids.

Radius of tetrahedral void
= 0.225 r = 0.225 x 181 pm = 40.725 pm
Cation having radius 40.725 pm will just fit into tetrahedral void.

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