Solid State

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New answer posted•a year ago
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1.25 Amorphous solids have short-range order with irregular shapes of constituent particles. They have isotropic nature and melt over a range of temperatures. They do not have a definite enthalpy of fusion. Examples of amorphous solids are glass, rubber, plastic, etc.

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1.24 Ferromagnetic substances would make better permanent magnets because when the ferromagnetic substance is placed in a magnetic field, all domains get oriented in the direction of magnetic field and strong a magnetic effect is produced.

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1.23 An n-type semiconductor conducts because of the presence of extra electrons. Therefore, a group 14 element can be converted to n-type semiconductor by doping it with a group 15 element.

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1.22 The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited. For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the

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1.21 Two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence some cation vacancies are created. For example : In an ionic solid 'NaCl', is impurity of Sr2+ is added (as SrCl2) then two Na+ ions left thus lattice sited. To maintain elec

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1.20 ZnS shows Frenkel Defect.
AgBr shows Frenkel Defect and Schottky Defect. Frenkel Defect: It is a kind of defect in crystalline solids in which atoms are displaced from their lattice position 
to interstitial site creating vacancy at the lattice point. It usually occurs in ionic solid with la

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1.19 When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant. Vacancydefect leads to a decrease in the density of the solid.

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1.18 Molar mass of the element = 2.7*10-2 kg mol-1
Edge length, a = 405 pm
Density, d = 2.7*103 kg m-3
Using the formula, d=? *? / ? 3NA
Putting the values given at their appropriate place, we get

(2.7 X 103 )X (405 X 10-12)3 X 6.022 X 10 / 2.7*10-2 = 3.99 which is approximately equal to 4

Therefore

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1.17 Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centered cubic lattices are 52.4% and 68% respectively

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1.16 The atoms of element M occupy 1/3rd of the tetrahedral voids.

Therefore, the number of atoms of M is equal to 2 1/3 = 2/3rd of the number of atoms of N. Therefore, ratio of the number of atoms of M to that of N is M : N = (2/3):1 = 2:3 Thus, the formula of the compound is M2N3.

ccp= fcc = 6 * 1/2

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