Solid State
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New answer posted
6 months agoContributor-Level 10
1.47 The energy gap between the valence band and conduction band in an insulator is very large while in a conductor, the energy gap is very small or there is overlapping between valence band and conduction band.
(ii) In a conductor, the valence band is practically filled or there is overlapping between valence band and conduction band while in semiconductor, there is always a small energy gap between them.
New answer posted
6 months agoContributor-Level 10
1.45 For fcc unit cell, a=2√2?.
Here, a is the edge length and r is the atomic radius (0.144 nm).
a = 2√2 *0.144 = 0.407 nm
Hence, the length of a side of a cell is 0.407 nm.
New answer posted
6 months agoContributor-Level 10
1.44 Ge is group 14 element and In is group 13 element. Hence an electron deficient hole is created and therefore, it is p–type.
2. B is group 13 elements and Si is group 14 elements, there will be a free electron. Hence, it is n-type
New answer posted
6 months agoContributor-Level 10
1.43 There is one octahedral hole for each atom in hexagonal close packed arrangement. If the number of oxide ions (O2−) per unit cell is 1, then the number of
New answer posted
6 months agoContributor-Level 10
1.42 The ratio less than 2:1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu+2) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu+2 ion, thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p -type semi conductor
New answer posted
6 months agoContributor-Level 10
1.41 These solids have conductive in the intermediate range from 10−6 to 104ohm−1m−1. As there is rise in
the temperature, conductivity also increases because electrons from the valence band jump to
conduction band.
Types of semiconductors
(a) n - type semiconductor when silicon or germanium crystal is doped with group 15 element like P or
As, the dopant atom forms four covalent bonds like a Si or Ge atom but the fifth electron no used in
bonding, becomes delocalised and contribute its share towards electrical conduction. Thus, silicon or
germanium doped with P or As is called n-type semiconductor (negative
New answer posted
6 months agoContributor-Level 10
Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?
1.40 It is given that nickel oxide has the formula as Ni0.98 O1.00.
As per the formula, there are 98 Ni ions for 100 oxide ions.
Out of 98 Ni ions, let x ions be in +2 oxidation state
98−x ions will be in +3 oxidation state.
Oxide ion has −2 charge.
To maintain electrical neutrality, total positive charge on cations = total negative charge on anions.
2x+3(98−x)+100(−2)=0
x=94
Fraction of Ni2+ ions = 94/98 = 0.96
Fraction of Ni2+ ions = 98-94/98 = 0.04
Hence, the fractions of nickel that exists as Ni2+ and Ni3+ are 0.
New answer posted
6 months agoContributor-Level 10
Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?
1.41 It is given that nickel oxide has the formula as Ni0.98 O1.00.
As per the formula, there are 98 Ni ions for 100 oxide ions.
Out of 98 Ni ions, let x ions be in +2 oxidation state
98? x ions will be in +3 oxidation state.
Oxide ion has ?2 charge.
To maintain electrical neutrality, total positive charge on cations = total negative charge on anions.
2x+3 (98? x)+100 (?2)=0
x=94
Fraction of Ni2+ ions = 94/98 = 0.96
Fraction of Ni2+ ions = 98-94/98 = 0.04
Hence, the fractions of nickel that exists as Ni2+ and Ni3+ are 0.96 a
New question posted
6 months agoNew answer posted
6 months agoContributor-Level 10
1.38 We apply pythagoras theorem AC2= AB2+ BC2
(2R)2= (R+r)2 +(R+r)2 = 2(R+r)2
4R2 = 2(R+r)2
(2R)2= (R+r)2
√2(R)2= √(R+r)2 = √2r = R+r
r = √2 R – R
r = (√2-1) R
r = (1.4114-1)R
r= 0.414 R
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