Chapter 1 The Solid State Solutions
Question - 41 : - What is a semiconductor?Describe the two main types of semiconductors and contrast their conductionmechanism.
Answer - 41 : -
Semiconductors aresubstances having conductance in the intermediate range of 10-6 to 104 ohm−1m−1.
The two main types ofsemiconductors are:
(i) n-type semiconductor
(ii) p-type semiconductor
n-type semiconductor: The semiconductor whose increased conductivityis a result of negatively-charged electrons is called an n-typesemiconductor. When the crystal of a group 14 element such as Si or Ge is dopedwith a group 15 element such as P or As, an n-type semiconductor isgenerated.
Si and Ge have fourvalence electrons each. In their crystals, each atom forms four covalent bonds.On the other hand, P and As contain five valence electrons each. When Si or Geis doped with P or As, the latter occupies some of the lattice sites in the crystal.Four out of five electrons are used in the formation of four covalent bondswith four neighbouring Si or Ge atoms. The remaining fifth electron becomesdelocalised and increases the conductivity of the doped Si or Ge.
p-type semiconductor: The semiconductor whose increased inconductivity is a result of electron hole is called a p-typesemiconductor. When a crystal of group 14 elements such as Si or Ge is dopedwith a group 13 element such as B, Al, or Ga (which contains only three valenceelectrons), a p-type of semiconductor is generated.
When a crystal of Si isdoped with B, the three electrons of B are used in the formation of threecovalent bonds and an electron hole is created. An electron from theneighbouring atom can come and fill this electron hole, but in doing so, itwould leave an electron hole at its original position. The process appears asif the electron hole has moved in the direction opposite to that of theelectron that filled it. Therefore, when an electric field is applied,electrons will move toward the positively-charged plate through electron holes.However, it will appear as if the electron holes are positively-charged and aremoving toward the negatively- charged plate.
Question - 42 : - Non-stoichiometriccuprous oxide, Cu2O can be prepared in laboratory. In this oxide,copper to oxygen ratio is slightly less than 2:1. Can you account for the factthat this substance is a p-type semiconductor?
Answer - 42 : -
In the cuprous oxide (Cu2O) prepared in the laboratory, copper to oxygen ratio is slightlyless than 2:1. This means that the number of Cu+ ions is slightly less than twice the number of O2− ions. This is because some Cu+ ions have been replaced by Cu2+ ions. Every Cu2+ ion replaces twoCu+ ions, thereby creating holes. As a result,the substance conducts electricity with the help of these positive holes.Hence, the substance is a p-type semiconductor.
Question - 43 : - Ferric oxidecrystallises in a hexagonal close-packed array of oxide ions with two out ofevery three octahedral holes occupied by ferric ions. Derive the formula of theferric oxide.
Answer - 43 : -
Let the number of oxide(O2−) ions be x.
So, number of octahedralvoids = x
It is given that two outof every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+) ionsTherefore, ratio of the number of Fe3+ ions to the number of O2− ions,
Fe3+ : O2− = 2 : 3
Hence, the formula ofthe ferric oxide is Fe2O3.
Question - 44 : - Classify each of thefollowing as being either a p-type or an n-type semiconductor:
(i) Ge doped with In (ii) Bdoped with Si.
Answer - 44 : -
(i) Ge (a group 14 element) is doped with In(a group 13 element). Therefore, a hole will be created and the semiconductorgenerated will be a p-type semiconductor.
(ii) B (a group 13 element) is doped with Si (agroup 14 element). Thus, a hole will be created and the semiconductor generatedwill be a p-type semiconductor.
Question - 45 : - Gold (atomic radius =0.144 nm) crystallises in a face-centred unit cell. What is the length of aside of the cell?
Answer - 45 : -
For aface-centred unit cell:
It is given that theatomic radius, r = 0.144 nm
So,
= 0.407 nm
Hence, length of a sideof the cell = 0.407 nm
Question - 46 : - In terms of band theory,what is the difference
Answer - 46 : -
(i) Between a conductor and an insulator
(ii) Between a conductor and a semiconductor
Answer
(i) The valence band of a conductor ispartially-filled or it overlaps with a higher energy, unoccupied conductionband.
On the other hand, inthe case of an insulator, the valence band is fully- filled and there is alarge gap between the valence band and the conduction band.
(ii) In the case of a conductor, the valenceband is partially-filled or it overlaps with a higher energy, unoccupiedconduction band. So, the electrons can flow easily under an applied electricfield.
On the other hand, thevalence band of a semiconductor is filled and there is a small gap between thevalence band and the next higher conduction band. Therefore, some electrons canjump from the valence band to the conduction band and conduct electricity.
Question - 47 : - Explain the followingterms with suitable examples:
Answer - 47 : -
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres
Answer
(i) Schottky defect: Schottkydefect is basically a vacancy defect shown by ionic solids. In this defect, anequal number of cations and anions are missing to maintain electricalneutrality. It decreases the density of a substance. Significant number ofSchottky defects is present in ionic solids. For example, in NaCl, there areapproximately 106 Schottky pairs per cm3 at room temperature. Ionic substances containingsimilar-sized cations and anions show this type of defect. For example: NaCl,KCl, CsCl, AgBr, etc.
(ii) Frenkel defect: Ionic solidscontaining large differences in the sizes of ions show this type of defect.When the smaller ion (usually cation) is dislocated from its normal site to aninterstitial site, Frenkel defect is created. It creates a vacancy defect aswell as an interstitial defect. Frenkel defect is also known as dislocationdefect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.
(iii) Interstitials: Interstitialdefect is shown by non-ionic solids. This type of defect is created when someconstituent particles (atoms or molecules) occupy an interstitial site of thecrystal. The density of a substance increases because of this defect.
(iv) F-centres: When the anionic sites of a crystal are occupiedby unpaired electrons, the ionic sites are called F-centres. These unpairedelectrons impart colour to the crystals. For example, when crystals of NaCl areheated in an atmosphere of sodium vapour, the sodium atoms are deposited on thesurface of the crystal. The Cl ions diffuse from the crystal to its surface andcombine with Na atoms, forming NaCl. During this process, the Na atoms on thesurface of the crystal lose electrons. These released electrons diffuse intothe crystal and occupy the vacant anionic sites, creating F-centres.
Question - 48 : - Aluminium crystallisesin a cubic close-packed structure. Its metallic radius is 125 pm.
Answer - 48 : -
(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
(i) For cubic close-packed structure:
= 353.55 pm
= 354 pm (approximately)
(ii) Volume of one unit cell = (354 pm)3
= 4.4 × 107 pm3
= 4.4 × 107 × 10−30 cm3
= 4.4 × 10−23 cm3
Therefore, number ofunit cells in 1.00 cm3 =
= 2.27 × 1022
Question - 49 : - If NaCl is doped with 10−3 mol % of SrCl2, what is theconcentration of cation vacancies?
Answer - 49 : -
It is giventhat NaCl is doped with 10−3 mol% of SrCl2.
This means that 100mol of NaCl is doped with 10−3 mol of SrCl2.
Therefore, 1 mol of NaCl isdoped withmol of SrCl2= 10−5 mol of SrCl2
Cation vacanciesproduced by one Sr2+ ion = 1
Hence, the concentrationof cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.
Question - 50 : - Explain the followingwith suitable examples:
Answer - 50 : -
(i) Ferromagnetism
(ii)Paramagnetism
(iii)Ferrimagnetism
(iv)Antiferromagnetism
(v)12-16 and 13-15 group compounds.
Answer
(i) Ferromagnetism: The substancesthat are strongly attracted by a magnetic field are called ferromagneticsubstances. Ferromagnetic substances can be permanently magnetised even in theabsence of a magnetic field. Some examples of ferromagnetic substances areiron, cobalt, nickel, gadolinium, and CrO2.
In solid state, themetal ions of ferromagnetic substances are grouped together into small regionscalled domains and each domain acts as a tiny magnet. In an un-magnetised pieceof a ferromagnetic substance, the domains are randomly-oriented and so, theirmagnetic moments get cancelled. However, when the substance is placed in amagnetic field, all the domains get oriented in the direction of the magneticfield. As a result, a strong magnetic effect is produced. This ordering ofdomains persists even after the removal of the magnetic field. Thus, theferromagnetic substance becomes a permanent magnet.
Schematic alignment of magnetic moments in ferromagnetic substances
(ii) Paramagnetism: The substances that are attracted by a magnetic field arecalled paramagnetic substances. Some examples of paramagnetic substances are O2,Cu2t, Fe3t, and Cr3t.
Paramagnetic substances get magnetised in amagnetic field in the same direction, but lose magnetism when the magneticfield is removed. To undergo paramagnetism, a substance must have one or moreunpaired electrons. This is because the unpaired electrons are attracted by amagnetic field, thereby causing paramagnetism.
(iii) Ferrimagnetism: The substances in which the magnetic moments of the domainsare aligned in parallel and anti-parallel directions, in unequal numbers, aresaid to have ferrimagnetism. Examples include Fe3O4 (magnetite),ferrites such as MgFe2O4 and ZnFe2O4.
Ferrimagnetic substances are weakly attracted bya magnetic field as compared to ferromagnetic substances. On heating, thesesubstances become paramagnetic.
Schematic alignment of magnetic moments in ferrimagnetic substances
(iv) Antiferromagnetism: Antiferromagneticsubstanceshave domain structures similar to ferromagnetic substances, but areoppositely-oriented. The oppositely-oriented domains cancel out each other’smagnetic moments.
Schematic alignment ofmagnetic moments in antiferromagnetic substances
(v) 12-16 and 13-15group compounds: The 12-16 groupcompounds are prepared by combining group 12 and group 16 elements and the13-15 group compounds are prepared by combining group 13 and group15 elements.These compounds are prepared to stimulate average valence of four as in Ge orSi. Indium (III) antimonide (IrSb), aluminium phosphide (AlP), and galliumarsenide (GaAS) are typical compounds of groups 13-15. GaAs semiconductors havea very fast response time and have revolutionised the designing ofsemiconductor devices. Examples of group 12-16 compounds include zinc sulphide(ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride(HgTe). The bonds in these compounds are not perfectly covalent. The ioniccharacter of the bonds depends on the electronegativities of the two elements.