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Classify each of the following as being either a p-type or an n-type semi conducor:
(i) Ge doped with In
(ii) B doped with Si.


(i) Ge is an element of 14th group (like Si) and has configuration 4s24p2. It has been doped with In, a 13th group element having 5s25p1 configuration i.e., element of 14th group has been doped with 13th group element. All three valence electrons of impurity atom (In) gets bonded with three out of four eectrons of Ge and one electron of Ge remains unbonded. Conductivity is due to unbonded electron of insulator, Ge. Therefore, it is a p-type semi-conductor.

(ii) Boron, B is an element of 13th group and has 2s22p1 configuration. It is doped with Si, an element of 14th group having 3s23p2 configuration. All three electrons of boron gets bonded with 3 out of 4 electrons of Si and 4th electron of impurity atom (i.e., Si) is responsible for conductivity. Thus it is a n-type semiconductor.

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Ferric oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three tetrahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Let us Suppose the number of oxide (O2–) ions = N.
Number of octahedral void = number of anions
So that the number of octahedral voids = N

We have Given that

Two out of every three octahedral holes are occupied by ferric ions.

So that the number of ferric (Fe3+) ions = 2N/3
The ratio of the number of Fe3+ ions to the number of O2− ions,

Fe3+ : O2− = 2N/3 : N

Multiply 3 and divide by N we get

Fe3+ : O2− = 2 : 3

Hence, the formula of the ferric oxide is Fe2O3.

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If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

An octahedral void is shown in Fig. Though an octahedral void is surrounded by six spheres, only four are shown. The spheres present above and below the void are not shown. Let us assume that the edge length of the unit cell is a cm and raius of octahedral void is r and the radius of sphere is R.

An octahedral void is shown in Fig. Though an octahedral void is surr

Fig. A cross-section of an octahedral void.
In right angle triangle ABC
           AB = BC = acm
The diagonal AC is:
              AB space equals space square root of AB squared plus BC squared end root
space space space space space space space space space space space space equals space square root of straight a squared plus straight a squared end root space equals space square root of 2 straight a

and also AC over AB equals fraction numerator square root of 2 straight a over denominator straight a end fraction equals fraction numerator square root of 2 over denominator 1 end fraction
            Substituting
                  AB = 2R,  AC = R +2r+R
                       = 2R + 2r
∴    fraction numerator 2 straight R plus 2 straight r over denominator 2 straight R end fraction space equals space fraction numerator square root of 2 over denominator 1 end fraction  or   straight r over straight R equals fraction numerator square root of 2 over denominator 1 end fraction

or         straight r over straight R space equals space square root of 2 minus 1 space equals space 0.414
or                r  = 0.414 R.
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Non-stoichiometric cuprous 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 fact that this substance is a p-type semi-conductor ?

In the sample of cuprous oxide, Cu2O prepared in laboratory, copper to oxygen ratio is less than 2 : 1 and this deficiency of copper causes the metal deficiency defects. In these defects the positive ions are less in number as compared to anions. These defects are caused in two ways:
(i) By Cation Vacancies : Some Cu+ may be missing from lattice sites and their positive charges are balanced by presence of extra charge on adjacent cations i.e., some Cu2+ are present and some cation vacancies are present.
(ii) By presence of extra anions at interstitial sites: The type of defect can arise where some oxide ion, O2– are present at lattice sites and their charge is balanced by neighbouring cations in higher oxidation states i.e., some Cu2+ ions are present instead of Cu2+ ions.

In the sample of cuprous oxide, Cu2O prepared in laboratory, copper t
Metal deficiency defect due cation vacancy.

In the sample of cuprous oxide, Cu2O prepared in laboratory, copper t
Metal deficiency defect due to presence of extra anions.

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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= 6.02 x 1023 mol–1).

Edge length of unit cell,

Edge length of unit cell,
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