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Solid State Test - 13

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Solid State Test - 13
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  • Question 1
    1 / -0

    Schottky defect is observed in crystals when

    Solution

    Equal number of cations and anions are missing from the lattice.

  • Question 2
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    Which of the following statements are not true?

    (i)  Vacancy defect results in a decrease in the density of the substance.

    (ii)  Interstitial defects results in an increase in the density of the substance.

    (iii)  Impurity defect has no effect on the density of the substance.

    (iv)  Frankel defect results in an increase in the density of the substance.

    Solution

    In impurity deffect there is a replacement of ions by the ions of other compound.. For example if molten NaCl containing a little amount of \(SrCl_2\) is crystallised some of the sites of \(Na^+\) are occupied by \(Sr^{2+}\) .

    Each \(Sr^{2+}\) replaces two \(Na^+\) ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of \(Sr^{2+}\) ions.Thus, even though the electrical neutrality of the solid is maintained, a vacancy deffect is caused due to creation of vacant sites in the ionic solid. This results into change in density of the solid.The statement at option (iii) Impurity defect has no effect on the density of the substance, therefore not true.

    In Frankel defect the ion (usually cation) is dislocated (moved) from its normal lattice site to an interstitial site. So it is also called dislocation defect. It does not change the density of the crystal. It only creates cationic vacancies. Frenkel defect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it does not change the density of the solid.Thus the statement at option (iv) is not true.

  • Question 3
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    Which of the following statements are true about metals?

    (i)  Valence band overlaps with conduction band.

    (ii)  The gap between valence band and conduction band is negligible.

    (iii)  The gap between valence band and conduction band cannot be determined.

    (iv)  Valence band may remain partially filled.

    Solution

    Gap between valence band and conduction band determines the conductivity of the metals .The electrical conduction through metals depend on -

    (i) the type of valence band & also

    (ii) the gap between the valence band and conduction band .

    The statements at option (i), (ii ) and (iv) satisfy the conditions for electrical conductance of metals.

  • Question 4
    1 / -0

    Match the items given in Column I with the items given in Column II.

    Column I Column II
    (i)  Mg in solid state (a)  p-Type semiconductor
    (ii)  \(MgCl_2\) in molten state (b)  n-Type semiconductor
    (iii)  Silicon with phosphorus (c)  Electrolytic conductors
    (iv)  Germanium with boron (d)  Electronic conductors
    Solution

    (i)  Mg in solid state show electronic conductivity due to presence of free electrons hence, they are known as electronic conductors.

    (ii)   \(MgCl_2\) in molten state show electrolytic conductivity due to presence of electrolytes in molten state.

    (iii)  Silicon doped with phosphorus contain one extra electron due to which it shows conductivity under the influence of electric field and known as p-type semiconductor.

    (iv)  Germanium doped with boron contains one hole due to which it shows conductivity under the influence of electric field and known as n-type semiconductor.

  • Question 5
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    Amorphous solid can also be called .............

    (i)  Pseudo solids

    (ii)  True solids

    (iii)  Super cooled liquids

    (iv) Super cooled solids

    Solution

    The structure of amorphous solids is similar to that of the liquids. They have a tendency to flow like liquids, though very slowly.Therefore, some times they are called pseudo solids or super cooled liquids.

  • Question 6
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    Assertion : Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four.

    Reason : Besides the body centre there is one octahedral void present at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.

    Solution

    "Besides the body centre there is one octahedral void at the centre of each12 edges which is shared between four adjacent unit cells."

    Thus , Octahedral voids present at the body centre of the cube = 1

    12 ocahedral voids located at each edge and shared between four unit cells  \(=(12\times 1/4) = 3\)  

    Total number of octahedral voids = 4.

  • Question 7
    1 / -0

    Assertion : The packing efficiency is maximum for the fcc structure.

    Reason : The cordination number is 12 in fcc structures. 

    Solution

    " In fcc structure no. of atoms present = 4 per unit cell which provides a maximum packing efficiency as 74%."

  • Question 8
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    Under the influence of electric field, which of the following statements is true about the movement of electrons and holes in a p-type semi conductor?

    (i)  Electron will move towards the positvely charged plate through electron holes.

    (ii)  Holes will appear to be moving towards the negatively charged plate.

    (iii)  Both electrons and holes appear to move towards the positively charged plate.

    (iv)  Movement of electrons is not related to the movement of holes.

    Solution

    Silicon or Germanium (group-14) doped with electron defficient impurity like group-13 element (viz. B , Al or Ga ) containing only three valence electrons is called p-type semiconductors. Electron holes are created when an atom of such trivalent metal gets bonded with three of its neighbouring tetravalent silicon or germanium atoms. The fourth valence electron of Ge or Si moves through these electron holes . Under the influence of electric field electrons would move towards the positively charged plate through electronic holes ; but it would appear as the electron holes are positively charged and are moving towards negatively charged plate.

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