The Solid State

  • Question 141
    CBSEENCH12005489

    How would you account for the following:
    Schottky defects lower the density of related solids.

    Solution
     It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.

    Thus there is equal number of positive and negtive ion leave correct lattice point and go outside the lattice creating  a pair of vacancy. So there are less number of ions thannbefore whichresult in the decrease in its density.
    Question 142
    CBSEENCH12005490

    How would you account for the following:
    Impurity doped silicon is a semiconductor.

    Solution
    Silicon forms four covalent bonds thus two types of impurities can be added to silicon.
    for example : boron can be added or phosphorus can be added.
    In case of boron : boron form three bonds with silicon, it will result in an electron deficient bond andwill create a hole. These holes can move through crystal like positive charge giving rise electrical conductivity.
    In case of phosphorus: when silicon forms four bonds with phosphorus one electron of phosphorous atom will remain unbonded due to which it become delocalizedn and contributes to electrical conductivity.
    Question 143
    CBSEENCH12005491

    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).

    Solution

    We have,
    Edge of length of cell a = 4.07x10–8cm
    Density p = 10.5 g /cm3
    Number of atoms in unit cell of fcc lattice = 4
    Avogadro number NA = 6.022x1023
    By using formula,
    Density space equals straight p space equals fraction numerator ZM over denominator straight a cubed straight N subscript straight A end fraction
straight M space equals space fraction numerator Pa cubed straight N subscript straight A over denominator straight Z end fraction

Putting space the space value space in space the space above space equation space we space get comma

fraction numerator 10.5 space straight x left parenthesis 4.07 right parenthesis cubed space straight x 6.022 space space straight x 10 to the power of 23 over denominator 4 end fraction
solving space the space equation space we space get comma

straight M equals space 107.09 straight g space Mol to the power of negative 1 end exponent

    Question 144
    CBSEENCH12005492

    Niobium crystallises in body centered cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

    Solution

    We have give that,
    Density (d)=8.55 g Cm-3
    Atomic mass (M) =93u =93 gMol-1
    Atomic radius (r) = ?

    We know that, Avogadro number Na =6.022 x 10 23 mol-1

    since given lattice is bcc

    therefore
    Number of atoms per unit cell (z) =2

    we know that

     

    straight d equals space fraction numerator zM over denominator straight a cubed straight N subscript straight A end fraction
8.55 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space 93 space over denominator straight a cubed space straight x 6.022 space straight x 10 to the power of 23 space mol to the power of negative 1 end exponent end fraction

straight a cubed space equals space fraction numerator 2 space straight x 93 over denominator 8.55 space straight x space 6.022 space straight x space 10 to the power of 23 end fraction cm cubed

straight a cubed space equals 3.6124 space straight x space 10 to the power of negative 23 end exponent

straight a cubed space equals 36.124 space straight x space 10 to the power of negative 24 end exponent
straight a equals space 3.3057 space straight x 10 to the power of negative 8 end exponent space cm

For space bb space unit space cell space radius space left parenthesis straight r right parenthesis space equals fraction numerator square root of 3 over denominator 4 end fraction straight a

straight r equals space fraction numerator 1.732 over denominator 4 end fraction space straight x space 3.3057 space straight x 10 to the power of negative 8 end exponent

straight r equals fraction numerator 5.725 over denominator 4 end fraction space straight x space 10 to the power of negative 8 end exponent cm
straight r equals space 14.31 space straight x space 10 to the power of negative 9 end exponent space cm

straight r equals 14.31 nm

     

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