The Solid State

  • Question 253
    CBSEENCH12010159

    (a) What type of semiconductor is obtained when silicon is doped with boron?

    (b) What type of magnetism is shown in the following alignment of magnetic moments?

    bold upwards arrow bold space bold space bold space bold upwards arrow bold space bold upwards arrow bold space bold upwards arrow bold space bold upwards arrow

    (c) What type of point defect is produced when AgCl is doped with CdCl2?

    Solution

    (a) When silicon is doped with boron, a p-type semiconductor is obtained.

     (b) The magnetism shown in the alignment of magnetic moments is ferromagnetism.

    Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets, or are attracted to magnets.

     (c) Impurity defect is produced when AgCl is doped with CdCl2.

    Question 254
    CBSEENCH12010183

    What is meant by ‘doping’ in a semiconductor?

    Solution

    Doping is the process of increasing the conductivities of the intrinsic semiconductors by adding suitable impurity.

    Question 255
    CBSEENCH12010203

    Tungsten crystallizes in the body-centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom?

    Solution
    It space is space body space centered space cubic space cell space for space which space radius space
straight a space equals space 316.5 space pm space left parenthesis given right parenthesis space
Radius equals space fraction numerator square root of 3 over denominator 4 end fraction space straight x space 316.5

fraction numerator 1.732 over denominator 4 end fraction space straight x space 316.5 space equals space 137.04
    Question 256
    CBSEENCH12010204

    Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro's number (At. Mass of Fe = 55.845 u)

    Solution

    a = 286.65 pm

    a= 286.65 x 10-10cm

    Density ( ) = 7.874 g cm-3
    At mass of Fe = 55.845 u

    Z = 2 (For body centred cubic unit cell)

    Avogadro number (N0) =?
    straight rho space equals space fraction numerator straight Z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript 0 end fraction

7.875 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space left parenthesis 55.845 straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent space cm right parenthesis space space straight x space left parenthesis 7.874 cm to the power of negative 3 end exponent right parenthesis space straight x space straight N subscript 0 end fraction

straight N subscript 0 space equals space fraction numerator 2 space straight x space left parenthesis 55.845 space straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent cm right parenthesis cubed space straight x space left parenthesis 7.874 space cm to the power of negative 3 end exponent right parenthesis end fraction space equals space 6.022 space straight x space 10 to the power of 23 space mol to the power of negative 1 end exponent

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