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Explain the crystalline structure by writing the electronic configuration of elementary semiconductors. |
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Answer» Solution :Elementary semiconductors are Si and Ge. Atomic number of Si is Z = 14. Its ELECTRONIC configuration is `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(2)`. SHELL K and L are filled completely by `1s^(2), 2s^(2)2p^(6)` and the shell for n = 3 is incomplete. It contains `3s^(2)3p^(2)` four valence electron. Atomic number of Ge is Z = 32. Its electronic configuration is `1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(6)3d^(10), 4s^(2)4p^(2)` shell K and L and are filled completely by `1s^(2), 2p^(2)2s^(2), 2p^(6), 3s^(2)3p^(6)3d^(10)` but for n = 4 shell N is incomplete there are `4s^(2)4p^(2)` valence electrons in it. Hence Si and Ge are tetravalent elements (with four valency) In its crystalline structure , every Si or Ge atom tends to share one of its four valence electrons with each of its four nearest neighbour atoms and also to take share of one electron from each such neighbour. These shared electron pairs forming a covalent BOND or a valence bond. Each bond contains two-two electrons.  Figure depicts one atom of Ge or Si covalent bonding with the four atoms in its neighbour atom in that solid dot are valence electrons. Here figure is in two-dimensional and +4 symbol INDICATES inner CORES of Si or Ge. Each of the bonded electrons bonds tightly to the atom they are attach to. Three-dimensional diamond like crystal structure for carbon, silicon and germanium with respective lattice spacing is shown in the figure.  Here not a single bond is broken so it shows the ideal situation which is at low temperature (means at absolute zero temperature). |
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