The ionisation energy of isolated pentavalent phosphorous atom is very large. How is it possible that when it goes into silicon lattice position to ralease its fifth electron at room temperature so that n-type semiconductor is obtained?
The ionisation energy of isolated pentavalent phosphorous atom is very large. How is it possible that when it goes into silicon lattice position to ralease its fifth electron at room temperature so that n-type semiconductor is obtained?
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To understand how phosphorus can release its fifth electron in a silicon lattice to form an n-type semiconductor despite its high ionization energy, we can break down the explanation into several steps:
### Step 1: Understanding Phosphorus and Silicon
- Phosphorus (P) is a pentavalent element, meaning it has five valence electrons. Silicon (Si) is tetravalent with four valence electrons.
- When phosphorus is introduced into the silicon lattice, it replaces a silicon atom.
**Hint:** Remember that the valency of an element indicates how many electrons it can share or bond with other atoms.
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Read the following text and answer the following questions on the basis of the same: Band theory of solid: Consider that the Si or Ge crystal contains N atoms. Electrons of each atom will have discrete energies in different orbits. The electron energy will be same if all the atoms are isolated, i.e., separated from each other by a large distance. However, in a crystal, the atoms are close to each other (2 Å to 3 Å) and therefore the electrons interact with each other and also with the neighbouring atomic cores. The overlap (or interaction) will be more felt by the electrons in the outermost orbit while the inner orbit or core electron energies may remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we need to consider the changes in the energies of the electrons in the outermost orbit only. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available 6N p-states. Obviously, some p-electron states are empty. This is the case of well separated or isolated atoms. The maximum possible electrons in an orbit is:
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