Physics 1 Marks Question

A p-type semiconductor is formed by doping a group 13 element with group 14 element (Si or Ge). As the group 13 element has only 3 electrons in its valence shell and the group 14 element has 4 electrons in its valence shell, when the group 13 element, say, Al, replaces one Si in the silicon crystal, only 3 covalent bonds are formed by it. And the fourth covalent bond is left in need of one electron. So, it creates a hole. Since the atom as a whole is electriclly neutral, the p-type semiconductor is also neutral.
In a p‒n junction, when the diffusion of holes takes place across the junction because of the difference in the concentration of charge carriers from p to n sides, these holes neutralise some of the electrons on the n side. So, the atom attached with that electron becomes one electron deficient and hence positively charged. This makes the n side of the p‒n junction positively charged and the p side of the p‒n junction negatively charged.

When the emitter of a transistor is forward biassed, the majority carriers move from the emitter to the collector through the base. As the base is thin and lightly doped, only small amount of a combination of electrons and holes takes place, leading to weak base current. This makes the collector current equal to the emitter current.
If we make any other choice for least concentration of impurity, we will have a low value of collector current. Thus, the purpose of a transistor will not be fulfilled.