Read the following passage and answer the questions.
A thin wafer of p-type silicon is made by adding boron, a trivalent element, to pure silicon. About one boron atom is added for every million silicon atoms. Pure silicon has about \(5\times10^{28}\) atoms per cubic metre. Its intrinsic carrier concentration \(n_i\) at room temperature is about \(1.5\times10^{16}\ \mathrm{m^{-3}}\). Next, a larger and carefully measured amount of phosphorus, a pentavalent element, is added to one part of the wafer. This part becomes n-type. The wafer now has a p-region and an n-region with a junction between them. As soon as the junction forms, charge carriers start to cross it. After a short time, the net flow of carriers stops and a potential difference appears across the junction. Finally, metal contacts are fixed at the two ends, and the device is joined to a dc supply.
Consider the following statements about the doped wafer before any carriers cross the junction:
(A) Each boron atom forms covalent bonds with only three neighbouring silicon atoms, leaving a vacancy in the bond with the fourth.
(B) In the n-region, the holes are fewer than \(n_i\) because many of them recombine with the extra electrons.
(C) The phosphorus atoms must be much smaller than the silicon atoms so that they fit into the lattice.
(D) Each doped region, taken as a whole, is electrically neutral.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Doping replaces a very small number of silicon atoms with dopant atoms. It changes the number of each kind of carrier, but the crystal gains no net charge.
(A) Each boron atom forms covalent bonds with only three neighbouring silicon atoms, leaving a vacancy in the bond with the fourth. — True. Boron has one valence electron less than silicon. It can bond with three neighbours only, so the bond with the fourth neighbour has a hole.
(B) In the n-region, the holes are fewer than \( n_i \) because many of them recombine with the extra electrons. — True. Phosphorus adds a large number of electrons. A hole now has a much greater chance of meeting an electron, so the number of holes falls below its intrinsic value.
(C) The phosphorus atoms must be much smaller than the silicon atoms so that they fit into the lattice. — False. A dopant must not distort the silicon lattice. For this, its atoms must be nearly the same size as the silicon atoms.
(D) Each doped region, taken as a whole, is electrically neutral. — True. The charge of the extra carriers is equal and opposite to the charge of the ionised dopant cores. So each region has no net charge.
NCERT (2026-27):
p-type semiconductor – “… The dopant has one valence electron less than Si or Ge and, therefore, this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to offer to the fourth Si atom.”
Extrinsic Semiconductor – “… But the rate of recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get reduced further.”
Extrinsic Semiconductor – “… The dopant has to be such that it does not distort the original pure semiconductor lattice. … A necessary condition to attain this is that the sizes of the dopant and the semiconductor atoms should be nearly the same.”
p-type semiconductor – “Note that the crystal maintains an overall charge neutrality as the charge of additional charge carriers is just equal and opposite to that of the ionised cores in the lattice.”