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Consider a n-type semiconductor in which $n_e$ and $n_h$ are number of electrons and holes, respectively.
(A) Holes are minority carriers
(B) The dopant is a pentavalent atom
(C) $n_e n_h \neq n_i^2$
(where $n_i$ is number of electrons or holes in semiconductor when it is intrinsic form)
(D) $n_e n_h = n_i^2$
(E) The holes are not generated due to the donorsChoose the correct answer from the options given below :

The correct answer is
(A), (B), (E) only

Semiconductor Statements Analysis

  • (A) Holes are minority carriers: In an n-type semiconductor, electrons are the majority charge carriers due to doping. Therefore, holes, being fewer in number, are the minority carriers. This statement is correct.
  • (B) The dopant is a pentavalent atom: To create an n-type semiconductor, the semiconductor material is doped with impurity atoms that have five valence electrons (pentavalent atoms). These atoms, called donors, contribute extra electrons to the conduction band. This statement is correct.
  • (C) $n_e n_h \neq n_i^2$: For any semiconductor material in thermal equilibrium, the product of the electron concentration ($n_e$) and hole concentration ($n_h$) is equal to the square of the intrinsic carrier concentration ($n_i^2$). This relationship is known as the mass action law. Therefore, this statement is incorrect.
  • (D) $n_e n_h = n_i^2$: This statement represents the mass action law, which holds true for semiconductors in thermal equilibrium. The product of electron and hole concentrations equals the square of the intrinsic carrier concentration. This statement is correct.
  • (E) The holes are not generated due to the donors: Donor atoms in n-type semiconductors contribute electrons to the conduction band. They do not create or generate holes. Holes are primarily generated through thermal excitation across the band gap, and their concentration is reduced in n-type materials compared to intrinsic ones. This statement is correct.

Conclusion

Based on the analysis, the correct statements regarding the n-type semiconductor are (A), (B), and (E). These correspond to Option 1.

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