The concentration Pn of holes in n-type semiconductor is: 

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  1. \(\rm P_n = \frac{n_i}{N_D}\)
  2. \(\rm P_n = \frac{n_i^2}{N_A}\)
  3. \(\rm P_n = \frac{n_i^2}{N_D}\)
  4. \(\rm P_n = \frac{n_i}{N_A}\)

Answer (Detailed Solution Below)

Option 3 : \(\rm P_n = \frac{n_i^2}{N_D}\)
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The concentration of minority carriers in an extrinsic semiconductor is given by the law of mass action, according to which:

n.p = ni2

n = concentration of electrons in the conduction band

p = concentration of holes in the valence band

ni = intrinsic carrier concentration

Cases:

In an n-type semiconductor, the minority hole concentration is given by:

\(p = \frac{{n_i^2}}{{{N_D}}}\)

ND = Concentration of Donor impurity

In a p-type semiconductor, the minority electron concentration is given by:

\(n = \frac{{n_i^2}}{{{N_A}}}\)

NA = Concentration of Acceptor impurity

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