The Fermi level EF in an intrinsic semiconductor, if effective masses of holes and electrons are same, is: 

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  1. EC + EV
  2. \(\rm \frac{E_C+E_V}{2}\)
  3. \(\rm \frac{E_C-E_V}{2}\)
  4. EC - EV

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{E_C+E_V}{2}\)
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Detailed Solution

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Concept:

Fermi Level:

  • The Fermi level in an intrinsic semiconductor is nearly midway between the conductive and valence band.
  • Fermi level is the highest energy state occupied by electrons in a material at absolute zero temperature.

 

F1 S.B Madhu 24.04.20 D1

  • For a p-type semiconductor, there are more holes in the valence band than there are electrons in the conduction band i.e. n < p.
  • This implies that the probability of finding an electron near the conduction band edge is smaller than the probability of finding a hole at the valence band edge.
  • Therefore, the Fermi level is closer to the valence band in a p-type semiconductor.


Explanation:

  • The Fermi level for a p-type lie near the valence band.
  • As the temperature increases above zero degrees, the extrinsic carriers in the conduction band and the valence band increases.
  • Since the intrinsic concentration also depends on temperature, ni also increases. But for small values of temperature, the extrinsic concentration dominates in comparison to the intrinsic concentration.
  • As the temperature continues to increase, the semiconductor starts to lose its extrinsic property and becomes intrinsic, as ni becomes comparable to the extrinsic concentration.

 

As the temperature increases the p type semiconductor become a simple semiconductor.

Hence, the Fermi level comes back to midway in forbidden energy gap.

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