In an extrinsic semiconductor, the concentration of charge carriers (either electrons or holes) is influenced by the type of doping (either p-type or n-type) and the temperature. Let's consider the scenarios for nh (number of holes) and ne (number of conduction electrons):
(a) nh > ne:
This condition would imply that there are more holes (positively charged carriers) than conduction electrons (negatively charged carriers) in the semiconductor. This could occur in a p-type semiconductor, where the majority carriers are holes, and the number of holes exceeds the number of electrons.
(b) nh = ne:
This condition would mean that the number of holes is equal to the number of conduction electrons. This balance is generally not the case in most extrinsic semiconductors because their doping levels usually result in an imbalance of either electrons or holes.
(c) nh < ne:
This condition suggests that there are more conduction electrons (negatively charged carriers) than holes (positively charged carriers) in the semiconductor. This could occur in an n-type semiconductor, where the majority carriers are electrons, and the number of electrons exceeds the number of holes.
(d) nh ≠ ne:
This condition indicates that the number of holes is not equal to the number of conduction electrons. In most cases of extrinsic semiconductors, this condition holds true due to the imbalance of charge carriers introduced by doping.
The correct answer depends on the type of extrinsic semiconductor and the specific doping used. In a general case, **(d) nh ≠ ne** would be the most accurate statement because the number of holes and conduction electrons is typically not equal in extrinsic semiconductors.
In an extrinsic semiconductor, the concentration of either holes or conduction electrons can be significantly larger than the other, depending on whether the semiconductor is p-type or n-type.
For p-type semiconductors, the concentration of holes (nh) is typically dominant, so nh > ne. So, option (a) nh > ne is correct for p-type semiconductors.
For n-type semiconductors, on the other hand, the concentration of conduction electrons (ne) is typically dominant, so ne > nh. So, option (c) nh < ne is correct for n-type semiconductors.
Therefore, the correct answer is (d) nh ≠ ne to include both possibilities for p-type and n-type semiconductors.
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Answer:
In an extrinsic semiconductor, the concentration of charge carriers (either electrons or holes) is influenced by the type of doping (either p-type or n-type) and the temperature. Let's consider the scenarios for nh (number of holes) and ne (number of conduction electrons):
(a) nh > ne:
This condition would imply that there are more holes (positively charged carriers) than conduction electrons (negatively charged carriers) in the semiconductor. This could occur in a p-type semiconductor, where the majority carriers are holes, and the number of holes exceeds the number of electrons.
(b) nh = ne:
This condition would mean that the number of holes is equal to the number of conduction electrons. This balance is generally not the case in most extrinsic semiconductors because their doping levels usually result in an imbalance of either electrons or holes.
(c) nh < ne:
This condition suggests that there are more conduction electrons (negatively charged carriers) than holes (positively charged carriers) in the semiconductor. This could occur in an n-type semiconductor, where the majority carriers are electrons, and the number of electrons exceeds the number of holes.
(d) nh ≠ ne:
This condition indicates that the number of holes is not equal to the number of conduction electrons. In most cases of extrinsic semiconductors, this condition holds true due to the imbalance of charge carriers introduced by doping.
The correct answer depends on the type of extrinsic semiconductor and the specific doping used. In a general case, **(d) nh ≠ ne** would be the most accurate statement because the number of holes and conduction electrons is typically not equal in extrinsic semiconductors.
Explanation:
Answer:
In an extrinsic semiconductor, the concentration of either holes or conduction electrons can be significantly larger than the other, depending on whether the semiconductor is p-type or n-type.
For p-type semiconductors, the concentration of holes (nh) is typically dominant, so nh > ne. So, option (a) nh > ne is correct for p-type semiconductors.
For n-type semiconductors, on the other hand, the concentration of conduction electrons (ne) is typically dominant, so ne > nh. So, option (c) nh < ne is correct for n-type semiconductors.
Therefore, the correct answer is (d) nh ≠ ne to include both possibilities for p-type and n-type semiconductors.