Question:
For an n-type semiconductor, the relation between the number density '\(n_{h}\)' and '\(n_{e}\)' is
For an n-type semiconductor, the relation between the number density '\(n_{h}\)' and '\(n_{e}\)' is
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Electrons are the majority carriers in an n-type semiconductor.
Updated On: Nov 12, 2026
- \(n_{e}=n_{h}\)
- \(n_{h}\lt \lt n_{e}\)
- \(n_{h}>>n_{e}\)
- \(n_{h}=n_{e}=0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
An n-type semiconductor is made by doping with a pentavalent donor impurity. Each impurity atom gives one free electron.
Step 2: Carrier Densities
Electrons become the majority carriers. Holes are produced only thermally and are the minority carriers.
Step 3: Relation
So the hole density is far less than the electron density, \(n_h\ll n_e\).
Final Answer:
For an n-type semiconductor, \(n_h\ll n_e\). \[ \boxed{\text{(B) }n_h\ll n_e} \]
An n-type semiconductor is made by doping with a pentavalent donor impurity. Each impurity atom gives one free electron.
Step 2: Carrier Densities
Electrons become the majority carriers. Holes are produced only thermally and are the minority carriers.
Step 3: Relation
So the hole density is far less than the electron density, \(n_h\ll n_e\).
Final Answer:
For an n-type semiconductor, \(n_h\ll n_e\). \[ \boxed{\text{(B) }n_h\ll n_e} \]
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