Question

Semi-conductor has phosphorus as impurity then it will have (A) \(\mathrm{n}_{\mathrm{e}}>\mathrm{n}_{\mathrm{h}}\) (B) \(\mathrm{n}_{\mathrm{e}}<<\mathrm{n}_{\mathrm{h}}\) (C) \(\mathrm{n}_{\mathrm{e}}=\mathrm{n}_{\mathrm{h}}\) (D) \(n_{e}=n_{h}=n_{i}\)

Short answer

In an n-type semiconductor, doped with phosphorus as an impurity, the number of free electrons will be greater than the number of holes. So, the relationship between electron concentration (\(n_e\)) and hole concentration (\(n_h\)) is given by \(n_e > n_h\). Therefore, the correct answer is (A).

Step-by-step solution

Understand n-type semiconductors

When a semiconductor is doped with phosphorus, it becomes an n-type semiconductor. This is because phosphorus has 5 valence electrons, which means it provides an extra free electron to the semiconductor. These extra electrons become mobile and contribute to the conductivity of the material. As a result, the electron concentration in the semiconductor (\(n_e\)) will increase while the hole concentration (\(n_h\)) will decrease.

Write the relation for an n-type semiconductor

In n-type semiconductors, the number of free electrons is greater than the number of holes. Therefore, we have the following relationship: \(n_e > n_h\)

Choose the correct answer

Based on the relationship we derived above, we can now choose the correct answer from the given options: (A) \(n_{\mathrm{e}} > n_{\mathrm{h}}\) So, the correct answer is (A).