Question

For an intrinsic semiconductor having band gap 0.7 eV. Calculate the density of holes and electrons at 27 degrees Celsius​

Answer 1

Answer:

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Answer 2

Given:

Band gap = 0.7 eV

Temperature = 27 °C = 300K

To Find:

The density of holes and electrons

Solution:

Intrinsic carrier concentration = ni

= 2 [ 2πkT/h²]³/² (me X mh) ³/⁴ exp [ -Eg / 2 k T]

Here, h = Planck's constant = 6.626 X 10⁻³⁴

me and mh are masses of electrons and holes respectively

k = Boltzmann Constant = 1.38 X 10⁻²³

Eg = Energy gap in Volt (0.7eV = 0.7 X 1.6 X 10⁻¹⁹ V)

We know that at 300K, 2 [ 2πkT/h²]³/² = 2.88 X 10⁷⁰

(me X mh) ³/⁴ = 6.81 X 10 ⁻⁴⁷ (approx)

Now, exp [ -Eg / 2 k T] = exp [ - 0.7 X 1.6 X 10⁻¹⁹ / 2 X 300 X 1.38 X 10⁻²³ ]

= exp (- 13.51)

= 1.357 X 10⁻⁶

So, ni = 2.88 X 10⁷⁰ X  6.81 X 10 ⁻⁴⁷  X 1.357 X 10⁻⁶

= 1.32 X 10¹⁸ m³.

At room temperature density of holes (nh) and density of electrons (ne) are equal and related to ni as:

ni² = ne X nh

Hence, the density of holes and electrons is 1.32 X 10¹⁸ m³ each.