Question

In an intrinsic semiconductor, the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K?

Answer 1

The ratio between the conductivity of an intrinsic semiconductor at 600 K and 300 K can be calculated using the Arrhenius equation. The hole mobility can be ignored due to its much smaller value and because it's independent of temperature, simplifying the calculation to a ratio of the change in number of carriers with temperature.

Step-by-step explanation:

The question is concerned with the conductivity of an intrinsic semiconductor at different temperatures. The conductivity (σ) of a semiconductor can be expressed using the formula:

• σ = nq(μe + μh)

where μe is the electron mobility, μh is the hole mobility, q is the charge of an electron, and n is the number of charge carriers per unit volume. For intrinsic semiconductors, the number of electrons is equal to the number of holes, and both increase with temperature according to the Arrhenius equation:

• n(T) ≈ T3/2e(-Eg/(2kBT))

Here, Eg is the energy gap (1.2 eV in this case), kB is the Boltzmann constant, and T is the temperature in Kelvin. Since the hole mobility (μh) is much smaller than the electron mobility (μe) and independent of temperature, it can be ignored in the conductivity ratio calculation:

σ(600K)/σ(300K) ≈ (n(600K)/n(300K))

We can then compute the ratio between the number of carriers at 600 K and 300 K using the given values. The final result shows the ratio of conductivity at these two temperatures.