In the solution, the term $k$ represents the Boltzmann constant. Its value is:

$k = 1.38 \times 10^{-23} \, \text{J/K}$

The calculation of the energy term $kT$ involves multiplying the Boltzmann constant $k$ by the given temperature $T$ in kelvins (300 K).

Thus:

$kT = (1.38 \times 10^{-23}) \cdot (300) = 4.14 \times 10^{-21} \, \text{J}$

However, to express $kT$ in terms of electron volts (eV), you divide the value by the conversion factor $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$:

$kT = \frac{4.14 \times 10^{-21}}{1.6 \times 10^{-19}} = 0.0259 \, \text{eV}.$

This is the value of $kT$ used in the solution for the integral calculation.

Let me know if you'd like more details on any step!

Related Questions:

1. Why is the value of $kT$ commonly expressed in eV in semiconductor physics?
2. How does the temperature $T$ affect the number of quantum states calculated?
3. What role does the Boltzmann constant play in the derivation of state densities?
4. Can this method be generalized to other temperatures, and how would the result change?
5. Why is $h$, Planck's constant, critical in this computation?

Tip: When dealing with quantum states and densities, always double-check the units to ensure consistency (e.g., J vs. eV).