Question

Carry out the integration indicated in equation (10.10)

Short answer

The integration is obtained and proved.

Step-by-step solution

Determine the formulas

Consider the formulato determine the thermal energy as:

$\mathbf{E}\mathbf{=}{\mathbf{k}}_n\mathbf{T}$ ….. (1)

Here,$k_n$ is Boltzmann constant and the temperature is T.

Solve for the integration as:

Solve for the integration as:

$N_{excited}={\int }_{E_F+\frac{1}{2}{E}_{gap}}^{\infty }\frac{1}{e^{(E-E_F)/k_{n}T}+1}DdE$

Consider$x=\frac{E}{k_{n}T}$ then,$dx=\frac{dE}{k_{n}T}$ .

Consider $x_F=\frac{E_F}{k_{n}T}$and$x_{gap}=\frac{E_{gap}}{k_{n}T}$

Substitute the values in the integration and solve as:

$\begin{array}{c}{N}_{excited}=D{k}_{B}T{\int }_{x_F+\frac{1}{2}{x}_{gap}}^{\infty }\frac{1}{e^{x_F-x_{gap}}+1}dx\\ =D{k}_{B}T{\int }_{x_F+\frac{1}{2}{x}_{gap}}^{\infty }\frac{e^{x_F-x_{gap}}}{e^{x_F-x_{gap}}+1}dx\end{array}$

Solve for the derivative of the function as:

$\frac{d\ln (1+e^{x_f-x_{gap}})}{dx}=\frac{1}{1+e^{x_f-x}}{e}^{x_p-x}(-1)$

Solve the integration further as:

$\begin{array}{c}{N}_{excited}=-D{k}_{B}T\ln {(1+e^{x_F-x})}_{x_F+\frac{1}{2}gap}^{\infty }\\ =D{k}_{B}T\ln (1+e^{\frac{E_{gap}}{2k_{B}T}})\end{array}$

Hence, the integration is obtained and proved.