Question

(a) Show that the slope dP/dEof Eq. 41-6 evaluated at$\mathit{E}\mathbf{=}{\mathit{E}}_{\mathbf{F}}$is $\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{4}\mathit{k}\mathit{T}$. (b) Show that the tangent line to the curve of Fig. 41-7bevaluated at$\mathit{E}\mathbf{=}{\mathit{E}}_{\mathbf{F}}$intercepts the horizontal axis at$\mathit{E}\mathbf{=}{\mathit{E}}_{\mathbf{F}}\mathbf{+}\mathbf{2}\mathit{k}\mathit{T}$.

Short answer

1. The slope at is $dP/dE\mathrm{at}E=E_F\mathrm{is}-1/4kT$.
2. The tangent line to the curve evaluated at intercepts $E=E_F$the horizontal axis at $E=E_F+2kT$.

Step-by-step solution

The given data

• The given equations are Eq. 41-6 and Eq. 41-7b.
• The given condition is $E=E_F$.

Understanding the concept of probability

Using the probability equation from the Fermi-Dirac statistics, we can get the differential value of the occupied probability. Now, using the equation of the straight line, we can get the intercept value by considering the graph values.

Formula:

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics, $P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)}+1}$ (i)

a) Calculation of the slope value dP/dE

Differentiating the probability equation (i) with respect to energy, we get the following equation as follows:

$$\begin{gathered} \frac{d\left(P\right)}{dE}=\frac{1}{{\left(e^{\left(E-E_F\right)/kT}+1\right)}^2}\frac{d\left(e^{\left(E-E_F\right)/kT}\right)}{dE} \\ =\frac{-1}{{\left(e^{\left(E-E_F\right)/kT}+1\right)}^2}\frac{1}{kT}{e}^{\left(E-E_F\right)/kT} \end{gathered}$$

Now, the above equation at$E=E_F$ gives the required slope value as follows:

${\overline{)\frac{d\left(P\right)}{dE}}}_{E=E_F}=\frac{-1}{{\left(e^{\left(E-E_F\right)/kT}+1\right)}^2}\frac{1}{kT}{e}^{\left(E_F-E_F\right)/kT}$

Hence, it is proved that the slope value at $E=E_F$is $\frac{-1}{4kT}$.

b) Calculation of the tangent line to the curve

The equation of a line can be written as follows: $y=m\left(x-x_0\right)$

From the above calculations, the value of the slope is found to be $m=\frac{1}{4kT}$at $E=E_F$.

Thus, the intercept value at$E=E_F$ is $P\left(E\right)\left(ory\right)=\frac{1}{2}$.

Thus, substituting these values in the equation (a), we get the equation of the tangent line to the curve as follows:

$$\begin{gathered} 1/2=\left(-1/4kT\right)\left(E_F-x_0\right) \\ {x}_0=E_F+2kT \end{gathered}$$

Hence, the tangent line to the curve evaluated at intercepts$E=E_F$ the horizontal axis at $E=E_F+2kT$.