Question 3.2: Determine the width (in eV) of a forbidden energy band. Dete...

Combining Equations (3.29) and (3.30), we have

\begin{array}{c} f(\alpha a)=P^{\prime} \frac{\sin \alpha a}{\alpha a}+\cos \alpha a \\ \end{array}     (3.29)

\begin{array}{c}f(\alpha a)=\cos k a \\ \end{array}     (3.30)

\begin{array}{c}\cos k a=P^{\prime} \frac{\sin \alpha a}{\alpha a}+\cos \alpha a \end{array}

At k a=\pi and using P^{\prime}=8, we have

-1=8 \frac{\sin \alpha a}{\alpha a}+\cos \alpha a

We need to find the smallest values of \alpha a that satisfy this equation and then relate \alpha to the energy E to find the bandgap energy. From Figure 3.8, we see that, at one value of k a=\pi, we have \alpha a=\pi \equiv \alpha_{1} a. Then

\alpha_{1} a=\sqrt{\frac{2 m E_{1}}{\hbar^{2}}} \cdot a=\pi

or

E_{1}=\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}=\frac{\pi^{2}\left(1.054 \times 10^{-34}\right)^{2}}{2\left(9.11 \times 10^{-31}\right)\left(4.5 \times 10^{-10}\right)^{2}}=2.972 \times 10^{-19} \mathrm{~J}

From Figure 3.8, we see that, at the other value of k a=\pi, \alpha a is in the range \pi<\alpha a<2 \pi. By trial and error, we find \alpha a=5.141 \equiv \alpha_{2} a. Then

\alpha_{2} a=\sqrt{\frac{2 m E_{2}}{\hbar^{2}}} \cdot a=5.141

or

E_{2}=\frac{(5.141)^{2} \hbar^{2}}{2 m a^{2}}=\frac{(5.141)^{2}\left(1.054 \times 10^{-34}\right)^{2}}{2\left(9.11 \times 10^{-31}\right)\left(4.5 \times 10^{-10}\right)^{2}}=7.958 \times 10^{-19} \mathrm{~J}

The bandgap energy is then

E_{g}=E_{2}-E_{1}=7.958 \times 10^{-19}-2.972 \times 10^{-19}=4.986 \times 10^{-19} \mathrm{~J}

or

E_{g}=\frac{4.986 \times 10^{-19}}{1.6 \times 10^{-19}}=3.12 \mathrm{eV}

Comment

The results of this example give an order of magnitude of forbidden energy band widths.