Question

Assuming an interatomic spacing of 0.15 nm, obtain a rough value for the width (in eV) of the$\mathbf{n}\mathbf{=}\mathbf{2}$ band in a one-dimensional crystal.

Short answer

The estimated value of the band width is $50\text{\hspace{0.17em}eV}$.

Step-by-step solution

Determine the formulas

Consider the relation between the wave numbers and the energy as follows:

$\mathbf{E}\mathbf{=}\frac{{\mathbf{h}}^2{\mathbf{k}}^2}{2m}$

Here, E is the energy, m is the effective mass, h is the planck’s constant and k is the wave number.

Determine the value of the width as:

Consider the formula for the change in energy as:

$\begin{array}{c}\Delta E=E_2-E_1\\ =\frac{h^2}{2m}(k_2^2-k_1^2)\end{array}$

Substitute the values and solve as:

$\begin{array}{c}\Delta E=\frac{h^2}{2m}(\left(\frac{2\pi }{a^2}\right)-\left(\frac{\pi }{a^2}\right))\\ =\frac{h^2}{2m}\left(\frac{3\pi }{a^2}\right)\\ =\frac{3h^2{\pi }^2}{2m{a}^2}\end{array}$

Solve further as:

$\begin{array}{c}\Delta E=\frac{3(1.055\times {10}^{-34}\text{\hspace{0.17em}Js})\left({\pi }^2\right)}{2(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){(0.15\times {10}^{-9}\text{\hspace{0.17em}m})}^2}\\ =50\text{\hspace{0.17em}eV}\end{array}$