Chapter 10: Q50E (page 469)

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

Short Answer

Expert verified

The estimated value of the band width is $50 eV$ .

Step by step solution

Determine the formulas

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

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

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:

$\begin{matrix} Δ E = E_{2} - E_{1} \\ = \frac{h^{2}}{2 m} (k_{2}^{2} - k_{1}^{2}) \end{matrix}$

Substitute the values and solve as:

$\begin{matrix} Δ E = \frac{h^{2}}{2 m} ((\frac{2 π}{a^{2}}) - (\frac{π}{a^{2}})) \\ = \frac{h^{2}}{2 m} (\frac{3 π}{a^{2}}) \\ = \frac{3 h^{2} π^{2}}{2 m a^{2}} \end{matrix}$

Solve further as:

$\begin{matrix} Δ E = \frac{3 (1.055 \times 10^{- 34} Js) (π^{2})}{2 (9.11 \times 10^{- 31} kg) {(0.15 \times 10^{- 9} m)}^{2}} \\ = 50 eV \end{matrix}$

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

Short Answer

Step by step solution

Determine the formulas

Determine the value of the width as:

One App. One Place for Learning.

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Assuming an interatomic spacing of 0.15 nm, obtain a rough value for the width (in eV) of then=2 band in a one-dimensional crystal.

Short Answer

Step by step solution

Determine the formulas

Determine the value of the width as:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fields in Physics

Torque and Rotational Motion

Waves Physics

Solid State Physics

Fluids

Medical Physics

Study anywhere. Anytime. Across all devices.

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