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The gap between valence and conduction bands in diamond is 5.47 eV. (a) What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie? (b) Explain why pure diamond is transparent and colorless. (c) Most gem diamonds have a yellow color. Explain how impurities in the diamond can cause this color.

Short Answer

Expert verified
(a) Maximum wavelength: 227 nm, in the UV region. (b) Transparent due to high band gap. (c) Impurities introduce colors.

Step by step solution

01

Understand the Energy-Wavelength Relationship

The energy (E) of a photon is related to its wavelength (\( \lambda \)) by the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light. To find the maximum wavelength that can excite the electron, we need to isolate \( \lambda \).
02

Rearrange for Wavelength

To find the wavelength, we rearrange the formula: \( \lambda = \frac{hc}{E} \). This allows us to calculate the wavelength once we know the energy gap, which is given as 5.47 eV.
03

Convert Energy from eV to Joules

We need to convert the energy from electron volts to joules. The energy gap is 5.47 eV and 1 eV is equal to \( 1.602 \times 10^{-19} \) J, making the energy: \[ E = 5.47 \times 1.602 \times 10^{-19} \text{ J} \approx 8.77 \times 10^{-19} \text{ J}. \]
04

Input Constants and Calculate Wavelength

Use Planck's constant \( h = 6.626 \times 10^{-34} \) J\(\cdot\)s and the speed of light \( c = 3.00 \times 10^8 \) m/s and substitute them into the equation: \[ \lambda = \frac{(6.626 \times 10^{-34} \cdot 3.00 \times 10^8)}{8.77 \times 10^{-19}} \text{ m}. \] Calculate \( \lambda \): \[ \lambda \approx 2.27 \times 10^{-7} \text{ m} \text{ or } 227 \text{ nm}. \]
05

Determine the Electromagnetic Spectrum Region

The calculated wavelength of 227 nm lies in the ultraviolet (UV) region of the electromagnetic spectrum, which is not visible to the human eye.
06

Explain Diamond's Transparency and Colorlessness

Pure diamond is transparent and colorless because its band gap (5.47 eV) is too high to absorb visible light, which is why it doesn't show color under normal circumstances.
07

Discuss the Effect of Impurities

Impurities, such as nitrogen, can introduce energy levels within the band gap, allowing absorption of certain wavelengths of visible light. This can cause color in diamonds, such as yellow, when blue light is absorbed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Band
The valence band in a semiconductor is where the electrons are when they are bound to atoms and do not have enough energy to be involved in conduction. It's essentially the outer shell of electrons that can participate in bonding.
In the context of energy bands, the electrons in the valence band are tightly bound to the atomic nuclei and have lower potential energy.
When energy is applied, these electrons can jump to the conduction band. The energy needed is equal to the band gap, which, for a diamond, is 5.47 electron volts (eV).
  • Electrons here remain stable and attached to their parent atom.
  • The valence band itself does not contribute to the flow of electric current.
Understanding the role of the valence band helps explain the transparency and insulating properties of materials like diamond.
Conduction Band
The conduction band is the range of electron energy, higher than that of the valence band, where electrons are free to move and contribute to conduction. Thus, when electrons gain enough energy, they can move to the conduction band and free themselves from their atomic bonds.
In materials like semiconductors, electrons in the conduction band have enough energy to participate in electrical conductivity.
The energy gap between the valence band and conduction band, known as the "band gap," determines the amount of energy needed for an electron to jump to the conduction band.
  • Electrons are free to conduct electricity once in this band.
  • The minimal energy required to excite an electron into this band is the band gap energy.
In diamonds, this band gap is relatively large, contributing to its transparency.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, each defined by different wavelengths and frequencies. It ranges from gamma rays with very short wavelengths to radio waves that can be kilometers long.
Visible light, the light we can see, is a tiny portion of this spectrum.
In the exercise, you encountered a photon with a wavelength of 227 nm, calculated to lie in the ultraviolet region of the spectrum, which is beyond the visible range for humans.
  • The spectrum includes visible light, microwaves, infrared, ultraviolet, X-rays, and gamma rays.
  • Each type of radiation in the spectrum has different applications, from communications to medical imaging.
Understanding the electromagnetic spectrum is crucial in explaining which energies are absorbed or transmitted by different materials.
Photon Wavelength
Photon wavelength is the distance between consecutive peaks of a light wave. This measurement determines the energy of the photon since energy is inversely proportional to wavelength.
In the context of this exercise, the energy of a photon is calculated using the formula: \[E = \frac{hc}{\lambda}\]where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
A smaller wavelength corresponds to higher energy photons, such as those found in the ultraviolet region.
  • Ultraviolet light has shorter wavelengths and higher energy than visible light.
  • By calculating the maximum photon wavelength that bridges the band gap, we can determine whether a material will absorb certain light fields.
Understanding photon wavelengths helps illustrate why certain materials appear the way they do or why certain photon energies result in specific reactions.
Impurities in Crystals
Impurities in crystals can drastically change their physical properties. These impurities are atoms or molecules that are different from the primary constituent material and get incorporated into the crystal structure.
In the case of diamonds, common impurities include nitrogen, which can absorb certain wavelengths of light and thus impart color, like a yellow hue.
These impurities create additional energy levels within the band gap, allowing electrons to jump at different energies that correspond to visible light frequencies.
  • Impurities can be introduced during the crystal's formation or through treatment.
  • They often create localized energy states within the band gap, influencing optical and electrical properties.
Understanding how impurities affect crystals supports explanations for variations in color and electrical conductivity, important in both gemology and electronics.

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Most popular questions from this chapter

To determine the equilibrium separation of the atoms in the HCl molecule, you measure the rotational spectrum of HCl. You find that the spectrum contains these wavelengths (among others): \(60.4 \mu m\), \(69.0 \mu m\), \(80.4 \mu m\), \(96.4 \mu m\), and \(120.4 \mu m\). (a) Use your measured wavelengths to find the moment of inertia of the HCl molecule about an axis through the center of mass and perpendicular to the line joining the two nuclei. (b) The value of \(l\) changes by \(\pm 1\) in rotational transitions. What value of \(l\) for the upper level of the transition gives rise to each of these wavelengths? (c) Use your result of part (a) to calculate the equilibrium separation of the atoms in the HCl molecule. The mass of a chlorine atom is \(5.81 \times 10^{-26}\) kg, and the mass of a hydrogen atom is \(1.67 \times 10^{-27}\) kg. (d) What is the longest-wavelength line in the rotational spectrum of HCl?

Consider a system of \(N\) free electrons within a volume \(V\). Even at absolute zero, such a system exerts a pressure \(p\) on its surroundings due to the motion of the electrons. To calculate this pressure, imagine that the volume increases by a small amount \(dV\). The electrons will do an amount of work \(p\) \(dV\) on their surroundings, which means that the total energy \(E_{tot}\) of the electrons will change by an amount \(dE_{tot} = -p dV\). Hence \(p = -dE_{tot}/dV\). (a) Show that the pressure of the electrons at absolute zero is \(p = \frac{3^{2/3}\pi^{4/3}\hbar^{2}}{5m} \lgroup \frac{N}{V}\ \rgroup^{5/3}\) (b) Evaluate this pressure for copper, which has a freeelectron concentration of \(8.45 \times 10^{28} m^{-3}\). Express your result in pascals and in atmospheres. (c) The pressure you found in part (b) is extremely high. Why, then, don't the electrons in a piece of copper simply explode out of the metal?

An Ionic Bond. (a) Calculate the electric potential energy for a K\(^+\) ion and a Br\(^-\) ion separated by a distance of 0.29 nm, the equilibrium separation in the KBr molecule. Treat the ions as point charges. (b) The ionization energy of the potassium atom is 4.3 eV. Atomic bromine has an electron affinity of 3.5 eV. Use these data and the results of part (a) to estimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your estimate? Explain your reasoning.

Two atoms of cesium (Cs) can form a Cs \(_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) molecule is \(0.447 \mathrm{nm} .\) Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is \(2.21 \times 10^{-25} \mathrm{~kg}\).

Our galaxy contains numerous \(molecular\) \(clouds\), regions many lightyears in extent in which the density is high enough and the temperature low enough for atoms to form into molecules. Most of the molecules are H\(_2\), but a small fraction of the molecules are carbon monoxide (CO). Such a molecular cloud in the constellation Orion is shown in Fig. P42.42. The upper image was made with an ordinary visiblelight telescope; the lower image shows the molecular cloud in Orion as imaged with a radio telescope tuned to a wavelength emitted by CO in a rotational transition. The different colors in the radio image indicate regions of the cloud that are moving either toward us (blue) or away from us (red) relative to the motion of the cloud as a whole, as determined by the Doppler shift of the radiation. (Since a molecular cloud has about 10,000 hydrogen molecules for each CO molecule, it might seem more reasonable to tune a radio telescope to emissions from H\(_2\) than to emissions from CO. Unfortunately, it turns out that the H\(_2\) molecules in molecular clouds do not radiate in either the radio or visible portions of the electromagnetic spectrum.) (a) Using the data in Example 42.2 (Section 42.2), calculate the energy and wavelength of the photon emitted by a CO molecule in an \(l\) \(=\) 1\(\rightarrow\) \(l\) \(=\) 0 rotational transition. (b) As a rule, molecules in a gas at temperature \(T\) will be found in a certain excited rotational energy level, provided the energy of that level is no higher than \(kT\) (see Problem 42.39). Use this rule to explain why astronomers can detect radiation from CO in molecular clouds even though the typical temperature of a molecular cloud is a very low 20 K.

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