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(a) What maximum light wavelength will excite an electron in the valence band of diamond to the conduction band? The energy gap is \(5.50 \mathrm{eV}\). (b) In what part of the electromagnetic spectrum does this wavelength lie?

Short Answer

Expert verified
The maximum wavelength is 226 nm, which is in the ultraviolet region of the spectrum.

Step by step solution

01

Identify the Formula for Wavelength

The maximum wavelength \( \lambda \) that can excite an electron is given by the relation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy required to excite the electron (the energy gap), \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \), and \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s})\).
02

Convert Electron Volts to Joules

Convert the energy gap from electron volts to joules because the standard formula \( E = \frac{hc}{\lambda} \) requires energy in joules. Use the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). So, \( 5.50 \text{ eV} = 5.50 \times 1.602 \times 10^{-19} \text{ J} = 8.81 \times 10^{-19} \text{ J} \).
03

Substitute Values into the Wavelength Formula

Substitute the known values into the formula: \( \lambda = \frac{hc}{E} \). Thus, \( \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{8.81 \times 10^{-19}} \). Calculate the result to find \( \lambda \).
04

Calculate the Maximum Wavelength

Perform the calculation: \( \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{8.81 \times 10^{-19}} \approx 2.26 \times 10^{-7} \text{ m} \) or \( 226 \text{ nm} \).
05

Determine Electromagnetic Spectrum Region

Compare the calculated wavelength with standard electromagnetic spectrum regions to determine where it falls. A wavelength of \( 226 \text{ nm} \) falls within the ultraviolet (UV) range of the spectrum.

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Key Concepts

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

Wavelength
Understanding wavelength is crucial for analyzing how electrons get excited in materials. Wavelength is the distance between successive peaks of a wave. It is usually measured in meters or nanometers (nm). In the context of exciting electrons, the energy of the incident light is inversely related to its wavelength. This means that shorter wavelengths correspond to higher energy, while longer wavelengths correspond to lower energy.

When trying to excite an electron from the valence band to the conduction band, you need enough energy to bridge the energy gap. This energy can be provided by a photon, and the photon's energy is determined by its wavelength. The formula used to determine the wavelength needed to excite the electron is:

\[ E = \frac{hc}{\lambda} \]

Where:
  • \( E \) is the energy of the photon, also equivalent to the energy gap.
  • \( h \) is Planck's constant.
  • \( c \) is the speed of light.
  • \( \lambda \) is the wavelength.

To find maximum wavelength, rearrange to\( \lambda = \frac{hc}{E} \) and input values. Calculating this from our given example, where the energy gap is \( 5.50 \text{ eV} \), yields a wavelength of approximately \( 226 \text{ nm} \). This value indicates the longest wavelength capable of providing the needed energy to move that electron.
Planck's Constant
Planck's constant \( (h) \) is a pivotal constant in the field of quantum mechanics and plays a critical role in understanding energy quanta. It has a value of \( 6.626 \times 10^{-34} \text{ J s} \). Planck’s constant relates the energy contained in a photon to the frequency of its corresponding electromagnetic wave.

The relationship between a photon's energy and frequency is given by the formula:

\[ E = hf \]

But using the speed of light equation \( c = \lambda f \), we adapt this to find how energy relates to wavelength:

\[ E = \frac{hc}{\lambda} \]

This equation shows us that as wavelength decreases, the energy increases, and vice versa. Planck's constant is a tiny number, reflective of the extremely small energy levels dealt with in quantum mechanics, yet it is fundamental in calculations involving photons and electron excitation.

In the example of the diamond's energy gap, Planck’s constant is used to convert the energy to wavelengths so that we can locate them in the electromagnetic spectrum. In doing so, it highlights the quantum nature of how light interacts with matter.
Ultraviolet Spectrum
The ultraviolet (UV) spectrum is a part of the electromagnetic spectrum with wavelengths between approximately 10 nm and 400 nm. It lies beyond the visible range of light and is responsible for various phenomena, such as sunburns and the fluorescence in certain materials.

Ultraviolet light has more energy than visible light because it has shorter wavelengths. This makes UV light ideal for exciting electrons in materials and is why, for example, a wavelength of around \( 226 \text{ nm} \), like the one calculated in our example, falls squarely within this range.

In the context of the exercise, recognizing that a \( 226 \text{ nm} \) wavelength falls in the UV range helps us understand why certain materials, like diamond, have specific optical properties. This insight is useful in fields like electronics and materials science, where controlling electron excitation is crucial.
  • Shorter wavelengths mean higher frequencies, and thus, higher energies.
  • Applications of UV light include disinfection, studying chemical compounds, and in areas of research involving new materials.

Easily damaging to organic tissue, UV light demonstrates the powerful energy carried at these wavelengths, making it a significant component of the spectrum to understand.

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Most popular questions from this chapter

Copper, a monovalent metal, has molar mass \(63.54 \mathrm{~g} / \mathrm{mol}\) and density \(8.96 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the number density \(n\) of conduction electrons in copper?

A potassium chloride crystal has an energy band gap of \(7.6 \mathrm{eV}\) above the topmost occupied band, which is full. Is this crystal opaque or transparent to light of wavelength \(140 \mathrm{nm} ?\)

A state \(63 \mathrm{meV}\) above the Fermi level has a probability of occupancy of 0.090 . What is the probability of occupancy for a state \(63 \mathrm{meV}\) below the Fermi level?

A certain computer chip that is about the size of a postage stamp \((2.54 \mathrm{~cm} \times 2.22 \mathrm{~cm})\) contains about 3.5 million transistors. If the transistors are square, what must be their maximum dimension? (Note: Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than \(0.7 \mu \mathrm{m}\) are now commonly and inexpensively fabricated.)

In a simplified model of an undoped semiconductor, the actual distribution of energy states may be replaced by one in which there are \(N_{v}\) states in the valence band, all these states having the same energy \(E_{v},\) and \(N_{c}\) states in the conduction band, all these states having the same energy \(E_{c}\). The number of electrons in the conduction band equals the number of holes in the valence band. (a) Show that this last condition implies that $$\frac{N_{c}}{\exp \left(\Delta E_{c} / k T\right)+1}=\frac{N_{v}}{\exp \left(\Delta E_{v} / k T\right)+1}$$in which$$\Delta E_{c}=E_{c}-E_{\mathrm{F}} \quad \text { and } \quad \Delta E_{v}=-\left(E_{v}-E_{\mathrm{F}}\right)$$ (b) If the Fermi level is in the gap between the two bands and its distance from each band is large relative to \(k T,\) then the exponentials dominate in the denominators. Under these conditions, show that$$E_{\mathrm{F}}=\frac{\left(E_{c}+E_{v}\right)}{2}+\frac{k T \ln \left(N_{v} / N_{c}\right)}{2}$$and that, if \(N_{v} \approx N_{c}\), the Fermi level for the undoped semiconductor is close to the gap's center.

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