Problem 55
Question
The energy gap, \(\Delta E\), for silicon is \(110 \mathrm{kJ} / \mathrm{mol}\). What is the minimum wavelength of light that can promote an electron from the valence band to the conduction band in silicon? In what region of the electromagnetic spectrum is this light?
Step-by-Step Solution
Verified Answer
The minimum wavelength of light that can promote an electron from the valence band to the conduction band in silicon is found by substituting given values into Planck's equation. The region of the electromagnetic spectrum that this light falls in can be determined by comparing the obtained wavelength with known ranges for different regions of the electromagnetic spectrum.
1Step 1: Convert energy to appropriate units
First, let's convert the energy gap from kilojoules per mole to joules per mole by multiplying by \(1000\), so \(\Delta E = 110000 \, \mathrm{J} / \mathrm{mol}\). Next, since we know the energy is for one mole of photons, we can find the energy for a single photon by dividing by Avogadro's number \((6.022 \times 10^{23} \, \mathrm{mol}^{-1})\). Therefore, we get \(E_{\mathrm{photon}} = (110000 \, \mathrm{J} / \mathrm{mol}) / ( 6.022 \times 10^{23} \, \mathrm{mol}^{-1})\).
2Step 2: Use Planck's equation to determine wavelength
Next use Planck's equation, \(E=hc/\lambda\), rearranged to solve for wavelength: \(\lambda = hc/E\). Here, \(h = 6.626 \times 10^{-34} \, \mathrm{J} \cdot \mathrm{s}\) is Planck's constant and \(c = 2.998 \times 10^{8} \, \mathrm{m/s}\) is the speed of light. Substitute the values for \(h\), \(c\) and \(E_{\mathrm{photon}}\) to get the value for \(\lambda\) in meters.
3Step 3: Determine the region of the electromagnetic spectrum
The obtained wavelength will indicate the region of the electromagnetic spectrum this light falls in. For instance, if it falls between \(400-700 \, \mathrm{nanometers}\), it is in the visible light range.
Key Concepts
Energy Gap in SiliconPhoton Wavelength CalculationElectromagnetic Spectrum Regions
Energy Gap in Silicon
The energy gap, often symbolized as \(\Delta E\), in semiconductors like silicon is a crucial concept. It is the energy required to move an electron from the valence band, where it is bound to an atom, to the conduction band, where it can move freely and conduct electricity. In silicon, this energy gap is given as \(110,000 \, \text{J/mol}\).
This particular value happens to be a characteristic property of silicon, influencing its electrical behavior and application in devices such as diodes and transistors. Understanding this energy gap helps in designing materials suitable for electronics based on the energy required to excite electrons.
To find how much energy is needed for one electron, we convert this molar energy (for Avogadro's number of electrons) into energy per electron, using Avogadro's constant, \(6.022 \times 10^{23} \, \text{mol}^{-1}\). This conversion makes it applicable at the atomic scale, necessary for calculating photon-related transitions.
This particular value happens to be a characteristic property of silicon, influencing its electrical behavior and application in devices such as diodes and transistors. Understanding this energy gap helps in designing materials suitable for electronics based on the energy required to excite electrons.
To find how much energy is needed for one electron, we convert this molar energy (for Avogadro's number of electrons) into energy per electron, using Avogadro's constant, \(6.022 \times 10^{23} \, \text{mol}^{-1}\). This conversion makes it applicable at the atomic scale, necessary for calculating photon-related transitions.
Photon Wavelength Calculation
To determine the wavelength of light necessary to promote an electron across the energy gap, we use the relationship between energy and wavelength given by Planck's equation: \(E = \frac{hc}{\lambda}\).
In this equation, \(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\) is Planck's constant and \(c = 2.998 \times 10^{8} \, \text{m/s}\) is the speed of light in a vacuum. The energy \(E\) in this context is the energy of a single photon, calculated from the given energy gap for silicon.
By rearranging the formula to solve for wavelength \(\lambda\), we get \(\lambda = \frac{hc}{E_{\text{photon}}}\).
Substituting the known constants and the calculated photon energy into this equation gives the wavelength of light required to excite an electron from the valence band to the conduction band in silicon. This wavelength is key to identifying the type of light used in practical applications.
In this equation, \(h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\) is Planck's constant and \(c = 2.998 \times 10^{8} \, \text{m/s}\) is the speed of light in a vacuum. The energy \(E\) in this context is the energy of a single photon, calculated from the given energy gap for silicon.
By rearranging the formula to solve for wavelength \(\lambda\), we get \(\lambda = \frac{hc}{E_{\text{photon}}}\).
Substituting the known constants and the calculated photon energy into this equation gives the wavelength of light required to excite an electron from the valence band to the conduction band in silicon. This wavelength is key to identifying the type of light used in practical applications.
Electromagnetic Spectrum Regions
The electromagnetic spectrum spans a vast range of wavelengths, from very short gamma rays to long radio waves. Each type of radiation within the spectrum is categorized by its wavelength range.
Upon calculating the wavelength necessary to excite silicon's energy gap, we can determine which part of the electromagnetic spectrum it falls into. For example:
Upon calculating the wavelength necessary to excite silicon's energy gap, we can determine which part of the electromagnetic spectrum it falls into. For example:
- If the calculated wavelength is between 400 to 700 nanometers, the light is visible to the human eye, generally perceived as visible light.
- Wavelengths shorter than this range fall into ultraviolet or even X-ray regions.
- Longer wavelengths transition into the infrared region.
Other exercises in this chapter
Problem 52
Which of the following substances, when added in trace amounts to germanium, would produce an n-type semiconductor: (a) sulfur, (b) aluminum, (c) tin, (d) cadmi
View solution Problem 54
Explain why the electrical conductivity of a semiconductor is significantly increased if trace amounts of either donor or acceptor atoms are present, but is unc
View solution Problem 57
The Lewis structure of \(\mathrm{N}_{2}\) indicates that the nitrogento-nitrogen bond is a triple covalent bond. Other evidence suggests that the \(\sigma\) bon
View solution Problem 58
Show that both the valence-bond method and molecular orbital theory provide an explanation for the existence of the covalent molecule \(\mathrm{Na}_{2}\) in the
View solution