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For a certain semiconductor, the longest wavelength radiation that can be absorbed is 2.06 mm. What is the energy gap in this semiconductor?

Short Answer

Expert verified
The energy gap is approximately 0.000602 eV.

Step by step solution

01

Understand the Concept

The energy gap in a semiconductor is related to the wavelength of absorbed radiation through the concept of photon energy. Photons of light have an energy \( E \) that can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
02

Gather Constants and Values

Planck's constant \( h = 6.626 \times 10^{-34} \) J·s and the speed of light \( c = 3.00 \times 10^8 \) m/s are constants. The wavelength given is \( 2.06 \) mm. First, convert this wavelength to meters: \( \lambda = 2.06 \times 10^{-3} \) m.
03

Set Up the Formula for Photon Energy

Use the formula \( E = \frac{hc}{\lambda} \) to find the energy of the photon that can be absorbed in the semiconductor, where \( E \) is the energy gap.
04

Perform the Calculation

Substitute the known values into the equation: \( E = \frac{6.626 \times 10^{-34} \, \text{J·s} \times 3.00 \times 10^8 \, \text{m/s}}{2.06 \times 10^{-3} \, \text{m}} \). Calculate \( E \).
05

Solve for Energy Gap

Simplify and solve: \[E = \frac{6.626 \times 3.00}{2.06} \times 10^{-23} \, \text{J} = 9.65 \times 10^{-23} \, \text{J}\]Convert this to electron volts (eV) using \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\): \[E = \frac{9.65 \times 10^{-23} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 0.000602 \, \text{eV}\]
06

Conclusion

Therefore, the energy gap in this semiconductor is approximately \(0.000602 \text{ eV}\).

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Key Concepts

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

Photon Energy
Photon energy is the energy carried by a single photon, which is the elementary particle of light. It is directly associated with electromagnetic radiation, encompassing aspects like visible light, microwaves, and X-rays. The energy of a photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where:
  • \( E \) is the energy of the photon in joules.
  • \( h \) represents Planck's constant, approximately \( 6.626 \times 10^{-34} \) J·s.
  • \( c \) symbolizes the speed of light, about \( 3.00 \times 10^8 \) m/s.
  • \( \lambda \) is the wavelength of the photon, expressed in meters.
Understanding photon energy is crucial for evaluating how semiconductors behave under different wavelength illuminations. For instance, a photon with enough energy can excite an electron across the semiconductor's energy gap, facilitating electrical conduction.
This concept extends beyond semiconductors to cover photosynthesis, solar energy harvesting, and more.
Planck's Constant
Planck's constant is a fundamental constant in physics that plays a critical role in quantum mechanics. Denoted as \( h \), it links the energy of photons with their frequency and describes the quantized nature of light. In essence, it tells us that energy is not continuous but rather comes in "chunks" or "quanta."
  • Planck's constant has a value of \( 6.626 \times 10^{-34} \) J·s.
It was first introduced by Max Planck in the early 20th century and laid the groundwork for understanding atomic and subatomic processes. In our semiconductor example, Planck's constant is a key factor in calculating photon energy through the formula \( E = \frac{hc}{\lambda} \). Any deviation or manipulation involving this constant significantly alters our comprehension of energy interactions in microscopic systems.
Wavelength Conversion
Wavelength conversion is a vital process when working with the formula for photon energy. This is because wavelength can be expressed in various units, such as meters, nanometers, or millimeters. For accurate calculations, it's paramount to convert the wavelength into meters, the standard unit in scientific calculations.
  • To convert millimeters to meters, as in our exercise, use the conversion: \(1 \text{ mm} = 1 \times 10^{-3} \text{ m}\).
Conversion ensures precision in calculations involving different physical constants, like the speed of light and Planck's constant. Understanding how to convert wavelengths efficiently and accurately is crucial, especially in fields dealing with optics, material science, and semiconductor technology. In the given exercise, failing to convert the wavelength appropriately would lead to calculation errors, thereby disrupting the accurate determination of the semiconductor's energy gap.

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Most popular questions from this chapter

(II) The energy gap between valence and conduction bands in germanium is 0.72 eV. What range of wavelengths can a photon have to excite an electron from the top of the valence band into the conduction band?

(II) The band gap of silicon is 1.12 eV. (a) For what range of wavelengths will silicon be transparent? (See Example 29-5.) In what region of the electromagnetic spectrum does this transparent range begin? (b) If window glass is transparent for all visible wavelengths, what is the minimum possible band gap value for glass (assume \(\lambda =\) 400nm to 700nm)? [\(Hint:\) If the photon has less energy than the band gap, the photon will pass through the solid without being absorbed.]

Must we consider quantum effects for everyday rotating objects? Estimate the differences between rotational energy levels for a spinning baton compared to the energy of the baton. Assume the baton consists of a uniform 32-cm-long bar with a mass of 230 g and two small end masses, each of mass 380 g, and it rotates at 1.8 rev/s about the bar's center.

(II) The equilibrium distance \(r_0\) between two atoms in a molecule is called the \(\textbf{bond length}\). Using the bond lengths of homogeneous molecules (like H\(_2\), O\(_2\), and N\(_2\)), one can estimate the bond length of heterogeneous molecules (like CO, CN, and NO). This is done by summing half of each bond length of the homogenous molecules to estimate that of the heterogeneous molecule. Given the following bond lengths: H\(_2\) (= 74 pm), N\(_2\) (= 145 pm), O\(_2\) (= 121 pm), C\(_2\) (= 154 pm), estimate the bond lengths for: HN, CN, and NO.

In the ionic salt KF, the separation distance between ions is about 0.27 nm. (\(a\)) Estimate the electrostatic potential energy between the ions assuming them to be point charges (magnitude 1\(e\)). (\(b\)) When F "grabs" an electron, it releases 3.41 eV of energy, whereas 4.34 eV is required to ionize K. \ Find the binding energy of KF relative to free K and F atoms, neglecting the energy of repulsion.

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