The first LEDs were made from GaAs, which has a band gap of 1.43 eV. What wavelength of light would be emitted from an LED made from GaAs? What region of the electromagnetic spectrum does this light correspond to: ultraviolet, visible, or infrared?

Short Answer

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= 2.29 × 10⁻¹⁹ J #tag_title#Step 2: Calculate the wavelength#tag_content# Now we can use the photon energy equation to calculate the wavelength: E = h * c / λ Rearranging the equation to solve for λ: λ = h * c / E Plugging in the values for h (6.63 × 10⁻³⁴ Js), c (3 × 10⁸ m/s), and E (2.29 × 10⁻¹⁹ J): λ = (6.63 × 10⁻³⁴ Js) * (3 × 10⁸ m/s) / (2.29 × 10⁻¹⁹ J) λ ≈ 8.68 × 10⁻⁷ m #tag_title#Step 3: Determine the region of the electromagnetic spectrum#tag_content# The wavelength found is 8.68 × 10⁻⁷ m, which is approximately 868 nm. This corresponds to the infrared region of the electromagnetic spectrum, as visible light is usually between 400 nm and 700 nm, and ultraviolet light has shorter wavelengths than visible light. #tag_title#Short Answer#tag_content# The wavelength of light emitted from a GaAs LED is approximately 868 nm, which corresponds to the infrared region of the electromagnetic spectrum.

Step by step solution

01

Convert energy to Joules

To determine the wavelength, we first need to convert the energy from electron volts (eV) to Joules (J). To do this, we use the conversion factor 1eV = 1.6 × 10⁻¹⁹ J: E = 1.43 eV × (1.6 × 10⁻¹⁹ J/eV)

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Key Concepts

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

Band Gap Energy
The concept of band gap energy is fundamental to understanding how light-emitting diodes (LEDs) function. In the simplest terms, the band gap is the energy difference between the top of the valence band (where the electron normally resides) and the bottom of the conduction band (where the electron can move freely, allowing for conduction). This band gap energy determines the energy of the photons that the material can emit when an electron transitions back down from the conduction band to the valence band.

For instance, Gallium Arsenide (GaAs) is a common material used in the production of LEDs and has a band gap of 1.43 electron volts (eV). This value is crucial since it directly relates to the color of the light emitted by an LED. The lower the band gap energy, the longer the wavelength of the emitted light, and consequently, the closer the light is to the red end of the visible spectrum. Conversely, a higher band gap energy corresponds to shorter wavelengths and light closer to the violet end.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, which range from very short wavelengths (like gamma rays) to very long wavelengths (like radio waves). Visible light is just a small part of this spectrum, which also includes ultraviolet (UV) light, infrared (IR) light, and others. Each type of electromagnetic radiation is characterized by its wavelength or frequency.

When discussing LEDs, it's important to identify where the emitted light falls on the electromagnetic spectrum. For example, if the wavelength of light emitted from a GaAs LED is calculated, it is crucial to determine whether this wavelength is within the visible range (approximately 380 nm to 750 nm), the ultraviolet range (less than 380 nm), or the infrared range (greater than 750 nm). Understanding the placement of this light within the electromagnetic spectrum allows us to predict its visibility to the human eye and potential applications.
Energy to Wavelength Conversion
Energy to wavelength conversion is a piece of essential scientific knowledge used to describe the relationship between a photon's energy and its wavelength within the electromagnetic spectrum. According to Planck's equation, the energy of a photon (\( E \) in Joules) is directly proportional to its frequency (\( u \) in Hertz) and inversely proportional to its wavelength (\( u \) in meters), with the proportionality constant being Planck's constant (\( h \) approx. equal to 6.626 x 10⁻³⁴ J·s).

The equation is written as: \[ E = h u \]Since the speed of light (\( c \) = 3.00 x 10⁸ m/s) is equal to the wavelength times the frequency (\( c = \lambda u \)), we can rearrange the equation to solve for wavelength (\( \lambda \) in meters): \[ \lambda = \frac{hc}{E} \]
For an LED with a specified band gap energy, such as the GaAs LED with 1.43 eV mentioned earlier, we first convert the energy from eV to Joules (since Planck's constant is in Joule·seconds) and then apply the equation above to determine the wavelength of the emitted light. This enables us to understand the color of the light and its position in the electromagnetic spectrum.

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Most popular questions from this chapter

Red light-emitting diodes are made from GaAs and GaP solid solutions, GaP \(_{x} A s_{1-x}(\) see Exercise 12.79\() .\) The original red LEDs emitted light with a wavelength of 660 nm. If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1,\) estimate the composition (the value of \(x\) ) that is used in these LEDs.

(a) What are the \(\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}\) bond angles in diamond? (b) What are they in graphite (in one sheet)? (c) What atomic orbitals are involved in the stacking of graphite sheets with each other?

GaAs and GaP make solid solutions that have the same crystal structure as the parent materials, with As and Prandomly distributed throughout the crystal. GaP As \(_{1-x}\) exists for any value of \(x .\) If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1,\) estimate the band gap for GaP \(_{0.5} \mathrm{As}_{0.5}\) . (GaAs and GaP band gaps are 1.43 \(\mathrm{eV}\) and 2.26 \(\mathrm{eV}\) , respectively.) What wavelength of light does this correspond to?

Which element or elements are alloyed with gold to make the following types of "colored gold" used in the jewelry industry? For each type, also indicate what type of alloy is formed: (a) white gold, (b) rose gold, (c) green gold.

Which statement correctly describes a difference between graphene and graphite? (a) Graphene is a molecule but graphite is not. (b) Graphene is a single sheet of carbon atoms and graphite contains many, and larger, sheets of carbon atoms. (c) Graphene is an insulator but graphite is a metal. (d) Graphite is pure carbon but graphene is not. (e) The carbons are \(s p^{2}\) hybridized in graphene but \(s p^{3}\) hybridized in graphite.

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