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Chapter 12: Problem 71

The semiconductor GaP has a band gap of \(2.2 \mathrm{eV}\). Green LEDs are made from pure GaP. What wavelength of light would be emitted from an LED made from GaP?

Short Answer

Expert verified
The wavelength of light emitted by a GaP green LED can be calculated using the energy-wavelength formula: \( λ = \frac{h * c}{E} \). Given the band gap of GaP is 2.2 eV, we convert it to joules and plug all the values into the formula. This yields a wavelength of approximately \( 5.65 \times 10^{-7} \mathrm{m} \), which is equivalent to 565 nm.

Step by step solution

01

Convert the band gap energy to joules

We have to convert the given energy value from electron volts (eV) to joules (J) using the relation: 1 eV = 1.602 x 10^-19 J.
02

Relate energy and wavelength

We can relate the energy and wavelength of light using the formula, E = h * c / λ, where E is the energy, h is the Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
03

Calculate the wavelength

Now we will use the energy-wavelength formula to determine the wavelength: \( λ = \frac{h * c}{E} \)
04

Plug the values and compute

We have all the values we need, let's plug them in and calculate the wavelength: \( λ = \frac{(6.626 \times 10^{-34} Js) \times (3 \times 10^8 m/s)}{(2.2eV) \times (1.602 \times 10^{-19} J/eV)} \)
05

Simplify and find the final wavelength

Now we will simplify the equation and find the final value of the wavelength: \( λ \approx 5.65 \times 10^{-7} m \)
06

Convert the wavelength to nanometers

To make the value more convenient to work with, let's convert the wavelength from meters to nanometers (1 meter = 1 x 10^9 nanometers): \( λ \approx 5.65 \times 10^{-7} m \times \frac{1 \times 10^9 nm}{1 m} = 565 nm \) So the wavelength of light emitted by a GaP green LED is approximately 565 nm.

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Most popular questions from this chapter

For each of the following alloy compositions indicate whether you would expect it to be a substitutional alloy, an interstitial alloy, or an intermetallic compound: (a) \(\mathrm{Cu}_{0.66} \mathrm{Zn}_{0.34},\) (b) \(\mathrm{Ag}_{3} \mathrm{Sn}\) (c) \(\mathrm{Ti}_{0.99} \mathrm{O}_{0.01}\)

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation \((\lambda),\) the angle at which the radiation is diffracted \((\theta),\) and the distance between planes of atoms in the crystal that cause the diffraction \((d)\) is given by \(n \lambda=2 d \sin \theta .\) X-rays from a copper X-ray tube that have a wavelength of \(1.54 \AA\) are diffracted at an angle of 14.22 degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming \(n=1\) (first-order diffraction).

The electrical conductivity of titanium is approximately 2500 times greater than that of silicon. Titanium has a hexagonal close-packed structure, and silicon has the diamond structure. Explain how the structures relate to the relative electrical conductivities of the elements.

Unlike metals, semiconductors increase their conductivity as you heat them (up to a point). Suggest an explanation.

Red light-emitting diodes are made from GaAs and GaP solid solutions, \(\mathrm{GaP}_{x}\) As \(_{1-x}\) (see Exercise 12.73 ). The original red LEDs emitted light with a wavelength of \(660 \mathrm{nm}\). 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.
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