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Chapter 12: Problem 77

The semiconductor gallium nitride (GaN) has a band gap of \(3.4 \mathrm{eV}\). What wavelength of light would be emitted from an LED made from GaN? What region of the electromagnetic spectrum is this?

Short Answer

Expert verified
The wavelength of light emitted from an LED made from GaN is approximately \(365.3 \mathrm{nm}\), which falls within the ultraviolet region of the electromagnetic spectrum (10 nm - 400 nm).

Step by step solution

01

Converting energy from eV to Joules

First, we need to convert the given energy in electron volts (eV) to Joules (J). We use the conversion factor 1 eV = 1.60218 x 10^-19 J. E(eV) = 3.4 eV E(J) = 3.4 eV * (1.60218 x 10^-19 J/eV)
02

Using Planck's equation to find the wavelength

Now we will use Planck's equation to find the wavelength (λ) of the emitted light. The equation is E = h * c / λ, where h (Planck's constant) = 6.626 x 10^-34 Js, c (speed of light) = 2.998 x 10^8 m/s, and E is the energy in Joules. Rearranging the equation to solve for λ: λ = h * c / E Plugging in the values: λ = (6.626 x 10^-34 Js) * (2.998 x 10^8 m/s) / (3.4 * 1.60218 x 10^-19 J)
03

Calculating the wavelength

Now, we can calculate the wavelength by solving the equation: λ = (6.626 x 10^-34 Js) * (2.998 x 10^8 m/s) / (3.4 * 1.60218 x 10^-19 J) ≈ 3.653 x 10^-7 m To convert the wavelength to nanometers (nm), we multiply by 1 x 10^9: λ ≈ 3.653 x 10^-7 m * 1 x 10^9 nm/m = 365.3 nm
04

Determining the region of the electromagnetic spectrum

Now, we will compare the calculated wavelength (365.3 nm) with the electromagnetic spectrum ranges to determine the region it belongs to. The electromagnetic spectrum includes (approximate ranges): - Gamma-rays: < 1 nm - X-rays: 1 nm - 10 nm - Ultraviolet rays: 10 nm - 400 nm - Visible light: 400 nm - 700 nm - Infrared light: 700 nm - 1 mm - Microwaves: 1 mm - 1 m - Radio waves: > 1 m Since the wavelength we obtained (365.3 nm) falls within the ultraviolet rays range (10 nm - 400 nm), the light emitted from an LED made from GaN is in the ultraviolet region of the electromagnetic spectrum.

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Most popular questions from this chapter

Besides the cubic unit cell, which other unit cell(s) has edge lengths that are all equal to each other? (a) Orthorhombic, \((\mathbf{b})\) hexagonal, \((\mathbf{c})\) rhombohedral, (d) triclinic, (e) both rhombohedral and triclinic.

(a) What are the \(\mathrm{C}-\mathrm{C}-\mathrm{C}\) 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?

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 \(154 \mathrm{pm}\) 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).

Silver chloride (AgCl) adopts the rock salt structure. The density of \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\) is $5.56 \mathrm{~g} / \mathrm{cm}^{3}$. Calculate the length of an edge of the AgCl unit cell.

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