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Chapter 21: Problem 4

Briefly explain why metals are opaque to electromagnetic radiation having photon energies within the visible region of the spectrum.

Short Answer

Expert verified
Answer: Metals are opaque to electromagnetic radiation in the visible region of the spectrum because their free electrons absorb the energy of the photons, causing them to move to higher energy states. This absorption of energy prevents the transmission of visible light, making metals appear opaque.

Step by step solution

01

Understanding Electromagnetic Radiation and Metals

Electromagnetic radiation refers to the waves of the electromagnetic field which propagate through space and carry energy. The electromagnetic spectrum includes different wavelengths, among which the visible region ranges approximately from 400 to 700 nanometers (nm). Metals are solid materials with a characteristic arrangement of electrons. The key to understanding the opacity of metals is the behavior of their electrons when interacting with electromagnetic radiation.
02

Role of Free Electrons in Metals

Metals have a unique property known as the "sea of electrons." The valence electrons (outermost electrons) in a metal form a collective outer shell when extending beyond their parent atoms. These electrons are mobile and are referred to as free (or conduction) electrons. The presence of free electrons is responsible for many metallic properties, such as their high electrical and thermal conductivity.
03

Interaction of Photons and Electrons

When electromagnetic radiation, such as visible light, encounters a metal surface, the photons (energy particles) in the radiation interact with the free electrons present in the metal. The energy of the photons is transferred to these free electrons, causing them to become excited.
04

Absorption of Visible Light by Metals

The energy transfer from photons to electrons usually results in the absorption of the electromagnetic radiation. In case of metals, the free electrons are able to absorb the energy of the photons in the visible region of the spectrum. This absorption causes the free electrons to move to higher energy states, leaving behind the holes (unoccupied energy states) at lower energy levels.
05

Opaqueness of Metals Due to Absorption

Since the metals absorb the photons in the visible region of the spectrum, very little or no light is transmitted through them. As a result, metals appear opaque to electromagnetic radiation with photon energies in the visible region. When the radiation is absorbed, the metal surface may reflect some of it, which gives metals their characteristic shiny appearance.

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Most popular questions from this chapter

The index of refraction of corundum \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are \(1.757\) and \(1.779\) in the direction of light propagation.

The indices of refraction of fused silica and a soda-lime glass within the visible spectrum are \(1.458\) and \(1.51\), respectively. For each of these materials determine the fraction of the relative dielectric constant at \(60 \mathrm{~Hz}\) that is due to electronic polarization, using the data of Table 18.5. Neglect any orientation polarization effects. \(21.9\) Using the data in Table 21.1, estimate the dielectric constants for borosilicate glass,

Distinguish between materials that are opaque, translucent, and transparent in terms of their appearance and light transmittance.

Briefly explain how reflection losses of transparent materials are minimized by thin surface coatings.

Briefly explain why the magnitude of the absorption coefficient ( \(\beta\) in Equation \(21.18\) ) depends on the radiation wavelength.
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