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Chapter 39: Problem 39

Radiation has been detected from space that is characteristic of an ideal radiator at \(T=2.728 \mathrm{~K}\). (This radiation is a relic of the Big Bang at the beginning of the universe.) For this temperature, at what wavelength does the Planck distribution peak? In what part of the electromagnetic spectrum is this wavelength?

Short Answer

Expert verified
The Planck distribution peaks at a wavelength of \(1.062 mm\), corresponding to the microwave region of the electromagnetic spectrum.

Step by step solution

01

Apply Wien's displacement law

To find the peak wavelength of the Planck distribution for a given temperature, Wien's Displacement Law is used. This law is represented by the equation: \(\lambda_{max} = \frac{b}{T}\), where \(b\) is Wien's displacement constant (\(2.898 * 10^{-3} m.K\)) and \(T\) is the temperature in Kelvin (\(T = 2.728 K\)).
02

plug in the values

Substitute Wien's displacement constant (\(b = 2.898 * 10^{-3} m.K\)) and the temperature (\(T = 2.728 K\)) into the equation: \(\lambda_{max} = \frac{2.898 * 10^{-3}}{2.728}\).
03

Calculate the peak wavelength

Upon performing the calculation, the peak wavelength is found to be \(1.062 * 10^{-3} m\), or \(1.062 mm\).
04

Identify the part of the electromagnetic spectrum

The wavelength corresponds to the microwave region of the electromagnetic spectrum.

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Key Concepts

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

Planck Distribution
The Planck Distribution describes how electromagnetic radiation is emitted by a black body in thermal equilibrium at a given temperature. This distribution reveals that radiation is not emitted equally across all wavelengths. Instead, it shows a distinct peak at a particular wavelength, depending on the temperature of the body.

When a body is extremely hot, like stars, the peak wavelength is in the visible or even ultraviolet spectrum. Conversely, cooler bodies emit radiation peaking in the infrared or microwave regions, such as cosmic microwave background radiation from the Big Bang.

This peak can be calculated using Wien's Displacement Law, which provides a simple equation to find the wavelength at which the radiation is most intense.
Electromagnetic Spectrum
The electromagnetic spectrum is a continuum of all electromagnetic waves arranged according to frequency or wavelength. It ranges from gamma rays, with very short wavelengths, to radio waves, which have very long wavelengths.

Some common sections of the spectrum include:
  • Gamma Rays - extremely high energy, short wavelengths.
  • X-Rays - used in medical imaging, slightly longer wavelengths than gamma rays.
  • Ultraviolet Light - just beyond the violet end of visible light.
  • Visible Light - the part of the spectrum visible to the human eye, ranging from violet to red.
  • Infrared - emitted by warm objects, experienced as heat.
  • Microwaves - used in communication and for cooking via microwave ovens.
  • Radio Waves - used for broadcasting and communication.
Each type of radiation within the spectrum has a different source and application, heavily influenced by its wavelength and energy.
Microwave Radiation
Microwave radiation is part of the electromagnetic spectrum, found between infrared and radio waves. This type of radiation has wavelengths typically ranging from 1 millimeter to 1 meter. It is crucial in various technologies, including microwave ovens that make use of these wavelengths to heat food by exciting water molecules.

Aside from domestic applications, microwaves are also essential for communication. They are used in satellite transmissions, radar, and mobile phone networks. Furthermore, microwaves are vital in probing cosmic phenomena. The Cosmic Microwave Background Radiation, which peaks in the microwave range, provides insight into the early universe coming from the Big Bang itself.

This aspect of microwaves reflects their importance beyond mere culinary convenience, showcasing their role in both science and modern technology.

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Most popular questions from this chapter

An electron has a de Broglie wavelength of \(2.80 \times 10^{-10} \mathrm{~m}\) Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

DATA As an amateur astronomer, you are studying the apparent brightness of stars. You know that a star's apparent brightness depends on its distance from the earth and also on the fraction of its radiated energy that is in the visible region of the electromagnetic spectrum. But, as a first step, you search the Internet for information on the surface temperatures and radii of some selected stars so that you can calculate their total radiated power. You find the data given in the table. $$ \begin{array}{l|cccc} \text { Star } & \text { Polaris } & \text { Vega } & \text { Antares } & \alpha \text { Centauri B } \\ \hline \text { Surface temperature (K) } & 6015 & 9602 & 3400 & 5260 \\ \hline \begin{array}{l} \text { Radius relative to that of the } \\ \text { sun }\left(R_{\text {sun }}\right) \end{array} & 46 & 2.73 & 883 & 0.865 \\ & & & & \end{array}$$ The radius is given in units of the radius of the sun, \(R_{\text {sun }}=6.96 \times 10^{8} \mathrm{~m}\) The surface temperature is the effective temperature that gives the measured photon luminosity of the star if the star is assumed to radiate as an ideal blackbody. The photon luminosity is the power emitted in the form of photons. (a) Which star in the table has the greatest radiated power? (b) For which of these stars, if any, is the peak wavelength \(\lambda_{\mathrm{m}}\) in the visible range \((380-750 \mathrm{nm}) ?\) (c) The sun has a total radiated power of \(3.85 \times 10^{26} \mathrm{~W} .\) Which of these stars, if any, have a total radiated power less than that of our sun?

Why is it easier to use helium ions rather than neutral helium atoms in such a microscope? (a) Helium atoms are not electrically charged, and only electrically charged particles have wave properties. (b) Helium atoms form molecules, which are too large to have wave properties. (c) Neutral helium atoms are more difficult to focus with electric and magnetic fields. (d) Helium atoms have much larger mass than helium ions do and thus are more difficult to accelerate.

(a) An electron moves with a speed of \(4.70 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is its de Broglic wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops \(6.50 \times 10^{-14} \mathrm{~m}\) away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is \(6.64 \times 10^{-27} \mathrm{~kg} .\) (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in \(\mathrm{MeV}\). (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?
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