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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 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 peak wavelength is 1.062 mm, in the microwave part of the electromagnetic spectrum.

Step by step solution

01

Understand the Given

We are given the temperature of the radiator as 2.728 K. This temperature relates to cosmic microwave background radiation, which can be analyzed using Wien's displacement law to find the peak wavelength.
02

Apply Wien's Displacement Law

Wien's displacement law states that the wavelength at which the emission of a black body peaks is inversely proportional to its temperature. The law is expressed as \( \lambda_{max} = \frac{b}{T} \), where \( b \) is Wien's displacement constant \( (b = 2.897 \times 10^{-3} \text{ m K}) \) and \( T \) is the temperature in Kelvin.
03

Calculate the Peak Wavelength

Using Wien's displacement law, substitute the given temperature value \( T = 2.728 \) K into the formula:\[\lambda_{max} = \frac{2.897 \times 10^{-3} \text{ m K}}{2.728 \text{ K}} = 1.062 \times 10^{-3} \text{ m}\]Therefore, the peak wavelength \( \lambda_{max} \) is approximately 1.062 mm.
04

Determine the Spectrum Region

The calculated wavelength, 1.062 mm, falls within the microwave region of the electromagnetic spectrum. Microwaves are generally considered to span wavelengths from 1 mm to 1 meter.

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Key Concepts

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

Cosmic Microwave Background Radiation
The cosmic microwave background (CMB) radiation is an essential clue to our universe's past. Imagine it like the warm glow left over from the Big Bang. This ancient radiation fills the universe and reaches us from all directions. When scientists first detected it, they realized it was a snapshot of the very young universe. The temperature of the CMB is incredibly cold, about 2.728 Kelvin, which is just above absolute zero. Because of this low temperature, cosmic microwave background radiation is an ideal place to apply concepts like Wien's displacement law to understand the peak wavelength of radiation emitted.
  • The CMB radiation is smooth and uniform.
  • It tells us about the universe just 380,000 years after the Big Bang.
  • The study of CMB has helped confirm the Big Bang theory and our understanding of the universe's expansion.
Electromagnetic Spectrum
The electromagnetic spectrum is like a rainbow of energy types. It ranges from high-energy gamma rays to the low-energy radio waves. Each type of radiation is characterized by a different wavelength or frequency. The electromagnetic spectrum shows us the different types of electromagnetic waves, visible and invisible, around us. Within this spectrum, microwaves have longer wavelengths than visible light but shorter than radio waves. When you hear about a wavelength like 1.062 mm, as in the case of cosmic microwave background radiation, it falls right into the microwave region of the electromagnetic spectrum. That鈥檚 why the cosmic microwave background is called just that鈥攎icrowave.
  • Visible light is only a small part of the spectrum.
  • Different parts of the spectrum have different uses, from cooking (microwaves) to medical imaging (X-rays).
  • Understanding different parts of the spectrum is crucial for technologies such as WiFi and broadcasting.
Planck Distribution
The Planck distribution is a formula that describes how the intensity of radiation emitted by a black body varies with wavelength at a given temperature. Named after physicist Max Planck, this distribution is vital in understanding thermal radiation, like that from cosmic microwave background radiation. The distribution shows us not just the peak wavelength, but also how energy is distributed over various wavelengths. It essentially describes the spectrum of radiation emitted by a "perfect" radiator at any temperature. When we apply Wien's displacement law to the Planck distribution, we pinpoint the specific wavelength with the most energy, which is helpful for analyzing cosmic signals.
  • The Planck distribution provides insight into quantum mechanics and thermodynamics.
  • It's key to technologies that rely on light and heat, such as thermal imaging and laser design.
  • Understanding Planck's distribution equips scientists to predict behaviors of stellar objects.

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Most popular questions from this chapter

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda = h/p\) still holds, but we must use the relativistic expression for momentum, \(p = mv/\sqrt{1 - v^2/c^2}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is $$v = \frac{c}{\sqrt1+(mc\lambda/h)^2} $$ (b) The quantity \(h/mc\) equals 2.426 \(\times\) 10\(^{-12}\) m. (As we saw in Section 38.3, this same quantity appears in Eq. (38.7), the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h/mc\), the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to c. In this case it is convenient to write \(v = (1 - \Delta)c\) and express the speed of the electron in terms of rather than v. Find an expression for \(\delta\) valid when \(\lambda \ll h mc\). [\(Hint:\) Use the binomial expansion (1 + \(z)^n = 1 + nz + [n(n - 1)z^2/2] + \cdots\) g, valid for the case 0 z 0 6 1.4 (c) How fast must an electron move for its de Broglie wavelength to be 1.00 \(\times\) 10\(^{-15}\) m, comparable to the size of a proton? Express your answer in the form \(v =(1 - \Delta)c\), and state the value of \(\Delta\)

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) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (\(Hint\): In each case the initial kinetic energy is negligible.)

An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy \(K\) of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron?

How does the wavelength of a helium ion compare to that of an electron accelerated through the same potential difference? (a) The helium ion has a longer wavelength, because it has greater mass. (b) The helium ion has a shorter wavelength, because it has greater mass. (c) The wavelengths are the same, because the kinetic energy is the same. (d) The wavelengths are the same, because the electric charge is the same.
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