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The shortest visible wavelength is about 400 nm. What is the temperature of an ideal radiator whose spectral emittance peaks at this wavelength?

Short Answer

Expert verified
The temperature is approximately 7242.5 K.

Step by step solution

01

Apply Wien's Displacement Law

Wien's Displacement Law relates the peak emission wavelength (\( \lambda_{max} \)) of a blackbody to its temperature (\( T \)). The law is given by:\[\lambda_{max} T = b\]where \( b \) is Wien's displacement constant, approximately \( 2.897 \times 10^{-3} \) m K.
02

Convert Wavelength to Meters

First, convert the given wavelength from nanometers to meters. Since 1 nm = \( 10^{-9} \) m:\[ 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4.0 \times 10^{-7} \text{ m} \]
03

Solve for Temperature

Use Wien鈥檚 Displacement Law to solve for the temperature \( T \):\[T = \frac{b}{\lambda_{max}}\]Substituting \( b = 2.897 \times 10^{-3} \) m K and \( \lambda_{max} = 4.0 \times 10^{-7} \text{ m} \):\[T = \frac{2.897 \times 10^{-3}}{4.0 \times 10^{-7}} \approx 7242.5 \text{ K}\]
04

Validate Units and Calculation

Ensure that the units cancel correctly to give temperature in Kelvin and verify the calculation:The units for \( b \) are m K, and for wavelength \( \lambda_{max} \), they are meters. Thus, \( m/m \) leaves Kelvin:\[ T = \frac{2.897 \times 10^{-3} \, \text{m K}}{4.0 \times 10^{-7} \, \text{m}} = 7242.5 \, \text{K}\]The calculation confirms that the ideal radiator's temperature is approximately 7242.5 K.

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Key Concepts

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

Spectral Emittance
Spectral emittance refers to the energy radiated by a blackbody per unit area, per unit time, and per unit wavelength. It is a measure of how much energy is emitted at a specific wavelength. The concept helps us understand which wavelengths carry the most energy and how the emission is distributed across different wavelengths.

In the context of blackbody radiation, spectral emittance essentially tells us how strong the emission is at each part of the spectrum. Each object can emit energy over a range of wavelengths, but the intensity varies with each one. The sun, for example, emits most of its energy in the visible light spectrum, which is why it appears bright to us during the daytime.

Understanding spectral emittance is vital in fields like astronomy and climate science. It helps scientists predict how an object will look at different temperatures and what its spectral signature might be. This information is crucial when looking at stars or planets and trying to understand their temperature and composition without physically being there.
  • Measured in watts per square meter per nanometer (W m鈦宦 nm鈦宦).
  • Determines how colors appear in photography or display technology.
  • Important for estimating energy output of stars and planets.
Blackbody Radiation
Blackbody radiation is the emission of electromagnetic radiation by an idealized object that absorbs all incoming light, known as a "blackbody". It is characterized by a spectrum of wavelengths that only depends on the temperature of the body, not on its material or structure.

A perfect blackbody doesn't reflect or transmit any light; it only emits radiation. This radiation is in equilibrium with any corresponding absorption due to the object鈥檚 temperature. The spectrum of blackbody radiation is continuous and covers a wide range of wavelengths, with a distinct peak at a wavelength that depends on temperature. This is where Wien's Displacement Law is handy, as it articulates that the peak wavelength shifts with changes in temperature.

The sun is a great example of a blackbody, as it emits radiation over a wide range of wavelengths. By studying the blackbody radiation emitted, we can infer many properties about celestial objects, including temperature.
  • Characterized by its Planck's Law spectrum.
  • Temperature affects peak emission wavelength.
  • Helps scientists model heat emission from stars and planets.
Peak Emission Wavelength
The peak emission wavelength is the specific wavelength at which a blackbody's emission is maximized. In simple terms, it is the color or type of light that a body at a particular temperature emits most intensely. This peak shifts with temperature changes and is critical in understanding blackbody radiation's characteristics.

Wien's Displacement Law directly relates temperature to peak emission wavelength, as detailed in the solution above. Higher temperatures result in shorter peak wavelengths - meaning hotter objects tend to be bluer, because blue light has a shorter wavelength. Conversely, cooler objects peak at longer wavelengths and appear redder.

This concept is not only central to astronomy and physics but is also widely used in applications like thermal imaging and material sciences. Knowing the peak emission wavelength allows scientists to estimate the temperature of distant stars by simply analyzing the light they emit.
  • Moves to shorter wavelengths as temperature increases.
  • Provides an indication of an object's surface temperature.
  • Enables the study and comparison of stars and heated objects.

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Most popular questions from this chapter

(a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K/E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K/E\) for a hydrogen atom (mass 1.67 \(\times\) 10\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n\) = 3 level. Determine the wavelength and frequency of the photon.

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

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.

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