Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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’s 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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⁻² nm⁻¹).
  • 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’s 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the de Broglie wavelength of a 5.00-g bullet that is moving at 340 m/s. Will the bullet exhibit wavelike properties?

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 \(\Delta\)

An alpha particle (\(m = 6.64 \times 10^{-27} kg\)) emitted in the radioactive decay of uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?

What is the de Broglie wavelength of a red blood cell, with mass 1.00 \(\times\) 10\(^{-11}\) g, that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?

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

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free