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Find \(\lambda_{m}\) for blackbody radiation at \((a) T=3 \mathrm{~K},(b) T=300 \mathrm{~K},\) and \((c) T=3000 \mathrm{~K}\).

Short Answer

Expert verified
The peak wavelength \(\lambda_m\) for T = 3 K is approximately \(966 cm\), for T = 300 K it is approximately \(0.966 cm\), and for T = 3000 K it is approximately \(0.0966 cm\).

Step by step solution

01

Recall Wien's Displacement Law

To find the wavelength, \(\lambda_m\), at which the radiation from a blackbody peaks, use Wien's Displacement Law, which states that \(\lambda_m \times T = b\), where \(b\) is Wien's displacement constant approximately equal to \(2.898 \times 10^{-3} m\cdot K\).
02

Calculate \(\lambda_{m}\) for \(T=3 \text{K}\)

Insert the temperature \(T=3 \text{K}\) into the formula \(\lambda_m \times T = b\) and solve for \(\lambda_m\): \[\lambda_m = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{3} \, m\cdot K.\]
03

Calculate \(\lambda_{m}\) for \(T=300 \text{K}\)

Insert the temperature \(T=300 \text{K}\) into the formula \(\lambda_m \times T = b\) and solve for \(\lambda_m\): \[\lambda_m = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{300} \, m\cdot K.\]
04

Calculate \(\lambda_{m}\) for \(T=3000 \text{K}\)

Insert the temperature \(T=3000 \text{K}\) into the formula \(\lambda_m \times T = b\) and solve for \(\lambda_m\): \[\lambda_m = \frac{b}{T} = \frac{2.898 \times 10^{-3}}{3000} \, m\cdot K.\]

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Key Concepts

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

Blackbody Radiation
Imagine an object that perfectly absorbs all the electromagnetic radiation that lands on it; this theoretical concept is known as a blackbody. In reality, no object is a perfect blackbody, but many approximate the behavior closely. Blackbodies not only absorb radiation, but they also emit radiation uniformly in all directions. This emitted radiation is what we refer to as blackbody radiation.

Blackbody radiation has a characteristic spectrum that depends solely on the body's temperature. The warmer the object, the more radiation it emits at every wavelength, and the peak of the emitted spectrum shifts to shorter wavelengths. This phenomenon leads to the commonly known fact that hot objects — like stars or heated metal — not only radiate more intensely but also change color, moving from red to white to blue as the temperature increases.
Planck's Law
At the turn of the 20th century, Max Planck introduced a groundbreaking law that transformed our understanding of thermal radiation. Planck's law describes the radiation emitted by a blackbody in terms of its temperature. It tells us the energy radiated per unit surface area per unit time per unit solid angle, in a particular frequency range. This law can be encapsulated in a complex formula, but the essential takeaway is that it expresses how the energy of radiation varies with frequency and temperature.

Planck's law predicts the emission spectrum of a blackbody, leading to the conclusion that the energy peaks at a certain wavelength for a given temperature. This connects directly to Wien's Displacement Law, as it provides the theoretical foundation to why the peak wavelength shifts with temperature — a result Planck could derive from quantum considerations, long before quantum mechanics was fully understood.
Thermal Radiation
You've probably felt the warmth of the sun on your skin or the heat coming from a toaster oven — these sensations are due to thermal radiation. It is the process by which the movement of atoms and molecules in a material generates electromagnetic radiation. Every object emits thermal radiation, simply as a result of having a temperature above absolute zero.

Thermal radiation can occur at all wavelengths. However, for any given temperature, there's a certain wavelength where the intensity of the radiation is at its peak. This is crucial for understanding not only natural phenomena, such as why we get sunburned more intensely by UV radiation at the beach, but also for practical applications. For instance, engineers use this knowledge to design systems for heating, cooling, and lighting that are based on the principles of thermal radiation and take into account temperature to predict the behavior of the emitted radiation.

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Most popular questions from this chapter

Find the momentum of a photon in \(\mathrm{eV} / \mathrm{c}\) and in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) if the wavelength is \((a) 400 \mathrm{nm},\) (b) \(1 \AA=0.1 \mathrm{nm}\) (c) \(3 \mathrm{~cm},\) and \((d) 2 \mathrm{nm}\)

The x-ray spectrometer on board a satellite measures the wavelength at the maximum intensity emitted by a particular star to be \(\lambda_{m}=82.8 \mathrm{nm}\). Assuming that the star radiates like a blackbody, \((a)\) compute the star's surface temperature. (b) What is the ratio of the intensity radiated at \(\lambda=70 \mathrm{nm}\) and at \(\lambda=100 \mathrm{nm}\) to that radiated at \(\lambda_{m} ?\)

Under optimum conditions, the eye will perceive a flash if about 60 photons arrive at the cornea. How much energy is this in joules if the wavelength of the light is \(550 \mathrm{nm} ?\)

The work function of molybdenum is \(4.22 \mathrm{eV}\). ( \(a\) ) What is the threshold frequency for the photoelectric effect in molybdenum? (b) Will yellow light of wavelength \(560 \mathrm{nm}\) cause ejection of photoelectrons from molybdenum? Prove your answer.

When light of wavelength \(450 \mathrm{nm}\) is incident on potassium, photoelectrons with stopping potential of \(0.52 \mathrm{~V}\) are emitted. If the wavelength of the incident light is changed to \(300 \mathrm{nm}\), the stopping potential is \(1.90 \mathrm{~V}\). Using only these numbers together with the values of the speed of light and the electron charge, \((a)\) find the work function of potassium and \((b)\) compute a value for Planck's constant.

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