Question

Determine \(\lambda_m\) , the wavelength at the peak of the Planck distribution, and the corresponding frequency \(f\), at these temperatures: (a) 3.00 K; (b) 300 K; (c) 3000 K.

Short answer

The peak wavelength \( \lambda_m \) and frequency \( f \) for temperatures 3 K, 300 K, and 3000 K are:\( \lambda_m = 9.657\times10^{-4}, 9.657\times10^{-6}, \text{and } 9.657\times10^{-7} \) m; \( f \approx 3.1\times10^{11}, 3.1\times10^{13}, \text{and } 3.1\times10^{14} \) Hz respectively.

Step-by-step solution

Understand the Problem

We need to find the wavelength at which the Planck distribution peaks (known as Wien's Law) for objects at three different temperatures: 3 K, 300 K, and 3000 K. After finding the peak wavelength, we will then determine the corresponding frequency using the speed of light.

Apply Wien's Displacement Law

Wien's Displacement Law is given by: \( \lambda_m = \frac{b}{T} \), where \( \lambda_m \) is the peak wavelength, \( b = 2.897 \times 10^{-3} \text{ m}\cdot\text{K} \) is Wien's displacement constant, and \( T \) is the temperature in Kelvin. We will use this formula to calculate \( \lambda_m \) for each temperature.

Calculate \( \lambda_m \) for 3 K

Substitute \( T = 3 \text{ K} \) into Wien's Law: \( \lambda_m = \frac{2.897 \times 10^{-3}}{3} \text{ m} = 9.657 \times 10^{-4} \text{ m} \).

Calculate \( \lambda_m \) for 300 K

Substitute \( T = 300 \text{ K} \) into Wien's Law: \( \lambda_m = \frac{2.897 \times 10^{-3}}{300} \text{ m} = 9.657 \times 10^{-6} \text{ m} \).

Calculate \( \lambda_m \) for 3000 K

Substitute \( T = 3000 \text{ K} \) into Wien's Law: \( \lambda_m = \frac{2.897 \times 10^{-3}}{3000} \text{ m} = 9.657 \times 10^{-7} \text{ m} \).

Calculate Frequency from Wavelength Formula

Use the equation \( c = \lambda_m f \), where \( c \) is the speed of light (~299,792,458 m/s) to find frequency \( f \). Rearranging gives \( f = \frac{c}{\lambda_m} \).

Determine Frequency for 3 K

Substitute \( \lambda_m = 9.657 \times 10^{-4} \text{ m} \) into \( f = \frac{c}{\lambda_m} \): \( f = \frac{299,792,458}{9.657 \times 10^{-4}} \approx 3.1 \times 10^{11} \text{ Hz} \).

Determine Frequency for 300 K

Substitute \( \lambda_m = 9.657 \times 10^{-6} \text{ m} \): \( f = \frac{299,792,458}{9.657 \times 10^{-6}} \approx 3.1 \times 10^{13} \text{ Hz} \).

Determine Frequency for 3000 K

Substitute \( \lambda_m = 9.657 \times 10^{-7} \text{ m} \): \( f = \frac{299,792,458}{9.657 \times 10^{-7}} \approx 3.1 \times 10^{14} \text{ Hz} \).

Key concepts

Planck's Distribution

Planck's distribution describes how electromagnetic radiation is emitted by a black body in thermal equilibrium at a given temperature. This concept is the foundation for understanding how objects emit radiation and the specific wavelengths they do so. When an object is heated, it radiates energy in the form of electromagnetic waves. The intensity of this radiation varies with wavelength, forming a characteristic spectral distribution.

• Each temperature has a unique emission curve.
• The distribution peaks at a certain wavelength that depends on the temperature.
• Lower temperatures result in longer peak wavelengths, while higher temperatures cause the peak to shift towards shorter wavelengths.

Understanding Planck's distribution is essential for applications like astrophysics, where it helps in determining the temperature of stars and other celestial bodies by analyzing the light they emit.

Peak Wavelength

The peak wavelength is a critical aspect of the Planck distribution, and it shifts depending on the temperature of the emitting body. Wien's Displacement Law precisely describes this relationship. It states that the wavelength \( \lambda_m \) where the emission is most intense is inversely proportional to the temperature of the object, following the equation: \[\lambda_m = \frac{b}{T}\]where \( b \) is Wien's constant, approximately \( 2.897 \times 10^{-3} \; \text{m}\,\text{K} \). The peak wavelength helps us understand the color of the emitted light:

• Cooler objects emit light at longer wavelengths, such as infrared.
• Hotter objects emit light at shorter wavelengths, like visible or ultraviolet light.

By calculating the peak wavelength using Wien's Law for various temperatures, one can predict which part of the electromagnetic spectrum the radiation falls into.

Frequency Calculation

To understand the full behavior of radiation emitted at different temperatures, it's essential to connect wavelength to frequency. The frequency \( f \) of a wave is related to its wavelength \( \lambda \) by the speed of light \( c \), using the formula:\[f = \frac{c}{\lambda}\]where \( c \) is approximately \( 299,792,458 \; \text{m/s} \). This formula makes it easy to calculate the frequency of the peak emission once the peak wavelength is known.

• Lower temperatures and longer wavelengths will result in lower frequencies.
• Higher temperatures and shorter wavelengths will lead to higher frequencies.

Understanding this relationship allows us to convert information about wavelength into frequency terms, which is useful for various practical applications like telecommunications and scientific research.

Temperature Effects on Wavelength

Temperature has a profound effect on the wavelength at which an object emits radiation. The higher the temperature, the shorter the peak wavelength. This is because higher temperatures increase the energy levels of molecules, shifting the emission towards shorter, more energetic wavelengths.

• At 3 K, emissions are typically in the microwave region.
• At 300 K, emissions start entering the infrared region.
• At 3000 K, emissions move into the visible light region, and can include ultraviolet light.

This principle is utilized in many scientific areas, from measuring the cosmic microwave background radiation to designing systems for infrared imaging. Knowing how temperature affects wavelength allows scientists and engineers to design accurate sensors and study the properties of materials under different thermal conditions. Additionally, it explains why stars of different colors (red, yellow, blue) have different surface temperatures.