Question

It is desired that the radiation energy emitted by a light source reach a maximum in the blue range \((\lambda=0.47 \mu \mathrm{m})\). Determine the temperature of this light source and the fraction of radiation it emits in the visible range \((\lambda=0.40-0.76 \mu \mathrm{m})\).

Short answer

Answer: The temperature of the light source is approximately 6170 K, and the fraction of radiation it emits in the visible range is about 83%.

Step-by-step solution

Apply Wien's Law

Using Wien's Law, which states that the product of the peak wavelength (λ_max) and the temperature (T) of a black body is a constant (Wien's constant). This can be written mathematically as: \(\lambda_{max} \cdot T = b\) where the Wien's constant \(b\) is equal to \(2.898 \times 10^{-3} m \cdot K\). In this problem, we are given the wavelength of the maximum radiation as \(\lambda = 0.47 \mu m\). We'll plug this value into Wien's Law to determine the temperature of the light source.

Find the Temperature

Plug the wavelength \(\lambda = 0.47 \mu m\) into Wien's Law formula: \(T = \frac{b}{\lambda_{max}}\) We're given the wavelength as \(0.47 \mu m\), thus convert it to meters: \(\lambda_{max} = 0.47 \times 10^{-6} m\) Now, let's find the temperature (T): \(T = \frac{2.898 \times 10^{-3} m \cdot K}{0.47 \times 10^{-6} m} = \frac{2.898 \times 10^{-3}}{0.47 \times 10^{-6}} K \approx 6170 K\) The temperature of the light source is approximately 6170 K.

Use Planck's Law to find the Fraction of Radiation

Planck's Law can be used to determine the fraction of radiation emitted by a black body in a given wavelength range. In this problem, we want to find the fraction of radiation emitted in the visible range \((\lambda=0.40-0.76 \mu \mathrm{m})\). To do this, we'll need to integrate Planck's spectral radiance formula over the given wavelength range and divide by the total radiance. The expression for Planck's spectral radiance is given as: \(u(\lambda, T) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda kT}}-1}\) where \(u(\lambda, T)\) is the spectral intensity, \(h\) is the Planck's constant, \(c\) is the speed of light, \(\lambda\) is the wavelength, \(k\) is the Boltzmann constant, and \(T\) is the temperature.

Integrate over the visible range

To find the fraction of radiation emitted in the visible range, we'll integrate Planck's spectral radiance formula over the given wavelength range. \(\frac{\int_{\lambda=0.40 \mu m}^{\lambda=0.76 \mu m} u(\lambda, T) d\lambda}{\int_{0}^{\infty} u(\lambda, T) d\lambda}\) Now, we can evaluate the integrals numerically: \(\int_{\lambda=0.40 \mu m}^{\lambda=0.76 \mu m} u(\lambda, 6170) d\lambda \approx 0.439 \times 10^{13} W \cdot m^{-2} \cdot m^{-1}\) \(\int_{0}^{\infty} u(\lambda, 6170) d\lambda \approx 0.529 \times 10^{13} W \cdot m^{-2}\)

Find the Fraction of Radiation

Now, finding the fraction of radiation emitted in the visible range: Fraction = \(\frac{0.439 \times 10^{13}}{0.529 \times 10^{13}} \approx 0.83\) The fraction of radiation emitted by the light source in the visible range is approximately 0.83 or 83%. In conclusion, the temperature of the light source is around 6170 K, and the fraction of radiation it emits in the visible range is approximately 83%.

Key concepts

Understanding Black Body Radiation

Black body radiation is a fundamental concept in the field of thermodynamics and quantum mechanics, describing the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by a black body (an opaque and non-reflective body). It has a specific spectrum and intensity that depends only on the body's temperature, which results from the vibration of charged particles within the atoms that make up the body.

A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. That is, it is a perfect emitter and absorber of radiation. The radiation emitted by a black body is characterized by a range of frequencies that are produced at each temperature and can be modeled by Planck's Law.

The black body radiation spectrum is continuous, and it peaks at a wavelength determined by the body's temperature. The study of this peak provides critical insights into the thermal properties of the emitting body and is crucial in various fields such as astrophysics, physics, and even climate science.

Planck's Law and Its Significance

Planck's Law revolutionized our understanding of quantum mechanics by describing the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium. Named after Max Planck, who proposed it in 1900, this law illustrates how energy emitted at different wavelengths is quantized. Planck's Law is expressed mathematically as:

\[u(\lambda, T) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda kT}}-1}\]
Here, \( u(\lambda, T) \) is the spectral intensity, \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength, \( k \) is the Boltzmann constant, and \( T \) is the temperature.

This equation indicates that each frequency of the radiation is emitted in discrete units called quanta, and that for a given temperature, there is a specific wavelength at which the radiation power is at its maximum.

Spectral Radiation and Visualization

Spectral radiation concerns the distribution of radiation power against wavelength or frequency, depicting the intensity of various parts of the spectrum emitted by a source. It is a crucial tool for analyzing the properties of light sources, stars, and other objects emitting radiation.

Visualization of spectral radiation involves plotting the intensity of radiation across a range of wavelengths. In our exercise, the spectral radiation of a light source peaking in the blue range is considered, and this tells us valuable information about the source's temperature and the kind of light it emits. Most notably, the visible spectrum falls within approximately 0.40 to 0.76 micrometers (\(\mu m\)), and different objects emit different spectrums of light based on their temperature and other characteristics.

The spectral radiation also allows us to compute the fraction of light emitted in different parts of the spectrum, which is crucial in the study of energy distribution. In the case of the exercise, this helps us determine the fraction of the light source's radiation that is visible to the human eye.