Question

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature \(T=300 . \mathrm{K}\), which is approximately the surface temperature of Earth.

Short answer

Based on the given Planck spectrum and the step-by-step solution provided: a) The peak frequency as a function of temperature is given by: $$f_p(T) = \frac{2.821 \cdot k_B T}{h}$$ b) The peak frequency at a temperature of T = 6000 K is approximately: $$f_p(6.00 \cdot 10^3 \mathrm{~K}) \approx 7.48 \times 10^{14} \mathrm{~Hz}$$ c) The peak frequency at a temperature of T = 2.735 K is approximately: $$f_p(2.735 \mathrm{~K}) \approx 3.34 \times 10^{11} \mathrm{~Hz}$$ d) The peak frequency at a temperature of T = 300 K is approximately: $$f_p(300 \mathrm{~K}) \approx 3.29 \times 10^{13} \mathrm{~Hz}$$

Step-by-step solution

a) Differentiate the function and find the critical points

First, we need to find the critical points of the function by taking its derivative with respect to frequency and setting it to zero. The function is given as: $$\epsilon_T(f) =\frac{2 \pi h}{c^2}\left(\frac{f^3}{e^{h f / k_B T} -1}\right)$$ Take the derivative with respect to \(f\): $$\frac{d\epsilon_T(f)}{df} = \frac{2 \pi h}{c^2}\left(\frac{3f^2}{e^{h f / (k_B T)} -1} - \frac{f^3 \cdot h}{k_B T} \cdot \frac{e^{hf/(k_BT)}}{(e^{hf/(k_BT)}-1)^2}\right)$$ Now, set the derivative to zero: $$\frac{d\epsilon_T(f)}{df} = 0$$

a) Solve for the peak frequency

To proceed further, we can simplify the equation by multiplying both sides by \((e^{hf/(k_BT)}-1)^2\) and rearranging the equation to get: $$3f^2\cdot (e^{hf/(k_BT)} -1)^2 = f^3\cdot h \cdot \frac{e^{hf/(k_BT)}}{k_B T}$$ To make it easier to solve, we can introduce a new variable, \(x = \frac{hf}{k_B T}\), then the equation becomes: $$3\left(\frac{k_B T \cdot x}{h}\right)^2 \cdot \left( e^x -1 \right)^2 = \left(\frac{k_B T \cdot x}{h}\right)^3 \cdot x \cdot \frac{e^x}{x}$$ By simplifying and rearranging the equation, we get: $$\frac{dx}{e^x-1} = 3x^2$$ This is a transcendental equation and cannot be solved analytically. However, we can use methods like the Lambert W function to find a numerical approximation for x: $$x \approx 2.821$$ Now, we can express the peak frequency as a function of temperature: $$f_p(T) = \frac{2.821 \cdot k_B T}{h}$$

b) Find the peak frequency at T=6000K

Substitute the value of T into the peak frequency expression: $$f_p(6.00 \cdot 10^3 \mathrm{~K}) = \frac{2.821 \cdot k_B (6.00 \cdot 10^3 \mathrm{~K})}{h}$$ Using the given values of Boltzmann's constant and Planck's constant, we can evaluate the expression: $$f_p(6.00 \cdot 10^3 \mathrm{~K}) \approx 7.48 \times 10^{14} \mathrm{~Hz}$$

c) Find the peak frequency at T=2.735K

Substitute the value of T into the peak frequency expression: $$f_p(2.735 \mathrm{~K}) = \frac{2.821 \cdot k_B (2.735 \mathrm{~K})}{h}$$ Evaluate the expression: $$f_p(2.735 \mathrm{~K}) \approx 3.34 \times 10^{11} \mathrm{~Hz}$$

d) Find the peak frequency at T=300K

Substitute the value of T into the peak frequency expression: $$f_p(300 \mathrm{~K}) = \frac{2.821 \cdot k_B (300 \mathrm{~K})}{h}$$ Evaluate the expression: $$f_p(300 \mathrm{~K}) \approx 3.29 \times 10^{13} \mathrm{~Hz}$$

Key concepts

Blackbody Radiation

Blackbody radiation plays a crucial role in understanding how objects emit and absorb electromagnetic radiation. In essence, a blackbody is an idealized object that absorbs all incoming light and emits radiation in a spectrum that depends solely on its temperature.

This phenomenon is described by the Planck spectrum, which reveals the intensity of radiation at different frequencies. At lower frequencies, the spectrum follows a cubic relationship with the frequency, but as we move higher, the intensity peaks and then gradually declines due to quantum effects. This creates a distinct 'hump' in the spectrum, characteristic of blackbody emitters.

The Planck spectrum can be applied to real-world objects, like stars and planets, allowing us to estimate their temperatures based on the radiation they emit. For example, the surface of the Sun has been estimated at approximately 6000K by observing its spectrum and finding the frequency at which it peaks, a direct application of the Planck spectrum.

Boltzmann Constant

The Boltzmann constant, denoted as \(k_B\), is a fundamental physical constant that establishes a bridge between macroscopic and microscopic physics. It shows up in many areas of physics but is most notably used in statistical mechanics and thermodynamics.

In terms of the Planck spectrum, \(k_B\) plays a pivotal role in relating the frequency of radiation emitted by a blackbody to its temperature. It helps in quantifying the distribution of energy among different possible microscopic states, which in turn, determines the spectral density of the emitted radiation.

Understanding \(k_B\) empowers students to delve into the microscopic world, translating temperature into energy per particle and grasping the essence of entropy. Hence, the Boltzmann constant is not just a number; it's a gateway to the microscopic nature of thermodynamically macroscopic quantities.

Cosmic Microwave Background

The Cosmic Microwave Background (CMB) is the afterglow radiation left over from the Big Bang. Studying it is like peering back in time to the early universe. It represents a nearly uniform blackbody spectrum, corresponding to a temperature of approximately \(2.735\text{K}\).

The CMB is a relic from an epoch when the universe was dense and hot. As the universe expanded, the radiation stretched into the microwave region of the electromagnetic spectrum, cooling it down to the remnant heat we detect today. The peak frequency of this radiation, determined through the Planck spectrum, tells us the temperature of the universe when it became transparent to radiation, which is an essential piece of evidence for the Big Bang theory.

The fact that the CMB closely follows the Planck spectrum serves as a significant verification for our understanding of the universe's origin and thermal history. It confirms the predictions of quantum physics and thermodynamics on a cosmic scale, illustrating the universal applicability of physics laws.