Thermal Radiation. Photon as the Quantum of Light

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19.1Thermal Radiation. Photon as the Quantum of Light

In this Physics tutorial, you will learn:

  • What is thermal radiation? What object(s) produce(s) it?
  • What is total emissivity?
  • What is spectral emissivity and how does it differ from total emissivity?
  • What are the factors affecting emissivity of objects?
  • What is a black body? How does it radiate EM waves?
  • How do reflection and absorption abilities of an object relate to each other?
  • What can we find using the Stefan-Boltzmann Law and Wien Law?
  • What are the drawbacks of classical approach in thermal physics?
  • What issues were present when comparing the theoretical and experimental graphs?
  • Why the findings of Max Planck were so important in modern physics?
  • What is the nature of light (continuous or discrete)?
  • How can we find the energy of radiation?

Introduction

In classical physics, light was considered as a (transverse) wave, as explained in Section 12 of this course. This is because light possesses all physical properties of waves such as reflection, refraction, interference, diffraction and polarization. However, this approach did not give satisfactory answers to a number of light-related phenomena observed in experiments. For example, if light is considered as a wave, we cannot use the formula

E = m ∙ A2 ∙ ω2/2

explained in tutorial 11.3 as light waves (unlike mechanical waves) have no mass of medium involved, as they mostly propagate in vacuum.

Likewise, if light is considered as a pure wave, we cannot use the relativistic equation

ε2 - p2 ∙ c2 = m20 ∙ c4

explained in tutorial 18.6 either, as a pure wave has no rest mass m0.

Therefore, it is obvious that a completely new approach must be used to explain any light-related phenomena, as light is different from all the other physical quantities discussed so far. This need for such a new approach gave birth to a new theory - the corpuscular (particle) theory of light, which we will explain gradually, starting from this tutorial.

What is Thermal Radiation?

In a given temperature above the absolute zero, a thermal motion produces various types of motion in atoms and molecules of matter such as electronic, vibrational and rotational (when spinning around themselves). When particles move to lower energetic levels, they produce EM radiation, otherwise known as thermal radiation.

The EM spectrum is a continuous spectrum, which theoretically includes wavelengths from 0 to infinity. Different wavelengths give different contributes in the transportation of energy produced by thermal radiation. However, there is a characteristic wavelength λm which gives the main contribution in this process for a given temperature. This characteristic wavelength decreases with the increase in temperature of material. Thus, in normal temperatures (around 300 K or 27° C), this wavelength is in the range of infrared part of EM spectrum. This is the reason why we cannot see objects radiate at normal temperature (IR radiation is invisible for humans). Objects start to glow at 800 K roughly. For example, the resistor of a heater does not glow immediately after turning on the switch; it needs a few seconds for the metal conductor to reach the above temperature and start glowing. However, the conductor has been radiating EM waves long before turning the switch on; this radiation was simply invisible for our sight.

The Sun is the main contributor of thermal (EM) radiation coming to the Earth. The surface of Sun is about 5800 K and for this temperature, the main contribution in thermal radiation is given by the wavelength λm ≈ 500 nm, which corresponds to the cyan colour - "the colour of the sky" in our perception. Obviously, without the presence of the Sun everything would dive into darkness.

Total and Spectral Emissivity

The energy radiated by the unit area of an object in every second from all possible wavelengths is known as the total emissivity E of an object. It is measured in [J/(s·m2)] or [W/m2]. All wavelengths of thermal radiation contribute in the total emissivity. There is another quantity which represents the contribution of a particular wavelength in thermal emissivity. It is known as spectral emissivity e(λ) and is a function of wavelength. Spectral emissivity depends from the type of object and its temperature. The graph showing the dependence of spectral emissivity from wavelength of light is given below.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Radiation. Photon as the Quantum of Light

The graph has the shape of a bell, where the highest point represents the maximum value of thermal emissivity reached for λ = λm. Other wavelengths close to λm also give a considerable contribution in thermal emissivity and transportation of energy of radiation. Wavelength that are 10 times greater or smaller than λm give little or no contribution in this regard.

Objects in normal temperature are not visible because they are incandescent (glowing) but because they reflect part of radiation falling on them. An object absorbs all the other wavelengths of solar radiation falling on it, except the wavelength representing the colour of object. Only this wavelength is reflected from the object's surface and eventually comes to our eyes. This is how we see the objects around us. In the next paragraph we will see how emissivity and absorptivity of a material relate with each other.

Radiation of a Black Body

In tutorial 12.3 "Reflection of Light", we have explained that when light rays fall on the surface of an object, it may reflected or absorbed depending on the physical features of that surface such as roughness, brightness or colour. Thus, rough and dark colour surfaces absorb more light radiation than smooth and bright surfaces. Moreover, a good absorber of radiation may be a good emitter in high temperatures. (Emission involves the release in the environment of some part of the absorbed radiation). From this condition, it is obvious that the emission (which is nothing else but the radiation of EM waves from an object to the surroundings), depends on the temperature of the object itself. Higher the temperature of the object, higher the rate of emission (radiation) of EM waves.

A completely black (and rough surface) object is the best absorber (and therefore the best emitter at high temperature) of light rays. However, an object is not necessary to be black and rough to be an ideal absorber of light (EM radiation). In practice, there is no need to have a black colour object to obtain an ideal absorber of light; we can achieve this using an irregularly shaped cavity with a very thin hole, just enough to allow a light ray enter in the cavity, as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Radiation. Photon as the Quantum of Light

If a light ray emitted from a light source (for example a torch) enters in the small hole of cavity, it experiences a number of reflections in the inner walls but it cannot escape from the cavity anymore. The light ray loses part of its energy after every collision with the internal walls of cavity until it is completely absorbed by these walls. If more and more rays enter in the cavity through the small hole, the cavity will behave like a perfect absorber of radiation, just like a black body.

Different objects have different reflecting abilities. This ability is mathematically represented through the reflection coefficient r, which in general depends on the incident wavelength and temperature of the object. On the other hand, the absorption coefficient of an object is denoted by a. They are dimensionless coefficients between 0 and 1. From the law of conservation of energy, we have

r + a = 1

An ideal mirror has r = 1 and a = 0, while an ideal black body has r = 0 and a = 1.

We denote the spectral emissivity of a black body by e0(λ). By the end of XIX century, the dependence of spectral emissivity of a black body, e0(λ) from the light wavelength, λ has been found experimentally. The graph of this dependence for two different temperatures T1 and T2 (T1 < T2) obtained through experiments has the shape shown below.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Radiation. Photon as the Quantum of Light

Experiments show that for an object in thermal equilibrium we have

e/a = e0

where e is the object's emissivity. The above equation is known as the Kirchhoff's Law of Emissivity. From this law, it is obvious that the emissivity of an object is less than the emissivity of a black body for the same wavelength of light and temperature.

In the graph shown above, it is evident that the position of maximum spectral emissivity shifts due left (towards smaller wavelengths) with the increase in object's temperature while the emissivity itself increases. In addition, the area under the graph increases with the increase in temperature.

These features of the e vs λ graph express the laws of thermal radiation of a black body discovered experimentally by Stefan, Boltzmann and Wien. Years later, Max Planck explained these laws theoretically.

Stefan-Boltzmann Law

This law expresses the relationship between the total emissivity of a black body and its temperature. It says:

"Total emissivity of a black body is proportional to the fourth power of absolute temperature."

Mathematically, we write:

E = σ ∙ T4

where σ = 5.67 × 10-8 J/K4m2s is a constant; it is known as the Stefan-Boltzmann constant.

Example 1

Calculate the power of radiation emitted by the window of an oven which has an area of 200 cm2. The oven is at 15230 C and its window is considered as a black body.

Solution 1

Clues:

A = 200 cm2 = 0.02 m2 = 2 × 10-2 m2
T = 1523° C = (1523 + 273) K = 1800 K = 1.8 × 103 K
(σ = 5.67 × 10-8 J/(K4m2s)
P = ?

Since the unit of total emissivity E is [J/s·m2] or [W/m2], we obtain

E = P/A = σ ∙ T4

Thus, the power of radiation emitted by the oven is

P = σ ∙ T4 ∙ A
= (5.67×10-8 W/m2 ∙ K4) ∙ (1.8 × 103 K)4 ∙ (2 × 10-2 m2 )
= 1.19 × 104 W
= 11.9 kW

Wien's Law

This law gives the relationship between the characteristic wavelength λm of thermal radiation emitted by a black body and the body temperature. This relationship is an inverse variation given by the equation

λm = b/T

where b = 2.9 × 10-3 m · K is known as the Wien's constant.

Wien's Law is otherwise known as the "law of displacement" to point out the fact that the value of characteristic wavelength λm "shifts" towards smaller values when the temperature of radiating body increases.

Example 2

Calculate the wavelength of EM radiation which gives the main contribution in the transportation of energy in the form of thermal radiation emitted by the human body, given that the normal body temperature of a healthy human is 37° C.

Solution 2

Clues:

T = 37° C = (37 + 273) K = 310 K = 3.1 × 102 K
(b = 2.9 × 10-3 m · K)
λm = ?

From the formula of Wien's Law, we have

λm = b/T
= 2.9 × 10-3 m ∙ K/3.1 × 102 K
= 0.9355 × 10-5 m
= 9355 nm

From the values of EM spectrum given in tutorial 11.6, it is clear that this wavelength belongs to infrared spectrum. We are not able to see this radiation but we only can "feel" it in the form of heat.

The Ultraviolet Catastrophe and Planck's Hypothesis

All attempts of scientists during the XIX century in finding a formula that corresponds to the graph of spectral emissivity, failed. This is because they relied on the classical approach, which stresses the continuous nature of electromagnetic radiation emitted by atoms and molecules. Based on this hypothesis, two scientists - Rayleigh and Jeans - found the mathematical expression below for the emission ability of a black body:

e0 (λ) = 2π ∙ c ∙ T/λ4

The graph below shows the curve found experimentally (the solid curve) and the theoretical curve obtained through the formula of Rayleigh-Jeans.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Radiation. Photon as the Quantum of Light

From the graph is evident that for long wavelength the curve obtained from the Rayleigh-Jeans formula represents a good fit for the experimental findings but for short wavelengths, this curve deflects too much from the curve found experimentally. Moreover, the emission ability of objects points towards infinity when wavelength points towards zero.

It was an absurdity to accept the idea that object with finite dimensions could have an infinite emission ability and they are able to radiate infinitely large amounts of energy in every second. This nonsense represents a notable failure of the classical theory of thermal radiation, which is known in the history of physics as the "ultraviolet catastrophe".

In 1900, Max Planck overcame this handicap by proposing a new hypothesis. According to him, the radiation emitted from atoms and molecules do not occur in a continuous way but with interruptions and in very small portions called quanta (quantum in singular). Moreover, the energy E of a quantum of light (known as photon - the particle of light) is proportional to the frequency of radiation, i.e.

E = h ∙ f

where h (which is a pseudo-letter; it is not the traditional h of our alphabet) represents the Planck constant. It has the value of 6.626 × 10-34 J/s.

Relying on his hypothesis on the quantum nature of light, Planck managed to find theoretically the true shape of the spectral emissivity curve for a black body and to explain the experimental findings (laws) of Stephan-Boltzmann and Wien. The compliance with the experimental data was complete. With the Planck hypothesis, a new era of modern physics represented by the quantum physics, did start. This new branch of physics today stands at the forefront of scientific development and technology.

The above equation represents the fundamental equation of quantum theory in Modern Physics.

Example 1

Stefan and Boltzmann found experimentally the value of constant σ that bears their name. As stated earlier, it is σ = 5.67 × 10-8 J/K4m2s, while Planck found theoretically that this constant is related to the other three universal constants h, k and c (Planck constant, Boltzmann constant and light speed in vacuum respectively) through the expression

σ = 5 ∙ k4/15c2 ∙ h3

Based on the above formula, calculate the value of Planck's constant h giving that k = 1.38 × 10-23 J/mol·K and c = 3 × 108 m/s.

Solution 1

This simple exercise involves only calculations. Thus, after rearranging, we obtain for the Planck's constant h:

σ = 5 ∙ k4/15c2 ∙ h3
h = ∛ 5 ∙ k4/15c2 ∙ σ
= ∛ 2 ∙ (3.14)5 ∙ (1.38 × 10-23 J/mol ∙ K)4/15 ∙ (3 × 108 m/s)2 ∙ (5.67 × 10-8 J/K4 ∙ m2 ∙ s)
= ∛ 289.24 × 10-102 J3/s3
= 6.613×10-34 J/s

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