Lecture 2 Blackbody radiation
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1 Lecture 2 Blackbody radiation Absorption and emission of radiation What is the blackbody spectrum? Properties of the blackbody spectrum Classical approach to the problem Plancks suggestion energy quantisation Bose-Einstein statistics
2 Objectives Learn how classical theories cannot account for the spectral distribution of light from blackbody objects such as stars etc. Show that by making the assumption that light consists of small packets of energy (photons) we can develop an expression which perfectly fits the experimental data.
3 Blackbody radiation Heated bodies radiate energy, but what is the mechanism? On an atomic scale, heat causes the molecules and atoms of a solid to vibrate. As atoms consist of electrical charges in the form of electrons and protons, it is the vibration of these charges which is responsible for the emission of electromagnetic radiation. A very hot object will emit visible light as the electrons vibrate. How do bodies absorb radiation? In order to radiate energy, an object must first absorb it. Suppose we shine a light on an object. If we shine it on glass Light passes through If we shine it on a metal Light is reflected If we shine it on carbon Light is absorbed In glass the electrons are tightly bound to atoms and only oscillate at certain frequencies outside the range of visible light. This makes glass appear transparent as very little of the visible light is absorbed.
4 Blackbody radiation II Metals conduct and have free electrons not bound to any particular atom. These electrons oscillate in response to the light and then radiate light themselves. This radiation is reflected light. Again, there is very little absorption of light, most of it is reflected. The electrons in carbon have a short mean free path, when they collide their energy is transferred to the lattice. They are efficient absorbers of the incident light, hence carbon appears black. Carbon and similar materials are effective at converting incident light into heat. In a reverse process, as the carbon atoms warm up and vibrate more vigorously, more of the lattice energy is transferred to the free electrons, thus carbon is also a good radiator of heat. It cools down much faster than a metal as it is more efficient at converting the lattice energy into radiation.
5 Measuring the distribution of emitted radiation A simple spectroscope If we pass white light through a prism we obtain a spectrum. By measuring the intensity of each wavelength of light in the spectrum we can plot the spectral distribution.
6 The blackbody spectrum Consider a box with a small hole in the side. Radiation entering the hole is scattered inside the box and only a small amount comes out again. If the temperature of the box is in equilibrium the spectrum of light coming out of the hole looks something like this. 2.0x10-8 U(f) (arb. units) 1.5x x x T = 3000K T = 6000K 0.5x x x x10 15 Frequency (Hz) Typical blackbody spectra U f = Intensity of radiation of frequency f. We find similar spectral distributions if we measure the light from stars.
7 From red hot to white hot The colour of a hot object changes as the temperature increases
8 Properties of the blackbody spectrum For small f, U is proportional to f 2. But at some value of f there is a peak before U falls to zero. If we double the temperature the position of the peak doubles in frequency ( f peak T ) (Wien s displacement law), however the height of the peak is multiplied eight times. Why? U f 2 so if the temperature is constant U 2 f =4 U f, but we doubled the temperature so in fact U 2T =8U T. The total energy radiated by the body, the area under the curve, increases by a factor of 16, i.e 2 4 times more, when the temperature is doubled. Stefans law of radiation: U = T 4, = W/m 2 /K 4 - Stefans constant. U is the energy radiated from 1m 2 of black surface at temperature T.
9 Wien s displacement law f peak =at or peak T =b, b = mk (Wien s displacement law). For frequency f peak =2.82 k B T / h. k B = Boltzmann s constant = J/K. Note that f peak peak c.
10 Relationship between emitted radiation and radiation inside the cavity The energy (electromagnetic radiation) comes out of the hole at a speed c. Some of the light comes out at an angle and the hole appears smaller. Emitted radiation = R, radiation inside the cavity = P c m 2 /2 R= 0 0 cp cos θ d 4, d =sin θ d d θ A = cos θ P θ 2 /2 R= cp cos θ sin θ d d θ= cp 0 8 d = cp 4 Energy incident at an angle θ on a hole of unit area. So R=cP /4. The energy emitted from the hole is representative of the energy inside the cavity.
11 Origins of the spectrum The energy spectrum is formed by a continuous process of absorption and re-emission of radiation by the atoms and molecules forming the walls of the box. In this way the energy can shift from one mode to another. When thermal equilibrium is reached the characteristic spectrum will be established. To form a theoretical description of the spectrum we need to determine how many modes of oscillation have frequencies in a given energy range. These oscillators are the electromagnetic waves inside the box, which can be thought of as standing waves.
12 Modes in a box The electric field at the cavity wall must be zero
13 Standing waves I Pluck a string n=1 n=2 n=3 n=4 n 2 =L The natural modes of vibration of the string are standing waves with nodes at the ends. In the same way, electromagnetic waves inside the box are also standing waves.
14 Standing waves II 1.0 Amplitude n=1 n=2 n= Position (x/a) (arb. units) Standing waves inside a cavity of length a. The standing waves have amplitude y=a sin(2 π x / λ)cos(2 π f t ). Let k=2 /, and =2 f then y=a sin(kx)cos(ωt). If the box has sides of length a then a=n /2 for n =1,2,3 The frequencies of these waves are f =c/ =n c/ 2 a. So k=n /a and =ck.
15 K space We can represent the standing waves in 3D space by a set of k vectors k= k x, k y, k z = a l,m,n. Each point can be associated with a cubic volume of space and represents a frequency of =ck=c k x 2 k y 2 k z 2, the volume of the associated space is 3 / a 3. For a fixed frequency,, we obtain a set of values for k x, k y and k z which lie on a spherical surface: 2 c =k 2 2 x k 2 y k 2 z with radius k= /c. To get the total number of vibrational modes in the frequency range zero to we count the number of cubes contained in the sphere with radius k= /c. N cubes =Volume between spheres / volume of cube.
16 K-space II Each mode occupies a discrete volume of K-space.
17 Mode counting The volume of the whole sphere is 4 k 3 /3, but we are limited to positive values of k, i.e. one octant of the sphere, so the volume becomes V s = k 3 = k Now k 3 = 3 c 3 and =2 f so k 3 = 8 3 f 3 c 3. If V cube = 3 / a 3, the total number of cubes, N, is V s /V cube, so N = 8 f 3 a 3. 6 c 3 Putting, V =a 3, differentiating with respect to f, and multiplying by two because we can have two orthogonal transverse electromagnetic waves at each frequency we get N f = 8 f 2 V. df c 3
18 Comparison with experiment In the classical approach we assign each mode an energy k B T. The total energy emitted at each frequency from a box of unit volume is given by N f k B T =U RJ f = 8 f 2 c 3 k B T. This is the Rayleigh-Jeans approximation (1900). k B is Boltzmann s constant = J/K. We find good agreement between the Rayleigh-Jeans equation and the observed results for low values of f, i.e. where U f f 2. The higher the temperature the bigger the frequency range over which the agreement is good. Doubling the temperature doubles the energy output at low frequencies, as expected from the proportionality to T. However, at higher frequencies the experimental and theoretical results diverge. We expect more energy to be output at higher frequencies, but in experiments the energy distribution falls to zero. This failure of the Reyleigh-Jeans model is sometimes called the UV catastrophe.
19 Planck s suggestion Planck suggested that energy could only be emitted in chunks that are multiples of hf, where h is Planck s constant ( Js). Experimentally, we can see this reflected in the fact that f peak T. By doubling the temperature the number of modes that can radiate freely is also doubled. From this explanation it is clear that if the average energy per mode is k B T and the value hf for a particular mode is, e.g., 5 k B T, that mode will be unlikely to radiate. As the frequency increases, the probability of radiation decreases. Replacing U =k B T with U =hf in the Rayleigh-Jeans equation gives U Planck f = 8 f 2 hf P BE f, where P BE f is the Bose-Einstein factor for c 3 the average number of photons per mode at frequency f. hf P BE f is the average energy, E, of photons with frequency f.
20 The Bose-Einstein factor We assume that the probability of occupying an energy level E is given by P E = e E /k B T. This was proved by Planck. N 3 E 3 = 3hf P 3 = Aexp(-3hf/k B T) N 1 =N 0 e hf / k B T N 2 N 1 E 2 = 2hf E 1 = hf P 2 = Aexp(-2hf/k B T) P 1 = Aexp(-hf/k B T) N 2 =N 0 e 2hf / k B T N 0 E 0 = 0 P 0 = A Energy levels of a quantum oscillator N n =N 0 e nhf / k B T Let x=hf / k B T, then N 1 =N 0 e x, N 2 =N 0 e 2x, so N Total =N 0 1 e x e 2x.... E n =N n nhf = N 0 e nx nhf = N 0 hf ne nx and E Total =N 0 hf 0 e x 2 e 2x....
21 The Bose-Einstein factor II The average energy is given by E = E N 0 hf Total n = N Total N 0 n ne nx e nx =hfp BE f ne nx n P BE f = n e = d nx dx log n e nx = d dx log 1 1 e x = 1 e x 1 = 1 e hf / k B T 1 So our final result is U Planck f = 8 f 2 c 3 hf e hf /k B T 1. Even though this is proportional to f 3 the Bose-Einstein term reduces the energy to zero at high frequencies. At low frequencies the term on the right approximates to 1, matching the Rayleigh-Jeans approximation. As f 0 or T the average energy approaches k B T, the same as the classical result.
22 Various blackbody spectra
23 Conclusions Classical physics, in which electromagnetic radiation is assumed to be a continuous wave, cannot account for the blackbody radiation spectrum. The assumption that electromagnetic radiation is emitted in quanta with energy E =hf allows us to develop an expression which accurately describes the spectral distribution of radiation emitted from a blackbody.
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