Lecture #8. Light-matter interaction. Kirchoff s laws
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1 1 Lecture #8 Light-matter interaction Kirchoff s laws
2 2 Line emission/absorption Atoms: release and absorb photons with a predefined set of energies (discrete). The number of protons determine the chemical element. Each element has a unique set of energy levels that its electrons can occupy. Electrons can only move between available energy levels Atom gets energy: absorb a photon, electron movs to a higher level Release energy: emit a photon, electron moves to a lower level Example: Hydrogen atom (1 proton, 1 electron) absorption spectrum emission spectrum
3 3 Line emission/absorption Atoms: release and absorb photons with a predefined set of energies (discrete). The number of protons determine the chemical element. Each element has a unique set of energy levels that its electrons can occupy. Each element has its own fingerprint of energy levels A spectrum is the number of photons (or total amount of energy) emitted at each wavelength. Optical spectra of different elements Optical spectra of a combination of elements Wavelength
4 4 Line emission/absorption Molecules: have additional energy levels because their molecules can vibrate and rotate This complicates their spectra: large numbers of very close vibrational and rotational levels. Emission from H2: Molecule formed by 2 hydrogen atoms Combination of lines from atoms and bands from molecule. Optical spectra of different elements Wavelength
5 5... and then... what? What happens to a photon after the atom or molecule releases it? 1. If the matter is transparent: Photons can travel freely in the matter and can flow freely out We observe an emission line spectrum A cloud of gas is transparent What can we learn from this spectrum?
6 6... and then... what? What happens to a photon after the atom or molecule releases it? 1. If the matter is opaque: The matter has high density, many atoms are close together. Then the photons bounce around, sharing their energy. They end up with a thermalized distribution of energies Lonely runner: can choose his speed, no one in his way. (transparent matter) Marathon: a runner needs to consider where the others are. Ends up going at an average speed. (opaque matter)
7 7 Thermal emission All opaque objects emit thermal radiation This means, you, me, plants, dust particles, stars, planets... but... why?? Because the atoms and molecules in all things have some kinetic energy (if the temperature is bigger than 0 Kelvin!), and will always be releasing photons An object s thermal radiation spectrum depends on only one property: The Kinetic energy of the matter: its temperature
8 8 Thermal emission Definitions: Thermal energy: the combined kinetic energy of many particles Temperature: the average kinetic energy of each particle. Less thermal energy More thermal energy
9 9 Thermal emission Definitions: Thermal energy: the combined kinetic energy of many particles Temperature: the average kinetic energy of each particle. More thermal energy Higher temperature
10 10 Thermal emission Thermal radiation has a very specific spectral shape Number of photons at each wavelength is very easy to predict The hotter the object, the more blue light in the spectrum The hotter the object, the more total photons it emits colder hotter
11 11 Question 1. We are dense and opaque. Do we glow in the dark? a) No, people do not emit light b) Yes, we do. We just cannot see it with our eyes. c) No,people are too small to emit light d) No, people do not contain enough radioactive material e) N.A.
12 12 Question 1. We are dense and opaque. Do we glow in the dark? a) No, people do not emit light b) Yes, we do. We just cannot see it with our eyes. c) No,people are too small to emit light d) No, people do not contain enough radioactive material e) N.A.
13 13 Properties of thermal radiation For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Units: Constant: [Flux]= Flux = σt 4 energy time area σ = J m 2 K 4 s Flux: energy flow into or out of a surface. = J m 2 s
14 14 Example: Stefan-Boltzmann law For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Flux = σt 4 The Sun has a surface temperature of about σ = J m 5800 K. How much TOTAL energy is it releasing each 2 K 4 s second?
15 15 Example: Stefan-Boltzmann law For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Flux = σt 4 The Sun has a surface temperature of about σ = J 5800 K. How much TOTAL energy is it releasing each m 2 K 4 s second? Flux = σt 4 Flux = J m 2 K 4 s (5800K)4 Flux = J m 2 s Energy that comes out from each patch of area 1m 2, each second Total Energy = (Total surface of the Sun) x (Energy per second per m 2 )
16 16 Example: Stefan-Boltzmann law For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Flux = σt 4 The Sun has a surface temperature of about σ = J 5800 K. How much TOTAL energy is it releasing each m 2 K 4 s second? Total Energy = (Total surface of the Sun) x (Energy per second per m 2 ) Energy total =Flux Area = R = m Energy total =Flux Area = J m 2 s 4πR2 J m 2 s 4π( m) 2 Energy total =Flux Area = J s
17 17 Example: Stefan-Boltzmann law For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Flux = σt 4 The Sun has a surface temperature of about σ = J 5800 K. How much TOTAL energy is it releasing each m 2 K 4 s second? Total Energy = (Total surface of the Sun) x (Energy per second per m 2 ) Energy total =Flux Area = R = m Energy total =Flux Area = J m 2 s 4πR2 J m 2 s 4π( m) 2 Energy total =Flux Area = J s Total energy coming out of the Sun each second: LUMINOSITY
18 18 Properties of thermal radiation For every patch (area, m 2 ) on the surface of the object 1. Stefan-Boltzmann law: Hotter means brighter Flux = σt 4 σ = J m 2 K 4 s Flux: energy flow into or out of a surface through a patch of area 1m2 Luminosity: TOTAL energy coming out of an object (Flux x Area) Units of Luminosity: [Luminosity] = [flux area] = J m 2 s m2 = J s = Watts
19 19 Properties of thermal radiation For every patch (area, m 2 ) on the surface of the object 2. Wien s law: Hotter means bluer The peak of the thermal radiation for a hotter object, is at a bluer wavelength λ peak T =2.9mmK
20 20 Example: Wien s Law For every patch (area, m 2 ) on the surface of the object 2. Wien s law: Hotter means bluer The surface temperature of the Sun is about 5800 K. At what wavelength does its thermal radiation peak? λ peak T =2.9mmK
21 21 Example: Wien s Law For every patch (area, m 2 ) on the surface of the object 2. Wien s law: Hotter means bluer The surface temperature of the Sun is about 5800 K. At what wavelength does its thermal radiation peak? λ peak T =2.9mmK λ peak = 2.9mmK 5800K λ peak = m λ peak = m λ peak T =2.9mmK = mk K λ peak = 500nm The sun looks yellow!
22 22 Kirchoff s laws (a) Continuous spectrum = hot opaque (dense) source (b) Absorption line spectrum = transparent source (cooler, not fully ionized) (c) Emission line spectrum = hot transparent (optically thin) source. y-axis: intensity x-axis: wavelength
23 23 Break Time (5 mins)
24 24 Spectra We can learn a lot by studying the details in a spectrum. What is this object?
25 25 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Let s play detectives, step by step
26 26 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Which are the absorption lines? a) A b) B c) D d) A,C,D e) C,D A B C D E
27 27 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Which are the absorption lines? a) A b) B c) D d) A,C,D e) C,D A B C D E
28 28 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Which letter better labels emission lines? a) A b) B c) C d) D e) E A B C D E
29 29 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Which letter better labels emission lines? a) A b) B c) C d) D e) E A B C D E
30 30 Spectra We can learn a lot by studying the details in a spectrum. What is this object? Three types of spectra coming from a single object (very common!) Emission line spectrum thermal spectrum Absorption line spectrum A B C D E
31 31 Spectra What is this object?? Reflected sunlight! Continuous thermal spectrum of visible light is like the Sun s (find the peak), except that some of the blue light is missing. It has been absorbed. A B C D E
32 32 Spectra What is this object?? Reflected sunlight Thermal radiation: Infra-red spectrum peaks at wavelengths corresponding to a temperature of 255 K (-18 C). High density material, not very cold for astronomical standards. A B C D E
33 33 Spectra What is this object?? Reflected sunlight Thermal radiation Carbon dioxide: These absorption lines are the fingerprints of cool CO2 gas A B C D E
34 34 Spectra What is this object?? Reflected sunlight Thermal radiation Carbon dioxide Ultraviolet emission lines: hot, low-density (transparent) gas. An atmosphere! The atmosphere has hot gas and cooler CO2 A B C D E
35 35 Spectra What is this object?? Reflected sunlight Thermal radiation Carbon dioxide Ultraviolet emission lines Is a planet around a star! In this case: Mars We can determine the content and temperature of the atmosphere. A B C D E
36 36 Quiz #8 Toolkit: Light λ f = c E = h f NO TALKING! Counts for double attendance if you get this one right. Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK c = m/s σ = J m 2 K 4 s
37 37 Quiz #8 1. Which object is hotter? left or right? why? 2. Which object is brighter? left or right? why? 3.Which object is brighter? left or right? why? Toolkit: Light λ f = c E = h f Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK c = m/s σ = J m 2 K 4 s NO TALKING! Counts for double attendance if you get this one right.
38 38 Quiz #8 1. Which object is hotter? left or right? why? 2. Which object is brighter? left or right? why? Please pass your quizzes to the hall 3.Which object is brighter? left or right? why? Toolkit: Light λ f = c E = h f Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK c = m/s σ = J m 2 K 4 s
39 39 Quiz #8 1. Which object is hotter? left or right? why? Bluer colors have smaller wavelengths. Wien s law tells us that if the wavelength decreases then the temperature (T) needs to increase to compensate. Toolkit: Light λ f = c E = h f Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK c = m/s σ = J m 2 K 4 s
40 40 Quiz #8 1. Which object is hotter? left or right? why? 2. Which object is brighter? left or right? why? Since the object on the right was hotter, then looking at Stefan-Boltzmann law, we see that if we increase the temperature, then the FLUX also increases. This means more light is coming out of the hotter object, i.e. it is brighter. Toolkit: Light λ f = c E = h f Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK c = m/s σ = J m 2 K 4 s
41 41 Quiz #8 3.Which object is brighter? left or right? why? Toolkit: Light λ f = c E = h f Flux = σt 4 Lum = Flux Area λ peak T=2.9mmK Both objects are the same color, according to Wien s law they have the same temperature. This means that the flux (per unit area) that comes out of these objects is the same. The object on the right has a bigger area than the object on the left, so it is brighter overall. c = m/s σ = J m 2 K 4 s
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