1. Stellar Spectra Homework 1: Due today Homework 2: Due Class time Tuesday, Feb. 11 Reading for today: Chapters 3.1 - 3.4 Reading for next lecture: Chapters 3.5 Topics for Today: Bohr Model of the Atom Wave Nature of Matter Emission and Absorption of Light Spectral Classification Hertzsprung-Russell Diagram

2. Bohr Model of the Hydrogen Atom To explain the spectral lines, Neils Bohr's in 1922 considered electrons orbiting in only certain allowed orbits. He also assumed that radiation in the form of a single discrete quantum (a photon) is emitted or absorbed as the electron changes orbit. Allowed orbits were those in which the electron had an angular momentum that were integer multiples of h/2π, therefore : m v r = n h/2π, where n is an integer. The radii of the allowed orbits (see extra slides) are: r n = n 2 h 2 /(4 π 2 m e e 2 ), where m e is the electron mass and e is the charge on the electron or proton

3. The electron in each allowed orbit has energy (both kinetic and potential energy). The energy of the electron in the n th orbital is E n given by (see extra slides): E n = -2π 2 e 4 m e /(n 2 h 2 ) = -2.17x10 -11 ergs/n 2 If the energy of the electron in the m th orbital is E m, then an atom can emit or absorb a photon that corresponds to the difference in energy between the two orbits (the n th and m th ): hν = E n – E m or ν = (E n – E m )/h

4. We often represent these energy levels by an energy level diagram. It is very common to arbitrarily set the n=1 level energy to 0, then n = ∞ will have energy of E = 13.6 eV. If we express energy in units of the eV (1 eV = 1.6x10 -12 ergs), then: E n = -13.6 eV/n 2

5. Spectral lines: hν = h c/λ = E n – E m = -13.6 eV (1/n 2 - 1/m 2 ) The Balmer series is from m = 2 to n = 3, 4, 5,... 2-3 E = 1.88 eV or λ = 656 nm 2-4 E = 2.54 eV or λ = 486 nm 2-5 E = 2.85 eV or λ = 434 nm The Lyman series is from m = 1 to n = 2, 3, 4,.... The Paschen series is from m = 3 to n = 4, 5, 6...

6. Wave Nature of Electrons: In 1923 de Broglie proposed that particles could behave as waves. Postulated that the orbit circumference had to be an integer number of electron wavelengths. The wavelength of an electron: λ = h/p (where p = momentum), thus: λ = h/m e v The de Broglie's criteria would be: nλ = n h/m e v = 2π r Remember Bohr's quantization assumed: m e vr = n h/2π, which is identical. This wave concept developed into today's modern quantum mechanics.

7. Wave or Quantum Mechanics In 1927, Heisenberg realized that this wave description of matter gave rise to inherent indeterminacy. This became known as the Heisenberg Uncertainty Principle. About this same time, Schrödinger developed a mathematical model of the atom using this new “wave mechanics”. He introduced the wave function that describes the probability of finding a particle at a particular location. Thus the deterministic view of classical physics is replaced by a probabilistic view of quantum mechanics – we can only predict where a particle is likely to be, but not exactly where it will be. Thus the spectral lines are not infinitely narrow, but are broadened by this process (but for hydrogen less than one part in 10 million).

8. (Top) An atom can absorb a photon if its energy E = hν = hc/λ matches the energy required for an electron to move to a higher energy state. (Bottom) An electron in at atom can spontaneously drop from an upper to a lower energy state. The energy difference is given to a photon of frequency: ν = ΔE/h, or λ = hc/ΔE. Emission & Absorption of Light

9. Thus, a bright line or emission spectrum is seen where a rarefied gas is undergoing spontaneous emission. A dark line, or absorption line spectrum is seen where a cooler rarefied gas is absorbing the light from a continuum source

10. Stellar Spectra Each element has a unique set of allowed energy levels, so each element produces a unique spectrum (either absorption or emission). Once you move beyond hydrogen, the energy level structure is much more complex.

11. In the 1800’s, it was found that different stars had different patterns of absorption lines (the letters are historical). Is this because they have different elements in their atmospheres?

12. A more quantitative way to represent the spectra of a star is to plot the flux density of the star as a function of wavelength (or frequency). The absorption lines are then dips in the flux density. The pattern of absorption lines is related to surface temperature of the star.

13. No, stars do not have different chemical elements. Annie Jump Cannon & Cecilia Payne-Gaposchkin found that the differences were caused by differences in the star’s surface temperature. Stars are all primarily composed of hydrogen and helium with traces of the other elements. Temperature determines the electronic excitation of the atoms and their ionization state, and thus which spectral lines are present.

14. Example: Hydrogen To get the Balmer series of absorption lines (the only hydrogen lines at visible wavelengths), many of the hydrogen atoms in the atmosphere of the star need to have their electron in the n = 2 energy state. What determines the energy state of the electron ??? In stellar atmospheres, collisions between particles are most important and can convert thermal energy into orbital energy and visa versa.

15. The average thermal energy of a particle is 3/2 kT. Thermal energy is energy of motion: 3/2 kT = ½ mv 2, Thus, average thermal velocity of a particle: v = (3kT/m) 1/2 The n = 2 level in H is 10.2 eV above the n = 1 level (ground state), for the average hydrogen atom to have this amount of thermal energy (to cause a collisional excitation) would require a temperature of T = 79,000 °K !!! This would suggest it would take extremely hot stars to produce the Balmer series of hydrogen absorption lines. However, the particles do not all have the same velocity. Instead, for any temperature, there is a broad velocity distribution for a thermal gas.

16. This velocity distribution is called the Maxwell- Boltzmann distribution and is shown to the right for He atoms at different temperatures. For any temperature, there is a tail of particles moving at much higher speeds than average. At temperatures where 3/2 kT is less than the difference in energy between two states, it is the high speed particles in this tail that can still produce collisional excitation.

17. Hydrogen Excitation and Ionization At a temperature of 10,000 K, can get significant excitation of the n=2 level and see prominent Balmer absorption lines. The different levels of hydrogen are populated according to the Boltzmann equation (see text). If the temperature becomes too large, then collisions can strip an electron from the atom (needs only 3.4 eV more energy). In stars, hydrogen becomes nearly fully ionized for temperatures above about 20,000 °K (governed by Saha equation). The dependence on temperature for both collisional excitation and ionization, means that the Balmer lines of hydrogen are only present in stars with a narrow range of surface temperatures around 10,000 - 20,000 °K,

18. The strength of the various lines of atoms and ions is extremely temperature sensitive producing the large changes in the absorption lines seen in stars of different temperatures. The energy level structures are different for each element, so the visible absorption lines have different temperature sensitivity.

19. The spectral classification sequence: O B A F G K M L T hottest stars -----------------coolest stars T = 50,000°K -------------------- 3,000 °K (The spectral classifications of L and T recently added to reflect the discoveries of extremely cool “stars”) The Sun is a G spectral type star. Surface temperature about 5800 °K

20. Hertzsprung-Russell Diagram: In the early part of the 20 th century, Hertzsprung and Russell independently realized that the stars did not have random properties, but star’s luminosity and surface temperature were correlated. Created the Hertzsprung-Russell diagram (or H-R diagram): A plot of a star’s surface temperature or spectral type (O, B, A...) or color (B-V) on the horizontal axis and the star’s luminosity or absolute magnitude on the vertical axis. Each point in the diagram, then represents the properties of a single star.

21. Star are found in well defined regions of the H-R diagram. 22,000 stars from Hipparchos catalog supplemented by 1,000 nearby faint stars with distances. Note: low luminosity stars are under represented in this H-R diagram.

22. Main Sequence: Most stars fall on a part of the diagram running from upper left to lower right. Stars on the main sequence range from hot, very luminous stars (upper-left) to cool, low luminosity stars (lower-right). The Sun is a main sequence star. We will see that these stars are in a common evolutionary phase.

23. The properties of the stars are connected by the Stefan-Boltzmann relation: L = 4 π R 2 σ T 4 We can express a star's properties in terms of the Sun: L/L ⊙ = (R/R ⊙ ) 2 (T/T ⊙ ) 4 or (R/R ⊙ ) = [ L/L ⊙ (T/T ⊙ ) -4 ] 1/2 For stars on the main sequence: M-star, T = 3000, L/L ⊙ = 10 -3, thus R/R ⊙ = 0.12 A-star, T = 10,000, L/L ⊙ = 100, thus R/R ⊙ = 3.4 O-star, T = 40,000, L/L ⊙ = 10 5, thus R/R ⊙ = 6.6

24. Main Sequence: The main sequence ranges from hot, luminous and large stars to the left to cool, low luminosity, small stars to the right. The sizes range from about 0.1 solar radii for M main sequence stars to about 10 solar radii for O main sequence stars.

25. Giant and Supergiant Stars Much rarer than main sequence stars. Giant and Supergiant stars have much more luminosity than main sequence stars, although they have similar surface temperatures. How is this possible ? Answer: Must be physically much larger.

26. Remember that: (R/R ⊙ ) = [ L/L ⊙ (T/T ⊙ ) -4 ] 1/2 For a M giant star: M-star, T = 3000, L/L ⊙ = 100, thus R/R ⊙ = 37 For a A supergiant star: A-star, T = 10,000, L/L ⊙ = 10 5, thus R/R ⊙ = 106 For a M supergiant star: M-star, T = 3000, L/L ⊙ = 10 5, thus R/R ⊙ = 1180 (This is a radius of about 5 AU !!!)

27. M-Giant star: Aldebaran B-Supergiant star: Rigel M-Supergiant stars : Betelgeuse and Antares

28. White Dwarf Stars Not as rare as giants and supergiants, but difficult to detect. White dwarf stars have much less luminosity, although they are the same temperature as main sequence stars. What does this imply about these stars ? White dwarf stars are much smaller than main sequence stars of same temperature.

29. Again remember that: (R/R ⊙ ) = [ L/L ⊙ (T/T ⊙ ) -4 ] 1/2 For a A-spectral type white dwarf star: A-star, T = 10,000, L/L ⊙ = 10 -3, thus R/R ⊙ = 0.01 The Earth has a radius of R/R ⊙ = 0.009, so a white dwarf star has about the same radius as the Earth !!!! As we will see later, these stars have a mass roughly that of the Sun. The relative sizes of the IK Peg A (left), its white dwarf binary companion IK Peg B (center) and the Sun (right).

30. Extra Slides: Allowed orbits for Bohr model of hydrogen The force that binds the electron to the nucleus is the electric force. The electric force in cgs units is given by: q 1 q 2 /r 2, where charge q is in esu (electrostatic units). The force on the electron is e 2 /r 2, where e is the charge on the electron and proton (where e = 4.8x10 -10 esu). Assume the electron moves at a uniform speed in a circular orbit, then the centripetal acceleration is given by v 2 /r (we will derive this later) and therefore the centripetal force is m e v 2 /r (remember Newton's law: F = ma).

31. Thus to orbit the nucleus, the centripetal force must equal the electric force: m e v 2 /r = e 2 /r 2 or v 2 = e 2 /(r m e ) Bohr's quantization: m e v r n = nh/2π, or v = nh/(2π r n m e ). If we apply this quantization to the general expression for v 2, gives: v 2 = e 2 /(r n m e ) = n 2 h 2 /(4 π 2 r n 2 m e 2 ) We can now solve for the radius of the n th orbit: r n = n 2 h 2 /(4 π 2 m e e 2 ), note that r 1 = 5.3x10 -9 cm

32. Need the Energy: The energy is the sum of its kinetic and potential energy: E = KE + PE = ½ m e v 2 - e 2 /r Potential energy: PE = 0 at r = infinite. We know that v 2 = e 2 /(r m e ), inserting in above equation yields: E n = ½ m e v 2 - e 2 /r = ½ e 2 /r – e 2 /r = -½ e 2 /r However we know the allowed radii (r n = n 2 h 2 /(4 π 2 m e e 2 ), so we can solve for the allowed energy levels: E n = -½ e 2 /r = -2π 2 e 4 m e /(n 2 h 2 ) = -2.17x10 -11 ergs/n 2