1. The flames of Romance The flames of Romance Candlelight and Chemistry Molecular spectroscopy and reaction dynamics Arnar Hafliðason April 10 th 2015

2. The beauty of science Lets begin with Richard Feynman The story about the flower and his artist friend

3. You, light up my life What is a candle made of..? What is needed for it to burn..? What is, “to burn”..? Why is fire, yellow.. Why is fire, blue? Let’s gaze into the flames and see what’s cookin’?

4. What is candle made of? Paraffin wax C 31 H 64 Or actually a mixture of long hydro- carbon molecules, i.e. C n H 2n+2, ranging from n=25-40. Paraffin wax is a white or colorless soft solid derivable from petroleum Candle wick, a braided cotton that holds the flame of a candle

5. What is needed for it to burn? Prerequisite is OXYGEN Oxygen around is usually enough, though you might want to add more oxygen to spice things up You need the “spark” for the chemistry to happen The process needs to be exothermic to sustain itself The rate of the chemical reactions needs to be fast enough to keep the process going Some source of hydrocarbons (gas, wax) for the oxygen to react with

6. What is, “to burn”? Exothermic Reaction Energy

7. Energy from exothermic reaction Divided into 2 main groups 1.Kinetic Energy (Vibration, Rotation, translation) E kin = nk B T (Increase in temperature => more energy) HEAT 2.Radiation Energy (emission following e - transfer) E = h  = h(c/  (shorter wavelength => more energy) LIGHT

8. What is, “to burn”? Exothermic Reaction ????? Could this explain the different colors in the fire?? Energy Reactive radicals are formed

9. Why is fire, yellow.. Why is fire, blue? … it depends on what you’re burning … and what the heat is when it’s burning

10. 1)Just gas, no extra oxygen (candle) 2)Gas, and little bit of oxygen 3)Gas, and oxygen 4) Gas, and a lot of oxygen Lets connect a gas- burner: propane gas cylinder and oxygen cylinder

11. Heat from steps 1) – 4) estimated around 1000 K – 3000 K. 1)Yellow because of incomplete com- bustion caused by lack of oxygen. What we see as yellow/white is soot (C n (s)) that is glowing Less combustion – Less heat – less blue Propane/Oxygen 4)Blue because of “complete” combustion caused by abundance of oxygen. Notice the flame is almost clear above the blue inner core More combustion – More heat – more blue

12. Gaze into the flames and see what’s cookin’ Monochromator separates light into wavelengths

13. Experimental setup /Mono- chromator / gas burner /PMT/inlet slit

14. Diffraction grating Source about 4 cm from slit Opening is 5x5 mm Slit settings: 10, 30 and 50 μm

15. Measured emission spectra

16. Radicals and radiation C 2 and OH radicals Emission at 516 nm, the C 2 radical is in excited electronic state, it relaxes to a lower energy state, d 3 Π → a 3 Π, giving of radiation equal to that energy-difference Emission at 308.6 nm, the OH radical is in excited state and is relaxed to the ground state, A 2 Σ + → X 2 Π, giving of radiation equal to that energy-difference

17. Radicals and radiation CH radicals Emission at 430.7 nm, the CH radical is in excited electronic state and relaxes to the ground state, A 2 Δ → X 2 Π, giving of radiation equal to that energy-difference Emission at 389.1 nm, transition: B 2 Σ - → X 2 Π Emission at 314.7 nm, transition: C 2 Σ + → X 2 Π

18. A = 430.7 nm What do I mean by “energy- difference”? h = 430.7 nm h: Planck constant c: speed of light constant : wavelength of light CH Potential curves for the CH molecule B = 389.1 nm C = 314.7 nm = 389.1 nm = 314.7 nm

19. Radicals and radiation Emission observed is found at: 314.7, 389.1 and 430.7 nm for CH, 308.6 for OH and 516 nm for C 2 Human color vision

20. Measured emission spectra

21. Vibrational and rotational structure v’’ 0 1 2 3 4 0 1 2 3 v’ A2ΔA2Δ X2ΠX2Π Measurement

22. Location of vibrational bands nm Measured Calculated Total v’=3  v’’=3 v’=2  v’’=2 v’=0  v’’=0 v’=1  v’’=1

23. Rotational structure for v’=0  v’’=0 transition. 4 6 8 10 14 J’=20 Measurement

24. Spectral simulation at T=3000K Simulation with PGOPHER Measurement Simulation Lambda doubling e-e-

25. Simulation on A 2 Δ → X 2 Π for v‘=0 → v‘‘=0 B’’=14.17 A’’=29.75 D’’=0.00142 B’=14.56 A’=-1.1 D’=0.00152 B’’=14.19 A’’=27.95 D’’=0.00148 B’=14.57 A’=-1.1 D’=0.00146 CalculatedConstants from NIST Minimal adjustment of constants from NIST (National Institute of Standards and Technology). B: Rotational constant D: Centrifugal distortion A: Spin-orbit coupling cm -1 X2ΠX2Π A2ΔA2Δ

26. Few measurements under different circumstances 1.Different quantity of oxygen burned with propane gas, 5 different settings 2.6 different height settings of the slit from the source of the flame

27. 3 cm 3 mm Slit

28. 2x O 2 1900 K 4x O 2 3000 K 3x O 2 2200 K Difference in rotational distribution. More heat is observed with increase in use of oxygen. Simulations:

29. Comparing population distribution. Experiment vs. Boltzmann distribution J=7 J=6 T = 2200 KJ max = 6.74

30. 3x O 2 0 mm 50 mm 15 mm

31. 3x O 2 0 mm 50 mm 15 mm

33. Thank you all for coming

34. Some equations Boltzmann distribution Hönl-London factors

35. Frank-Condon factors. Can calculate transition probability between vibrational states v’ and v’’ v’’ v’ 0 0 1 1 2 2 3 3 4 4

36. Voltage and slit Intensity for different settings was measured Slit settings: 10, 30 and 50 μm Voltage settings: 900 and 1000 V

37. 156 cm-1