1. Emission Series and Emitting Quantum States: Visible H Atom Emission Spectrum Experiment 6

2. #6 Emission Series and Emitting Quantum States: Visible H Atom Emission Spectrum Goal: To determine information regarding the quantum states of the H atom Method: Calibrate a spectrometer using He emission lines Observe the visible emission lines of H atoms Determine the initial and final quantum states responsible for the visible emission spectrum, as well as the Rydberg constant

3. Electromagnetic Radiation Oscillating electric and magnetic fields

4. Light Energy  Wavelength distance peak-to-peak  Frequency oscillations per second  EnergyE  faster oscillation = more E

5. Electromagnetic Spectrum

6. Visible Emission Wavelengths,, increase Energies decrease Electronic transitions (“e - jumps”) 400 nm 500 nm 600 nm 700 nm

7. Dual Nature of Light/Relationships h Planck’s constant = 6.626×10 -34 J. s Units J = (J. s) (s -1 ) c speed of light = 2.998×10 8 m. s -1 Units s -1 = (m. s -1 )/(m) 1.Wavewavelength, frequency, 2.Particlephoton = “packet” E = h

8. Using the Equations (a)Calculate the frequency of 460nm blue light. (b)Calculate the energy of 460 nm blue light.

9. Spectroscopy Spectroscopy:study of interaction of light with matter h :photon 1.Absorption:matter + h → matter* 2.Emission:matter* → matter + h Energy change in matter:  E matter = E h

10. Discrete Energy Levels Observed energy level changes:  E = E h  = E final – E initial Ground state atomAbsorptionEmission

11. “Discrete” Atomic Emission Atomic absorption:electrons excited to higher energy levels Atomic emission:excited electrons lose energy Incandescent Hot Gas Cold Gas Continuous Discrete Emission Discrete Absorption

12. Quantized Energy Levels E h =  E levels  E = E f – E i Absorption: E f > E i Emission: E f < E i

13. Hydrogen Emission Spectrum ©The McGraw-Hill Companies. Permission required for reproduction or display

14. H atom emission 1)Electrical energy excites HH + energy  H* initial quantum state n i = 2, 3, 4, 5, 6, … 2)H* loses energyH*  H + h final quantum staten f = 1, 2, 3, … n f < n i You observe several  E transitions  visible s n i ’slevels > n f n f end at same n f You determine n i ’s and n f

15. Hydrogen Atom and Emission Lyman Balmer Paschen Ground State: n = 1 Excited States: n = 2, 3, 4, …

16. Rydberg Equation A “series” is associated with two quantum numbers: Lyman:n i = 2, 3, 4, …n f = 1 Balmer:n i = 3, 4, 5, … n f = 2 Paschen:n i = 4, 5, 6, … n f = 3 R H = 1.096776 ×10 7 m -1 = 2.180×10 -18 J = 2  e 4 m/h 3 c General transition eq’n: Hydrogen atomic emission lines fit (Rydberg eq’n):

17. Increasing  (decreasing E, smaller  E) n = principal E states (principal quantum #s) Hydrogen Atomic Emission Energy 

18. Part 1 Correlate color with wavelength Use lucite rod 20 nm intervals, 400–700 nm, color Boundary s short, long of max. intensity max Observe Hg atomic emission (handheld specs)

19. Part 2 Calibrate Spectrometer Determine if measured wavelengths are “true”  Use He emission  Record msr for lines  Plot true vs msr  7 or 8 lines

20. Calibration Plot H atom emission: Multiply:  msrd by slope Converts: measured true

21. Part 3 Record H emission s  Record color, msr (3 or 4 lines) color, msr  Determine true true  Calculate E h from true E h Units: E in J h in J. s c in m/s in m

22. Questions/Data Analysis 1)Which set of lines? Balmer or Paschen (n final = ?) 2)What is n initial for each line? 3)What is your experimental R H ?

23. Hydrogen Lines / Analysis Color (nm)  E (J) red6603.0×10 -19 blue-green4904.1×10 -19 blue-violet4304.6×10 -19 violet4104.8×10 -19

24. One way to think about the data Which series are we observing – Balmer or Paschen? Balmer:n f = 23  2, 4  2, 5  2 Paschen:n f = 34  3, 5  3, 6  3 These would be the three lowest energy transitions Example data:

25. Compare calculated ΔE to observed ΔE E H atom  1/n 2 = R H /n 2 so calculate E n=1, E n=2, etc. Find ΔE between levels and compare to observed E’s Experiment matches Balmer best

26. How? Plot  E atom vs. 1/n i 2 Rearranged Rydberg equation fits:

27. Example Balmer Rydberg Plot Slope (~R H ): 2×10 -18 J Close to R H 2.18×10 -18 J x-intercept: ~0.24 Close to 0.25 ~1/2 2

28. Balmer (n f = 2) – plot ΔE vs. 1/n i 2 Good: Slope  –R H x-intercept: 1 2 2 so n f = 2 ~ 0.25 = This plot verifies our data – we observed the Balmer series!

29. An alternative way to analyze the data 1) Data is for: Balmer (n f = 2) or Paschen (n f = 3) Be sure to correct wavelengths (measured  true) 2) Transitions are 3 lowest energy: Balmer (n i = 5, 4, 3) or Paschen (n i = 6, 5, 4)

30. Graphs Prepare two graphs (Balmer and Paschen) x-axis should extend to x-intercept (y = 0) y-axis should be appropriate Draw best-fit straight line Find slope (one should be close to –R H ) Find relative error in experimental R H Match and color to n i and n f

31. Paschen (n f = 3) Not too good Slope ≠ R H x-int. ≠ 1/3 2

32. Balmer (n f = 2) Good: Slope  –R H x-intercept: 1 2 2 so n f = 2 ~ 0.25 =

33. Example Balmer Rydberg Plot Slope (~R H ): 2×10 -18 J Close to R H 2.18×10 -18 J x-intercept: ~0.24 Close to 0.25 ~1/2 2

34. Data Experimental R H :2 ×10 -18 J

35. 1/ vs. 1/n i 2

36. Report Abstract Results 2a: Calibration data and plot 2b: Table Series plot (or Balmer and Paschen plots) depending on your analysis choice R H and error from literature Predicted wavelengths and error Sample calculations of: photon energy and Rydberg slope Discussion/review questions