# 8.13-14 Experimental Physics I & II Junior Lab

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Information Report
Category:

## Documents

Published:

Views: 0 | Pages: 3

Share
Related documents
Description
MIT OpenCourseWare http://ocw.mit.edu 8.13-14 Experimental Physics I & II Junior Lab Fall 2007 - Spring 2008 For information about citing these materials or…
Transcript
MIT OpenCourseWare http://ocw.mit.edu 8.13-14 Experimental Physics I & II Junior Lab Fall 2007 - Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.14 Junior Lab Data analysis Assignment #2 2/6/2008 Due: at the start of session #4 Objective: Distinguish Physical phenomena from resolution effects Analysis Exercise: Gaussian or Lorentzian ? In the experiments of Dopplerfree, Moessbauer, QIP, and Zeeman you will have to fit line-shapes (or dips). From the Uncertainty Principle we know: ΔE Δt ≥ h(bar)/2, for a wave-packet, which translates into: Γ τ ≥ h for a resonant line with width Γ (FWHM) from a de-excitation exponential lifetime τ. There is an interesting relation between Γ and τ, by the fact, that the energy Fourier transform of an exponential decay in time results into a Lorentzian non-relativisticaly (as mostly the case in 8.14). An easy derivation follows: Emission of a Spectral line is described as a damped oscillator with ω0 represented by a time dependent amplitude: ∞ ∞ C2 f (t ) = C ⋅ e iω 0 t e−γ t with ∫ f (t ) dt = C 2 ∫ e−2γt dt = 2 = 1 hence C = 2γ −∞ −∞ 2γ Complete Amplitude: f (t ) = 2γ e iω 0 t e−γ ⋅ t ∞ γ γ i Fourier transform: F(ω ) = ⋅ ∫ e− iωte iω te −γt dt = 0 π −∞ π ω 0 − ω + iγ γ ∫ 1 F (ω ) = ( ) 2 2 Spectral intensity I(ω) with F ω dt = 1 π (ω 0 − ω ) + γ 2 2 Inserting E= hν =h(bar)ω and Γ= 2γ h(bar) we obtain the probability/energy, the Γ /2π Lorentzian I (E ) = I 0 (E0 − E ) + Γ2 / 4 2 Compare this line shape to a Gaussian of equivalent: FWHM = 2.354 ⋅ σ 1 ⎡ 1 ⎛ E − E ⎞ 2⎤ I (E,E0,σ ) = exp⎢− ⎜ 0 ⎟⎥ you will see a substantial difference. σ 2π ⎣ 2⎝ σ ⎠ ⎦ Bevington, page 33. Also read chapter 9. What is the point? Well, the physically interesting quantity is the “Natural Line Width Γ”, whereas resolution effects like Doppler broadening and/or stochastic errors will make the line-shape appear more Gaussian; – recall the Central Limit Theorem. So, if you want to claim that you measured a Natural Line, better prove that the shape is right. Then you also may state the lifetime of the excited state. Problem: 1) Fit the dataset ‘lineshape1.txt’ available from http://web.mit.edu/8.13/www/handouts.shtml under Ulrich Becker to both a Lorentzian PDF and a Gaussian PDF. Compare the fit results and chi‐squared values to determine the correct fit hypothesis. 2)Fit the dataset ‘lineshapedata.txt’ also available from http://web.mit.edu/8.13/www/handouts.shtml to both a lorentzian and Gaussian PDF’s, this time with the possible addition of DC and linear background terms. What are the chi‐squared values? Which fit hypothesis is justified ? For each case: 1. Produce a plot of the data with error bars, assuming Poisson statistics. 2. Make an educated guess for initial values for a fit to Gaussians plus background. 3. Perform the fit and retain all values. Give a plot with the fit. 4. Make a subtraction plot Data – Fit, to show the residuals. Comments? 5. As a check repeat with different starting values. Compare the results. 6. Optional: Can you give confidence limits for each hypothesis?
Recommended

6 pages

12 pages

18 pages

4 pages

11 pages

15 pages

6 pages

5 pages

9 pages

17 pages

5 pages

20 pages

11 pages

11 pages

6 pages

View more...