hw6_solutions - PHY 251 Fall 2009: homework problem set 6,...

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Unformatted text preview: PHY 251 Fall 2009: homework problem set 6, due in the PHY 251 drop box in room A-129 by noon on Friday, Oct. 23. 1. Serway 5.20 Answer: If the wavelength is uncertain to Δ λ/λ = 10- 6 , that means that after 10 6 waves we don’t know if we have half a wave or a full wave left over. That means the position is defined to 10 6 λ or 10 6 · (6000 × 10- 10 m ) or 10 6 · (6 × 10- 7 m ) or 0.6 meters. 2. Serway 5.25 Answer: From the energy-time version of the uncertainty principle, we can associate a mean lifetime τ as giving an energy width of Δ E ≈ ¯ h 2 τ = h 4 πτ = (6 . 63 × 10- 34 J · s )(1 eV / 1 . 6 × 10- 19 J ) 4 π · 10- 10 s = 3 . 3 × 10- 6 eV . This is too small an energy blur to measure with a detector with 5 eV energy resolution. 3. Serway 5.26. Answer: The full-width half-maximum distribution (FWHM) is about 5 bins across, or FWHM= 5 · (25 MeV ) = 125 MeV /c 2 or in terms of energy E = mc 2 a FWHM value of 125 MeV. Now this looks like a Gaussian distribution of the form exp[ − ( x − x ) 2 / (2 σ 2 )] . The half-width at half max point is x HW such that the Gaussian has a value of 1/2, or (measuring distances x from the Gaussian center x ) we have exp( − x 2 HW 2 σ 2 ) = 1 2 ⇒ exp( x 2 HW 2 σ 2 ) = 2 x 2 HW 2 σ 2 = ln 2 ⇒ x HW = √ 2 ln 2 σ FWHM = 2 x HW = 2 √ 2 ln 2 σ = 2 . 35 σ so we’ll characterize the 1 σ uncertainty in energy as Δ E = σ = FWHM 2 . 35 = 125 MeV 2 . 35 = 53 MeV . The lifetime is thus Δ t = ¯ h 2Δ E = 6 . 58 × 10- 16 eV · sec 2(53 × 10 6 eV ) = 6 . 2 × 10- 24 seconds where this Δ t represents a Gaussian σ value; the FWHM lifetime is 2 . 35 σ or 14 . 57 × 10- 24 seconds or 1 . 5 × 10- 23 seconds when we use the sensible number of digits of precision based on our rough estimate of FWHM in mass. 4. Serway 5.32 Answer: The frequency comes from E = hf , so f = 1 . 8 eV 4 . 1 × 10- 15 eV · s = 4 . 4 × 10 14 Hz . 1 The wavelength is λ = hc/E = (1240 / 1 . 8) = 680 nm which is red. The energy uncertainty is Δ E ≈ h 4 πτ = 4 . 1 × 10- 15 4 π · 2 . × 10- 6 = 1 . 6 × 10- 10 eV 5. Serway 6.8 Answer: The velocity v is v = (0 . 100 × 10- 9 m yr ) · 1 yr 365 d · 1 d 24 h · 1 h 3600 s = 3 . 17 × 10- 18 m/s ....
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This note was uploaded on 05/28/2011 for the course PHY 251 taught by Professor Rijssenbeek during the Fall '01 term at SUNY Stony Brook.

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hw6_solutions - PHY 251 Fall 2009: homework problem set 6,...

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