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exam3_fall2009

# exam3_fall2009 - PHY 251 Fall 2009 Final Exam(Thursday Dec...

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PHY 251 Fall 2009: Final Exam (Thursday, Dec. 17). Some masses in u: 1 1 H 1.007 825 206 82 Pb 205.974 449 210 84 Po 209.982 857 2 1 H 2.014 102 208 83 Bi 207.979 727 211 84 Po 210.986 637 3 2 He 3.016 029 209 83 Bi 208.980 383 212 85 At 211.990 735 4 2 He 4.002 603 209 84 Po 208.982 416 214 86 Rn 213.995 346 1. Calculate the wavelengths associated with a 100 eV photon, a 20 eV electron, and a 200 keV electron. Answer: The wavelength of the 100 eV photon is λ = hc E = 1240 eV · nm 100 eV = 12 . 4 nm . The 20 eV electron is non-relativistic, so we have E k = p 2 / 2 m and p = 2 mE k giving a wavelength of λ = h p = h 2 mE k = hc 2 mc 2 E k = 1240 eV · nm r 2 · (511 × 10 3 eV ) · (20 eV ) = 0 . 274 nm . The 200 keV electron energy is close enough to the rest mass that we must use a relativistic approach. We know E k = ( γ - 1) mc 2 so γ = 1 + E k mc 2 = 1 + 200 keV 511 keV = 1 . 39 and then from γ 1 / 1 - β 2 we can find β = r 1 - 1 2 = r 1 - 1 / (1 . 39 2 ) = 0 . 695 We can then use λ = h/p and p = γmv = γβmc to find λ = h γβmc = hc γβmc 2 = 1240 eV · nm 1 . 39 · 0 . 695 · (511 × 10 3 eV ) = 0 . 0025 nm . 2. A He + ion has its electron undergo a transition from the 2 p to the 1 s state. What is the mean wavelength of the emitted photon? If the 2 p state were to have a lifetime of 2 × 10 - 14 seconds or 20 fsec, what would be the wavelength spread about the mean? Answer: We have E = - E 0 Z 2 /n 2 and a correction of m r /m e with m r m e = (4 · 939 · 0 . 511) / (4 · 939 + 0 . 511) 0 . 511 = 0 . 99986 which we will ignore. The photon energy is then Δ E = E 2 - E 1 = ( - 13 . 6) 2 2 2 2 - ( - 13 . 6) 2 2 1 2 = 13 . 6(4 - 1) = 40 . 8 eV and the photon wavelength is λ = hc/E = 1240 / 40 . 8 = 30 . 4 nm. The energy width is given by the uncertainty principle E )(Δ t ) = ¯ h/ 2 or Δ E = ¯ h t = 6 . 582 × 10 - 16 eV · sec 2 · (2 × 10 - 14 sec ) = 0 . 016 eV 1

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and since Δ E/E = Δ λ/λ we have Δ λ = λ Δ E E = hc E Δ E E = (1240 eV · nm )(0 . 016 eV ) (40 . 8 eV ) 2 = 0 . 012 nm as the 1 σ spread in wavelength. Note: it is incorrect to say
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exam3_fall2009 - PHY 251 Fall 2009 Final Exam(Thursday Dec...

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