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exam2_fall2009 - PHY 251 Fall 2009 Exam 2(Tuesday November...

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1. Calculate the de Broglie wavelength for an electron with a kinetic energy of 20 eV and 200 keV. Answer: 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. Consider an atom with a nucleus consisting of 2 protons, and instead of an electron a muon (207 times as heavy as an electron, or m μ = 207 m e ). Calculate the energy of the ground state and the first excited state, and the wavelength of light associated with the transition. Answer: The reduced mass is m r = 2 m p · m μ 2 m p + m μ = 2 m p · 207 m e 2 m p + 207 m e = 2 · (939 × 10 3 keV ) · 207 · (511 keV ) 2 · (939 × 10 3 keV ) + 207 · (511 keV ) = 1 . 00 × 10 5 keV . Since E 0 = me 4 / 8 ǫ 2 0 h 2 scales with m , we can scale E 0 with m e to the reduced mass by multiplying by m r /m e . The Bohr energy then becomes E n = Z 2 n 2 E 0 m r m e = 1 n 2 2 2 (13 . 60 eV ) 1 . 00 × 10 5 keV 511 keV = 1 n 2 1 . 065 × 10 4 eV so that the energy of the n = 1 state is 10,650 eV while the energy of the n = 2 state is 2,660 eV. The energy difference is 7,990 eV giving a photon wavelength of
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exam2_fall2009 - PHY 251 Fall 2009 Exam 2(Tuesday November...

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