hw5_solutions

# hw5_solutions - PHY 251 Fall 2009: homework problem set 5,...

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PHY 251 Fall 2009: homework problem set 5, due in the PHY 251 drop box in room A-129 by noon on Friday, Oct. 16. 1. A hydrogen atom is photoexcited from the n = 1 state to the n = 20 state. Calculate the photon energy needed to do this, and the photon energy needed to instead ionize the atom completely. Answer: The energy needed to go from state n i to state n f is Δ E = E n f E n i = Z 2 E 0 p 1 n 2 f 1 n 2 i P = Z 2 E 0 p 1 n 2 i 1 n 2 f P . With hydrogen ( Z = 1) and n i = 1 and n f = 20 , we have Δ E = 1 2 (13 . 60 eV ) ± 1 1 2 1 20 2 ² = (13 . 60 eV ) 399 400 = 13 . 57 eV . If instead n f → ∞ we have 1 /n 2 f 0 and Δ E = 13 . 60 eV. 2. The Lyman alpha line refers to the n = 1 to n = 2 transition in hydrogen, corresponding to the absorption of a photon. For light from a distant star, we see an absorption dip due to hydrogen in the periphery of the star at a wavelength of 421 nm. How fast is the galaxy moving away from us? Answer: In our reference frame, the wavelength of the Lyman alpha line in hydrogen ( Z = 1 ) is λ = hc Δ E = hc E 0 (1 /n 2 i 1 /n 2 f ) = hc E 0 (1 / 1 2 1 / 2 2 ) = 1240 eV · nm (13 . 60 eV )(3 / 4) = 121 nm . Now the relativistic Doppler shift for a receding source ( θ = 0 ) is ν = ν 0 / [ γ (1 + β )] . Let’s define x ν 0 ν = λ λ 0 = 421 nm 121 nm = 3 . 48 and solve for β : x = γ (1 + β ) = 1 + β 1 β 2 x 2 = (1 + β ) 2 (1 β )(1 + β ) = 1 + β 1 β x 2 x 2 β = 1 + β x 2 1 = β ( x 2 + 1) β = x 2 1 x 2 + 1 = (421 / 121) 2 1 (421 / 121) 2 + 1 = 0 . 847 3. Calculate the wavelengths of the ground state to first excited state transition for hydrogen, deuterium, and tritium. Do this with enough precision to show the differences accurately. Answer: The Bohr energy with a reduced mass of m r = m e M/ ( m e + M ) is E n = Z 2 n 2 m r e 4 8 h 2 ǫ 2 0 = Z 2 n 2 m e e 4 8 h 2 ǫ 2 0 m r m e = Z 2 n 2 E 0 m r m e . 1

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and find in turn that the wavelength corresponding to the transition from n i = 1 to n f = 2 with Z = 1 is given by λ = hc Δ E = hc E 0 (1 /n 2 i 1 /n 2 f ) = hc E 0 (1 / 1 2 1 / 2 2 ) = 1240 eV · nm (13 . 60 eV )(3 / 4) = 121 nm as we showed in problem 2. Now since λ 1 /E 0 , and we replace E 0 with E 0 ( m r /m e ) , we see that we need to scale the wavelengths by m e /m r . With nuclear masses M much larger than the electron mass m e , we can find m e m r = m e ( m e + M ) m e M = m e + M M = (1 + m e M ) and we can come up with numerical values for hydrogen, deuterium, and tritium using Appendix B on page A.2 of Serway and the atomic mass unit for an electron of m e = 5 . 486 × 10 4 u: Hydrogen: (1 + m e M ) = (1 + 5 . 486 × 10 4 1 . 007825 ) = 1 + 0 . 00544 so Δ λ = (121 nm ) · (0 . 00544) = 0 . 066 nm Deuterium: (1 + m e M ) = (1 + 5 . 486 × 10 4 2 . 014102 ) = 1 + 0 . 000272 so Δ λ = (121 nm ) · (0 . 00544) = 0 . 033
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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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hw5_solutions - PHY 251 Fall 2009: homework problem set 5,...

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