hw9_solution

Hw9_solution - E photon = Δ E =-13.6eV(1/3)^2 –(1/7)^2 = 1.23eV and = hc E photon = 1240 eV nm/1.23 eV = 1008 nm ~ 1010nm(not sure what’s up

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Physics 3C Webassign Homework 9 Solutions: E = -(13.6eV)/n 2 A) The electrons always end at n=2. Longest wavelength will be the lowest energy, or the shortest fall (from n=3) E photon = Δ E = 13.6eV( (1/2)^2 – (1/3)^2) = 1.89eV and λ = hc/ E photon = 1240 eV nm/1.89eV = 656nm B) Shortest wavelengths have the longest fall, from n=infinity E photon = Δ E = 13.6eV( (1/2)^2 – 0) = 3.4eV and λ = hc/ E photon = 1240 eV nm/3.4eV = 364nm
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Unformatted text preview: E photon = Δ E = -13.6eV( (1/3)^2 – (1/7)^2) = 1.23eV and λ = hc/ E photon = 1240 eV nm/1.23 eV = 1008 nm ~ 1010nm (not sure what’s up with the rounding in the answer key) b) The momentum of photon is p = h/ λ momenum is conserved, so momenum of atom is p = m v where m=1.67e-27 kg (proton mass) We have v = h/(m λ ) = (6.626e-34 J s)/((1.67e-27kg)*(1.01e-6m)) = 3.95e-1 m/s...
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This note was uploaded on 09/08/2008 for the course PHYS 3B taught by Professor Wu during the Spring '08 term at UC Irvine.

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Hw9_solution - E photon = Δ E =-13.6eV(1/3)^2 –(1/7)^2 = 1.23eV and = hc E photon = 1240 eV nm/1.23 eV = 1008 nm ~ 1010nm(not sure what’s up

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