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exam2_spring2009

# exam2_spring2009 - Physics 251 exam 2 April 1 2009 You may...

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Physics 251 exam 2, April 1, 2009. You may use a calculator, and the equation sheet that I have provided. Order of atoms in the periodic table: H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg ... 1. In a very hot plasma, a particular atom type is present only in a highly ionized form with just one electron bound per atom. Measurements of the absorption spectrum of the plasma shows wavelengths ranging from 2.53 to 3.38 nm. What type of atom is it? Answer: The shortest wavelength observed corresponds to the largest photon energy ob- served, which is for the transition from n = 1 to n → ∞ . Since in the Bohr model the energy of an electron’s state is E n = - E 0 Z 2 /n 2 , the energy of the n → ∞ state is zero so the photon energy is equal to the energy of the n = 1 state. Therefore E λ = hc λ = E 0 Z 2 1 2 Z = radicalBigg hc λE 0 = radicalBigg 1240 eV · nm (2 . 53 nm ) · (13 . 60 eV ) = 6 . 0 so this is a Z = 6 atomic number atom, or a carbon atom. By the way, I should have simply said a plasma, rather than a very hot one. Cold temperatures correspond to having all atoms in their ground state, while high temperatures correspond to having many atoms in excited states. Generally speaking one needs high temperatures to strip electrons off of atoms, however ... The longer wavelength corresponds to a n = 1 to n = 2 transition with E 2 - E 1 = - E 0 Z 2 ( 1 2 2 - 1 1 2 ) = - (13 . 60 eV )(6 2 )( 1 4 - 1) = 367 . 2 eV or λ = hc/E = 3 . 38 nm but of course there could also be n = 6 to n = 7 transitions, and n = 9 to n = 10 and so on; these transitions would have even lower energies/longer wavelengths ...

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