p40_036 - = hc E = 1240 eV nm 1 . 889 eV = 658 nm . (b) or...

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36. (a) The “home-base” energy level for the Balmer series is n = 2. Thus the transition with the least energetic photon is the one from the n =3leve ltothe n = 2 level. The energy diFerence for this transition is E = E 3 E 2 = (13 . 6eV) µ 1 3 2 1 2 2 =1 . 889 eV .
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Unformatted text preview: = hc E = 1240 eV nm 1 . 889 eV = 658 nm . (b) or the series limit, the energy diFerence is E = E E 2 = (13 . 6 eV) 1 2 1 2 2 = 3 . 40 eV . The corresponding wavelength is then = hc E = 1240 eV nm 3 . 40 eV = 366 nm ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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