p40_047 - n 1 = 2. So the transition is from the n = 4...

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47. (a) Since E 2 = 0 . 85 eV and E 1 = 13 . 6eV+10 . 2eV = 3 . 4 eV, the photon energy is E photon = E 2 E 1 = 0 . 85 eV ( 3 . 4eV)=2 . 6eV . (b) From E 2 E 1 =( 13 . 6eV) µ 1 n 2 2 1 n 2 1 =2 . 6eV we obtain 1 n 2 2 1 n 2 1 = 2 . 6eV 13 . 6eV ≈− 3 16 = 1 4 2 1 2 2 . Thus, n 2 =4and
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Unformatted text preview: n 1 = 2. So the transition is from the n = 4 state to the n = 2 state. One can easily verify this by inspecting the energy level diagram of Fig. 40-16....
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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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