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Published byHans-Petter Gundersen Modified over 2 years ago

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BALMERSERIEN

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n = 1 n = 2 n = 3 n = 4 n = 5 -13,6 eV -3,39 eV -1,51 eV -0,85 eV

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HH n = 1 n = 2 n = 3 n = 4 n = 5 -13,6 eV -3,39 eV -1,51 eV -0,85 eV

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HH n = 1 n = 2 n = 3 n = 4 n = 5 -13,6 eV -3,39 eV -1,51 eV -0,85 eV

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HH n = 1 n = 2 n = 3 n = 4 n = 5 -13,6 eV -3,39 eV -1,51 eV -0,85 eV

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HH n = 1 n = 2 n = 3 n = 4 n = 5 -13,6 eV -3,39 eV -1,51 eV -0,85 eV

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LymanBalmerPaschenLymanBalmerPaschen

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n = 1 n = 2 n = 3 n = 4 n = 5 2000 1000500 100 nm

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100 n = 1 n = 2 n = 3 n = 4 n = 5 2000 1000500 nm 100 Lyman

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100 n = 1 n = 2 n = 3 n = 4 n = 5 2000 1000500 nm 100 500 Lyman Balmer

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100 n = 1 n = 2 n = 3 n = 4 n = 5 1000500 nm 100 5001000 2000 Lyman BalmerPaschen

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100 n = 1 n = 2 n = 3 n = 4 n = 5 1000500 nm 100 500 Lyman BalmerPaschen 1000 2000

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Bohrs atommodell H - linjen

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E1E1 E2E2 E3E3 Graphically solve for E n ’s, now know α n ’s and k n ’s E 1 =0.0675eV –k 1 =1.366E9 –α 1 =4.9656E9 E 2 =0.291eV –k 2 =2.7789E9 –α 2 =4.3298E9.

E1E1 E2E2 E3E3 Graphically solve for E n ’s, now know α n ’s and k n ’s E 1 =0.0675eV –k 1 =1.366E9 –α 1 =4.9656E9 E 2 =0.291eV –k 2 =2.7789E9 –α 2 =4.3298E9.

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