p1046.pdf - Lyman series,ni = 1 in Equation 1.10 Thus 1 I =...

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Unformatted text preview: Lyman series,ni = 1 in Equation 1.10. Thus 1 I = 109 680 (1 — 37) cm" ”2 8 —I 10-91221—- = _ 1 J + 109 680cm" “2 1.03 x 10 » cm ’12 = 2.98 Q5 3 e n must be an integer. e find the value of ’12, the state of the h ydrogen atom that is obtained upon absorption, by quation 1.10 with n, = 1. , I -———~—_ = 109 680 (1 — 7) cm" ”2 1 I 1 M—109680 —--— cm" 97 2 X 10'7 cm (42 I122) r12 = 2 state of the hydrogen atom is n = 2. *‘\“__ that the Lyman series occurs between 91.2 nm and 121.6 nm, that the Balmer curs between 364.7 nm and 656.5 nm, and that the Paschen series occurs between and 1876 nm. Identify the spectral regions to which these wavelengths correspond. gth can be found by taking the smallest n2 allowed +- 1) and the minimum wavelength can be found by using the largest n2 allowed (/12 = 00). n series num wavelength mm wavelength 1 ( l ) —1 1 cm Balmer maximl minimL This C01 Pasche1 maximi minimu This co1 1—24. Calculate Lyman serie We have fou Calculate roton with atom. We use Equ: a. So ...
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