Question

-obtain wavelengths for the hydrogen emission lines and then calculate the energy levels according to bohr's theory

- determine wavelengths in nm associated with following transitions of the H atom:

nf = 1 with ni = 2, 3, 4

nf = 2 with ni = 3, 4, 5

nf = 3 with ni = 4, 5, 6

Hydrogen יויויין יויויויויויויוי 400 500 600 700 Fig e: Hydrogen emission spectra for calculations

Answer 1

to Bohr's theory According I = RH ( 11 where RA CRydberg constant) = 10973731 ma! So for ng =1, with mi=2 1 nm = 10973731 (1-4) 4x109 m 3x10973731 :. 4 3 X-10973731 25 = 121 502 nm Similarly for, with ni= 3 n=1 9 71 un 9x10 = 102.52 nm 8xRH ŹR (1-5) for mi = 4 168109 - 97.20 nm 15%RH Ž = RM (+-to): 1 = So the wavelengths of Lyman series are Ans. 121.502 nm, 102.52 nm. 97.20 nm Balmer Series . For inf = 2, with ni: 3 f = Rp. 4) = Rn Ans. 36x10 n'm 5655112 nm 5 RH for ng 4 t = R1 (+ 1/6 16x109. nm ... 3 = RH T6 3* RH 486.000 nm Ans 1 / = R4 (7-12) on =RH 21 25 25*109 21 * RH = 108.484 Ans. nm

PO Paschen Series with For nf ni = 4 w - RH lå - to D = RH x 0.0486 9. I X10 1875.03 mm. (Am) 10973731 x 0.0486 nm nf=3; ni= 5 ; S RH CH 25 ) = R* *0.0711 109 128149 ΟΥ i nm - 10973731 X0.0711 nm. (Am): for nf = 3; ni= 6. RHC 56 ) = RH * 0.08 33. . 9 109. nm 10973731 * 0.0833 109 3.56 nm (Am) Calculation of Energy Levels > Bohr Theory for 1st energy level (E.); nt he.ph lá 3-). So, E = According to ni = 1. he = 13.6 (0-1) er - 13.6 ev.cam) Sam similarly, E₂= hert (az - 22 ) h.c. RH -34 = 6.62x10 X (34100 X 10973731 J -18 = 2.1793 X10''J. 2.1793x1018 er 16*10-19 - 13.6 4 ev 3.4 ev. 13.6 eV (Am).

For, hert Cart - 1/2) 13.6 ev -1511er cam) heRH Eu R+ (a) - 13.6 16 0.85 ev (am] ev ev Es 13.6 25 hert ca 1 / 2 11 -0.544 ev.(10) E6 = herhi (21) = This is a 13.6 36 ev. =-0.378 ev can)