Question

atomic emission spectroscopy report: im having trouble finding the initial n value and the inverse of initial n^2 value. I was given a formula to use( i wrote it on the paper) but im not sure how to plug it in.

SECTION Analysis of the Hydrogen Spectrum. Actual wavelength of the bright green line from the fluorescent lights, (nm) Observed wavelength of the bright green line from the fluorescent lights, (nm) 546 Calibration factor, (nm) Hydrogen Observed Wavelength (nm) Corrected wavelength (nm) Initial n value Corrected wavelength (m) Color hed ABlue 486. Gneen 131.6 Vellow 410. O 10.0nm A10.0x 10 3.00 486.Inm 186.1x 10 431.On 134.0x10 Data to be plotted for graphical method Inverse of initial na value (in decimal form) Inverse of wavelength (m) 2510 1 43M.D10 R. (in m) from the slope of line plotted in the graphical method Percent error (%)

Answer 1

U bene voor I m R = 100 97 X102 nm =2] for Balmer series n=3 &m=2 = 1,097 x162 -1) A = 6560m) B). • n=4 &m=2 I=1.097X102 nm- = 48 6nm l a n=5 &m=2 1 =1.097 xlo ²1-116=ualul on=6 &m=2 t = 1.097 xlo2 (1-1) 7=ulonm | 7 nmi

(n) n=4 It is clear from calculateos that ential values ar ? Ya n=3 6560m allll 4860m 0.0625 nu5 434 nm 0.04 n=6 ulonm 0.0278 Theis initial values and / values were calculated.