Question

Item 16 A3.13 eV photon is emitted from a hydrogen atom Part A What is the Balmer formula n value corresponding to this emission? AE Submit My Answers Give Up

Please show work, and please be the correct answer. Thank you

Answer 1

The Balmer equation could be used to find the wavelength of the absorption/emission lines and was originally presented as follows (save for a notation change to give Balmer's constant as B):

n2-22

Transition of	3→2	4→2	5→2	6→2	7→2	8→2	9→2	$\infty$→2
Name	H-α / Ba-α	H-β / Ba-β	H-γ / Ba-γ	H-δ / Ba-δ	H-ε / Ba-ε	H-ζ / Ba-ζ	H-η / Ba-η
Wavelength (nm) [2]	656.3	486.1	434.1	410.2	397.0	388.9	383.5	364.6
Energy Difference (eV)	1.89	2.55	2.86	3.03	3.13	3.19	3.23	3.40

The energy transitions are known values from 7 to 2 the energy difference is 3.13 eV

Note that the energy in transitions is

E = 13.6 eV (1/m^2 - 1/n^2)

3.13=13.6 eV (1/2^2 - 1/7^2)

m=2 n=7