1) One of the lines of hydrogen’s emission spectrum results from electrons relax
ID: 946198 • Letter: 1
Question
1) One of the lines of hydrogen’s emission spectrum results from electrons relaxing from the n=3 energy levels to the n=2 energy levels.
a) Using the equation for energy levels in a hydrogen atom (also known as the Rydberg equation), calculate the amount of energy released with this energy jump.
b) Calculate the wavelength (in nm) of the light resulting from the energy jump. Does this match a line that you found in the hydrogen emission spectrum?
2) Within hydrogen’s emission spectrum, there are several spectral lines that you did not see in this experiment. These include the line representing electrons relaxing from the n=3 energy level to the n=1 energy levels. Why didn’t you see this line represented? (Show calculations as evidence.)
Element Visable Color Wavelength (nm) Color (of Peak) Hydrogen Pink 471.1 nm 481.9 nm Light Blue 645.6nm Red/ Orange 660.4nm Red Helium Peach 497.3nm Light Blue 584.1nm Yellow 669.3nm Red 710.3nm Red Krypton Purple 555.2nm Green 588.1nm Yellow 762.7nm Red 816.2nm Whte Mercury Light Blue 386.4nm Purple 418.0nm Indigo 544.7nm Green 579.3nm Yellow Neon Bright Red 588.9nm Orange 670.8nm Red 694.3nm Red 704.9nm RedExplanation / Answer
a)
We know that 1/ = R [ 1/n1^2- 1/n2^2]
= wavelength
R = Rydberg constant = 1.097 x 107 m-1
n1= 2 , n2 =3
E = hc/
= hcR [ 1/n1^2- 1/n2^2]
= (6.626 x 10-34 J.s) (3 x 108 m/s) (1.097 x 107 m-1) [ 1/4-1/9]
= 3.02 x 10-19 J
E = 3.02 x 10-19 J
b) 1/ = R [ 1/n1^2- 1/n2^2]
1/ = (1.097 x 107 m-1) [ 1/4-1/9]
= 6.563 x 10-7 m
= 656.3 x 10-9 m
= 656.3 nm
Therefore, wavelength = 656.3 nm
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