Atomic Emission Spectroscopy
You’ve done the experiment and you’ve collected the data. Now, how do you actually
find the Rydberg constant (RH) using your data?
1. Construct a calibration curve from your data for Hg and He:
Table 1: Dr J’s data for the Hg and He emission lines.
Atom
Color
Wavelength, λ, (nm)♣
Distance, x, (cm)*
Hg
Violet
435.8
2.70
He
Violet/Blue
447.3
2.90
He
Green
501.7
3.50
Hg
Green/Yellow
546.0
4.05
He
Yellow
587.8
4.50
He
Red
668.9
5.35
♣This column corresponds to the λ’s listed in the lab manual. *This column corresponds
to the distance data that you measured.
Now construct a plot of λ vs x for your data. There are instructions on using MS Excel to
construct a graph on the laboratory web page.
(http://classes.colgate.edu/jchanatry/chem101/week5.htm).
800.0
Calibration Curve for Monochromator
750.0
700.0
y = 87.91x + 194.11
R 2 = 0.9986
650.0
600.0
550.0
500.0
450.0
400.0
350.0
300.0
0.00
1.00
2.00
3.00
Distance, x, (cm)
4.00
5.00
6.00
Figure 1: Dr. J’s calibration curve for CW’s monochromator.
2. Verification of the Rydberg Constant (RH).
a. Solve for the λ’s for the Balmer series (i.e. hydrogen emission lines), using the
equation of the best-fit line from the calibration curve:
y = 87.91x + 194.11
where y = λ, x = distance that you measured for the Balmer series (remember, use the
equation from your calibration curve for your data).
Table 2: Dr J’s data for the Balmer series.
Distance, x, (cm)*
Wavelength, λ, (nm)
5.35
664.4
3.35
488.6
2.75
435.9
∗ Use your distances and solve for λ. You can check the accuracy of your calculated
wavelengths for the Balmer series by comparing your values with the literature values
(see your textbook).
Okay what’s next??? Let’s review what you’ve recently learned in class. Recall, the
energy levels of the hydrogen atom are given by:
−R
E = 2H ,n = 1,2,3,
n
Using the equation above to calculate the difference between two levels, we get:
−R H −R H
− 2
n 2f
ni
where ni and nf are the initial and final values of n, respectively.
−∆E =
For the Balmer series, the final state is always n = 2, so the equation can be written:
∆E Balmer = −R H
1
+C
n i2
where C is a constant. Yes! An equation for a straight line, we know how to deal with
this – you just need to construct a linear plot that uses all three of your measured lines to
estimate RH .
b. The y-axis: ∆E
Find ∆E using by the λ’s that you calculated for the Balmer series:
∆E = hc/λ
where h is Planck’s constant (6.6261 x 10-34 J•s) and c is the speed of light (2.9979 x 108
m•s-1). *Remember, λ is in nm, so you have to convert to m (1 nm = 1 x 10-9 m).
c. The x-axis: 1/n2
Note: n=3 corresponds to the lowest E transition (the most red λ) where as n=5
corresponds to the highest E transition (the most blue λ).
d. Construct a plot of ∆E vs 1/n2
Table 3: Dr. J’s data for ∆E and 1/n2
Wavelength, λ, (nm)
n
664.4
3
488.6
4
435.9
5
∆E (J)
2.9897 x 10-19
4.0655 x 10-19
4.5575 x 10-19
1/n2
0.1111
0.0625
0.0400
The Rydberg Constant
5.0000E-19
y = -2.21E-18x + 5.44E-19
R 2 = 1.00E+00
4.5000E-19
4.0000E-19
3.5000E-19
3.0000E-19
2.5000E-19
2.0000E-19
0.0300
0.0500
0.0700 2 0.0900
1/n
0.1100
0.1300
Figure 2: Graph of ∆E vs 1/n2So your estimation of the Rydberg constant is the slope of
the best-fit line. From my data, I found RH = 2.21 x 10-18 J. The literature value for RH is
2.178 x 10-18 J. Pretty good for an apparatus that cost only a few dollars and a little bit of
time to construct.