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Atomic Emission Spectra
Part I—Calibration of Spectroscope
Discussion Questions:
Question 1:
It can be said that the lines in the visible region can be deduced using the Balmer series, since it is only
series which can used to see where the lines belong too.
Using Balmer series when m = 2, the n
1
value for
the hydrogen is also 2. For the red spectral line the n
2
value should be n=3, when it is greenblue, n=4, for
the blue, n=5, and violet should be n=6.
Question 2:
Colour
Wavelength (nm)
Wavelength(cm)
Red
681.36
6.8136E5
Yellow
5.75
5.75E5
green/blue
459.74
4.5974E5
Violet
383.756
3.83756E5
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v R1m2 1n2
.
 = (

)
16 8136E 5
R 122 132
.

16 8136E 5
= ( .
)
R 0 1388888888
≈
.
R 105677 77
Question 3:
Rydberg Constant
Equation from the graph:
Y= 321000x + 33200
1/λ = R (1/n
2
– 1/ n
2
2
)
Y = R (0.25 –X)
(Here x= 1/ n
2
2
)
321000x + 33200 = R (0.25 –X)
Let x= 1/ n
2
2
= 1/ 7
2
= 1/49
Thus R = 321000x + 33200 / (0.25 –X)
Calculated R= 176395.55 cm
1
Theoretical R = 1.0973x 10^5 cm
1
% Error:
(1.0973x 10^5 176395.55)) * 100 / 1.0973x 10^5
= 60%
Question 4:
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View Full Document When hydrogen atoms emit light it gives off a bluish colour. There are four lines which the hydrogen
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This note was uploaded on 11/11/2010 for the course CHEMISTRY CHEM1800 taught by Professor Krista during the Winter '10 term at UOIT.
 Winter '10
 KRISTA
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