Atomic Hydrogen Lab Report
Chemistry 2500- Exercise C4
The Emission Spectrum of Atomic Hydrogen
Objective The purpose of this lab was to calculate an experimental value for the Rydberg constant and then the ionization energy for the hydrogen atom. These values will be obtained by using a prism spectrograph to measure the wavelength value for a section of the visible line spectrum of atomic hydrogen.
Theoretical Background When H+ combines with an electron it forms it’s excited state, H. This excited atom will release light with a photon of energy as it goes to the lowest possible energy state. The light and photon of energy that are released relate to the difference in energy levels between the ground-state and the excited state. The Rydberg constant …show more content…
is the limiting value for the highest possible wave-number of the photon that can be released from the hydrogen as it returns to it’s lowest possible energy state. The different energy levels that the atom will transition to are referred to as “n,” the quantum number where n=3,4,5… Rydberg was able to take an already developed equation (eq.1) and adjust it so that it can account for every line in the atomic hydrogen spectrum (eq.2). 109,680 represents the Rydberg Constant, RH.
After establishing the RH and the numerical value that is associated with it, another expression was obtained (eq.3) This expression was extremely important because it helped to create another equation (eq.4) that shows how to calculate the difference in energy between any two electronic energy levels. It also showed that it was possible to calculate an experimental value for the Rydberg Constant, this value is represented in the first set of brackets in equation 4.
In this experiment we were able to see the lines that occurred through the emission of light when hydrogen atoms relax to their ground state. This will be viewed using a hydrogen discharge lamp that contains hydrogen at a low pressure. Electrical discharge occurs in the tubes and excites the atoms, the light that they release when the excited atoms return to their lowest energy levels are what we will be measuring using the prism specto-graph. Three different bulbs were examined in this laboratory one containing Helium, one containing Hydrogen, and another containing Mercury. For each bulb at least three colors were identified and the angle was measured. Each line was measured twice and the average of both angles was taken and used as the final angle of diffraction measurement. The first angle was obtained by finding whichever number on the bottom of the scale matched up with the zero on the top of the scale. We then looked to find where the lines first match up perfectly on the scale, this value on the top was then divided by sixty, and added to the first value we found to make our first angle that we recorded. The exact same process was done to find the second angle, however, since the second angle is taken from the other hemisphere, 180 degrees needs to be subtracted from this value before it can be used. After finding the values of these measurements they were then put into two different plots. The first plot angle vs. wavelength yielded a polynomial line of best fit and the second equation, angle vs. 1/wavelength2 produces a linear line of best fit. The linear equation enables us to find the wavelengths from the measured angles because the angle is equal to the product of the slope and the wavelength squared inverse added to the intercept. Given that n1=2 we could assign values to for n2 and calculate the inverse of n22. The template provided in this experiment produced a plot of wave number vrs. 1/n22, the slope of the line of best fit was used to find the RH. Once this value was found the ionization energy can be determined for the ground state n=1 and the first excited state n=2 by solving for lambda (eq.5). Using the solved value for lambda we solved for E (eq.6).
5.) 1/lambda = R ( 1/n2 + 1/n2) 6.)E=hc/lambda
Results Using the wave length and angle data that I collected from appendix III(1) of helium and mercury two graphs were created. After examining this data it appeared that there had been an error occurring when the measurements were taken from helium. This error produced inaccurate graphs with slopes that did not provide sufficient information to calculate the wavelengths of hydrogen and then the Rydberg constant value. By omitting the angles for helium and producing a new chart (appendix III,2), the angle vs wavelength and angle vs 1/wavelength2 plots were much more useful in the experiment.
From the slope of the plot above, the wavelength for each visible line of hydrogen was calculated using the formula Angle=slope(wavelength-2) + intercept. (appendix I,3). The inverse of these values represents the wavenumber which was used in the plot wavenumber vs. 1/n22. The n2 values were assigned to wavelengths as 3,4, and 5 with n1 set as 2. This plot produced a graph where the slope was our experimental Rydberg constant. This value was 108098.35 m-1.
The value of ionization energy is determined by using equation 5 to find lambda (appendix I,4) first for the ground state, n1=1 n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/1 +0) lambda=9.25E-6 m
Equation 6 is then used to find the ionization energy
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/9.25-6 m =2.686E-20J
The same formula is used to find the ionization energy of the first excited state. It is calculated to be 5.372E-21J.
Discussion In the results section of this lab we can see that by measuring the angles of the visible lines of emission we can plot a graph of angle vs wavelength and then a plot of angle vs 1/wavelength2. The values of the mercury and helium wavelengths were provided in already known literature, however by using the equations stated above we were able to find the wavelengths of hydrogen for the angles that we found. A major error seemed to occur in this lab when we measured the angles for helium, so for the purpose of completely this lab successfully and producing relevant numbers the angles for helium were omitted. This source of error could have occurred for a variety of reasons. The first explanation is measuring inaccuracies, the numbers could have been read wrong on the scale or written down wrong after we measured. The second plausible explanation is that there was a problem with the equipment or it was not set properly before we entered the lab. Since helium was removed from data the two plots produced appropriate lines of best fit, the first being polynomial and the second being linear. The linear line of best fit which came from the plot of angle vs 1/wavelength2 was used to find the wavelengths of three visible lines of hydrogen. The violet is known to be 434 nm, and when calculated from the linear equation we found it to be 427nm, this yields a 1.6% error. Blue is known to be 486 nm, the experimental value was 491nm, this yields a 1.0% error. Red is known to be 656 nm and this lab produced a value of 659 nm which yields a 0.46% error. In this lab when observing the hydrogen lamp we could see light with wavelengths that ranged from approximately 420nm to 670nm, for this reason the deuterium atoms were not observed since they have wavelengths generally in the range of 180nm-350. If the commercial lamp that we were using was able to view these wavelengths, however, we would be able to resolve them from the hydrogen lines because they would appear far apart on the spectrum.
Since the values of percent error for the wavelengths that were found are within the acceptable range so these wavelengths are appropriate values to convert into wavenumbers and create a new plot of Wavenumber Vs. 1/n22. The plot of this slope yields the Rydberg constant of 108098.35 m-1. The known value of this is 109680m, giving a percent error of 1.44%. Since the experimental value calculated is within the expected uncertainty range, this value is acceptable. Ionization energy refers to the amount of energy required to remove an electron from an atom. If an electron is at ground state is much more stable and therefore requires more energy to remove from the nucleus. These ionization energies fit this generalization as the energy at ground state is much higher then the energy at the first excited level.
As seen in lab hydrogen lines fall in the visible spectrum from
Not visible 180-
Conclusion
To conclude, once the data of the helium angles was eliminated this lab was successful at calculating an experimental value for the Rydenberg Constant and the Ionization Energy. The Rydenberg Constant was calculate to be 108098.35 m-1, which when compared to the known value produced a percent error of 1.44%. The two values of ionization energy for the ground state and the first excited state were 2.686E-20J and 5.372E-21J.
Appendix I: Calculations
1.) Finding Angle for Each Emission line of Mercury angle+(minutes/30)=final angle 1 angle +(minutes/30)-180=final angle 2 (angle 1+angle 2)/2=average angle a.) 117.0+(9/30)=117.30 296.0+(25/30)-180=116.83 average angle=117.065
b.)122.0+(18/30)=122.60
302.0+(16/30)-180=122.53 average angle=122.565
c.)122.0+(14/30)=122.46 303.0+(18/30)-180=123.60 average angle=123.030
d.)122.0+(22/30)=122.73 303.0+(20/30)-180=123.67 average angle=123.200
2.)Finding Angle for Each Emission line of Hydrogen angle+(minutes/30)=final angle 1 angle +(minutes/30)-180=final angle 2 (angle 1+angle 2)/2=average angle a.) 116.0+(15/30)=116.50 296.0+(14/30)-180=296.47 average angle=116.485
b.) 120.0+(30/30)=121.0 300.0+(27/30)-180=120.90 average angle=120.950
c.) 125.0+(30/30)=126.0 305.0+(28/30)-180=125.93 average angle=125.965
3.) Finding Wavelength of Hydrogen Angle=slope(wavelength-2) + intercept
a.) 116.485= -2.697x10-8(wavelength-2) + 131.2 wavelength= 4.27E-05 cm
b.) 120.950= -2.697x10-8(wavelength-2) + 131.2 wavelength= 4.91E-05 cm c.) 125.965= -2.697x10-8(wavelength-2) + 131.2 wavelength= 6.59E-05 cm
4.)Finding Ionization Energy
a.) For ground State n1=1
n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/1 +0) lambda=9.25E-6 m
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/9.25-6 m =2.149E-20J
b.) For first excited State n1=2 n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/4 +0) lambda=3.7E-5 m
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/3.7E-5 m =5.372E-21J
Appendix II: List of Formulas and Equations Used
5.) 1/lambda = R ( 1/n2 + 1/n2) 6.)E=hc/lambda
Appendix III : Charts and Graphs
Ra 1.)Non-adjusted measured values
Exercise C4 - The Emission Spectrum of Atomic Hydrogen
Colour
Wavelength (cm)
Vernier 1
Final Angle 1
Vernier 2
Mercury
Angle
Minutes
Angle
Minutes
4.36E-05
116.0
26
117.29
296.0
25
5.46E-05
122.0
18
122.60
302.0
16
5.77E-05
122.0
14
122.46
303.0
18
5.79E-05
123.0
22
122.73
303.0
20
Helium
4.47E-05
130.0
29
130.97
310.0
4 5.02E-05
132.0
4
132.13
312.0
28
5.88E-05
134.0
11
134.36
315.0
9
Hydrogen
4.34E-05
116.0
15
116.50
296.0
14 4.86E-05
120.0
30
121.00
300.0
27
6.56E-05
125.0
30
126.00
305.0
28
Final Angle 2
Angle 2 - 180°
Average Angle
1/(wavelength)2
296.83
116.83
117.060
5.265E+08
302.53
122.53
122.565
3.354E+08
303.60
123.60
123.030
3.004E+08
303.67
123.67
123.200
2.983E+08
310.13
130.13
130.550
5.005E+08
312.93
132.93
132.530
3.976E+08
315.30
135.30
134.830
2.897E+08
Wavenumber (cm-1) n2 1/n22
296.47
116.47
116.485
5.309E+08
2.294E+04
3
0.111111111
300.90
120.90
120.950
4.232E+08
2.058E+04
4
0.0625
305.93
125.93
125.965
2.322E+08
1.524E+04
5
0.04
2.) Helium omitted from Data
Exercise C4 - The Emission Spectrum of Atomic Hydrogen
Colour
Wavelength (cm)
Vernier 1
Final Angle 1
Vernier 2
Mercury
Angle
Minutes
Angle
Minutes
4.36E-05
117.0
9
117.30
296.0
25
5.46E-05
122.0
18
122.60
302.0
16
5.77E-05
122.0
14
122.46
303.0
18
5.79E-05
123.0
22
122.73
303.0
20
Hydrogen
4.27E-05
116.0
15
116.50
296.0
14
4.91E-05
120.0
30
121.00
300.0
27
red
6.59E-05
125.0
30
126.00
305.0
28
Final Angle 2
Angle 2 - 180°
Average Angle
1/(wavelength)2
296.83
116.83
117.065
5.265E+08
302.53
122.53
122.565
3.354E+08
303.60
123.60
123.030
3.004E+08
303.67
123.67
123.200
2.983E+08
Wavenumber (cm-1) n2 1/n22
296.47
116.47
116.485
5.493E+08
2.344E+04
3
0.111111111
300.90
120.90
120.950
4.153E+08
2.038E+04
4
0.063
305.93
125.93
125.965
2.304E+08
1.518E+04
5
0.04
3.) Angle vs Wavelength without Helium
4.)Angle Vs. 1/Wavelength2 Without Helium
5.)Wavenumber Vs. 1/n22 Without Helium
References
Pavia, J. Introduction to Spectroscopy, 4th Edition, Nelson. Ottawa, 2009.
Zubrick, J.W. The Chem Lab Survival Manual: A Students Guide to Techniques, Wiley. New York, 1988
Adair, P.L.Molecular Hydrogen Emission. Massachusetts, 1982.
The Emission Spectrum of Atomic Hydrogen
Objective The purpose of this lab was to calculate an experimental value for the Rydberg constant and then the ionization energy for the hydrogen atom. These values will be obtained by using a prism spectrograph to measure the wavelength value for a section of the visible line spectrum of atomic hydrogen.
Theoretical Background When H+ combines with an electron it forms it’s excited state, H. This excited atom will release light with a photon of energy as it goes to the lowest possible energy state. The light and photon of energy that are released relate to the difference in energy levels between the ground-state and the excited state. The Rydberg constant …show more content…
is the limiting value for the highest possible wave-number of the photon that can be released from the hydrogen as it returns to it’s lowest possible energy state. The different energy levels that the atom will transition to are referred to as “n,” the quantum number where n=3,4,5… Rydberg was able to take an already developed equation (eq.1) and adjust it so that it can account for every line in the atomic hydrogen spectrum (eq.2). 109,680 represents the Rydberg Constant, RH.
After establishing the RH and the numerical value that is associated with it, another expression was obtained (eq.3) This expression was extremely important because it helped to create another equation (eq.4) that shows how to calculate the difference in energy between any two electronic energy levels. It also showed that it was possible to calculate an experimental value for the Rydberg Constant, this value is represented in the first set of brackets in equation 4.
In this experiment we were able to see the lines that occurred through the emission of light when hydrogen atoms relax to their ground state. This will be viewed using a hydrogen discharge lamp that contains hydrogen at a low pressure. Electrical discharge occurs in the tubes and excites the atoms, the light that they release when the excited atoms return to their lowest energy levels are what we will be measuring using the prism specto-graph. Three different bulbs were examined in this laboratory one containing Helium, one containing Hydrogen, and another containing Mercury. For each bulb at least three colors were identified and the angle was measured. Each line was measured twice and the average of both angles was taken and used as the final angle of diffraction measurement. The first angle was obtained by finding whichever number on the bottom of the scale matched up with the zero on the top of the scale. We then looked to find where the lines first match up perfectly on the scale, this value on the top was then divided by sixty, and added to the first value we found to make our first angle that we recorded. The exact same process was done to find the second angle, however, since the second angle is taken from the other hemisphere, 180 degrees needs to be subtracted from this value before it can be used. After finding the values of these measurements they were then put into two different plots. The first plot angle vs. wavelength yielded a polynomial line of best fit and the second equation, angle vs. 1/wavelength2 produces a linear line of best fit. The linear equation enables us to find the wavelengths from the measured angles because the angle is equal to the product of the slope and the wavelength squared inverse added to the intercept. Given that n1=2 we could assign values to for n2 and calculate the inverse of n22. The template provided in this experiment produced a plot of wave number vrs. 1/n22, the slope of the line of best fit was used to find the RH. Once this value was found the ionization energy can be determined for the ground state n=1 and the first excited state n=2 by solving for lambda (eq.5). Using the solved value for lambda we solved for E (eq.6).
5.) 1/lambda = R ( 1/n2 + 1/n2) 6.)E=hc/lambda
Results Using the wave length and angle data that I collected from appendix III(1) of helium and mercury two graphs were created. After examining this data it appeared that there had been an error occurring when the measurements were taken from helium. This error produced inaccurate graphs with slopes that did not provide sufficient information to calculate the wavelengths of hydrogen and then the Rydberg constant value. By omitting the angles for helium and producing a new chart (appendix III,2), the angle vs wavelength and angle vs 1/wavelength2 plots were much more useful in the experiment.
From the slope of the plot above, the wavelength for each visible line of hydrogen was calculated using the formula Angle=slope(wavelength-2) + intercept. (appendix I,3). The inverse of these values represents the wavenumber which was used in the plot wavenumber vs. 1/n22. The n2 values were assigned to wavelengths as 3,4, and 5 with n1 set as 2. This plot produced a graph where the slope was our experimental Rydberg constant. This value was 108098.35 m-1.
The value of ionization energy is determined by using equation 5 to find lambda (appendix I,4) first for the ground state, n1=1 n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/1 +0) lambda=9.25E-6 m
Equation 6 is then used to find the ionization energy
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/9.25-6 m =2.686E-20J
The same formula is used to find the ionization energy of the first excited state. It is calculated to be 5.372E-21J.
Discussion In the results section of this lab we can see that by measuring the angles of the visible lines of emission we can plot a graph of angle vs wavelength and then a plot of angle vs 1/wavelength2. The values of the mercury and helium wavelengths were provided in already known literature, however by using the equations stated above we were able to find the wavelengths of hydrogen for the angles that we found. A major error seemed to occur in this lab when we measured the angles for helium, so for the purpose of completely this lab successfully and producing relevant numbers the angles for helium were omitted. This source of error could have occurred for a variety of reasons. The first explanation is measuring inaccuracies, the numbers could have been read wrong on the scale or written down wrong after we measured. The second plausible explanation is that there was a problem with the equipment or it was not set properly before we entered the lab. Since helium was removed from data the two plots produced appropriate lines of best fit, the first being polynomial and the second being linear. The linear line of best fit which came from the plot of angle vs 1/wavelength2 was used to find the wavelengths of three visible lines of hydrogen. The violet is known to be 434 nm, and when calculated from the linear equation we found it to be 427nm, this yields a 1.6% error. Blue is known to be 486 nm, the experimental value was 491nm, this yields a 1.0% error. Red is known to be 656 nm and this lab produced a value of 659 nm which yields a 0.46% error. In this lab when observing the hydrogen lamp we could see light with wavelengths that ranged from approximately 420nm to 670nm, for this reason the deuterium atoms were not observed since they have wavelengths generally in the range of 180nm-350. If the commercial lamp that we were using was able to view these wavelengths, however, we would be able to resolve them from the hydrogen lines because they would appear far apart on the spectrum.
Since the values of percent error for the wavelengths that were found are within the acceptable range so these wavelengths are appropriate values to convert into wavenumbers and create a new plot of Wavenumber Vs. 1/n22. The plot of this slope yields the Rydberg constant of 108098.35 m-1. The known value of this is 109680m, giving a percent error of 1.44%. Since the experimental value calculated is within the expected uncertainty range, this value is acceptable. Ionization energy refers to the amount of energy required to remove an electron from an atom. If an electron is at ground state is much more stable and therefore requires more energy to remove from the nucleus. These ionization energies fit this generalization as the energy at ground state is much higher then the energy at the first excited level.
As seen in lab hydrogen lines fall in the visible spectrum from
Not visible 180-
Conclusion
To conclude, once the data of the helium angles was eliminated this lab was successful at calculating an experimental value for the Rydenberg Constant and the Ionization Energy. The Rydenberg Constant was calculate to be 108098.35 m-1, which when compared to the known value produced a percent error of 1.44%. The two values of ionization energy for the ground state and the first excited state were 2.686E-20J and 5.372E-21J.
Appendix I: Calculations
1.) Finding Angle for Each Emission line of Mercury angle+(minutes/30)=final angle 1 angle +(minutes/30)-180=final angle 2 (angle 1+angle 2)/2=average angle a.) 117.0+(9/30)=117.30 296.0+(25/30)-180=116.83 average angle=117.065
b.)122.0+(18/30)=122.60
302.0+(16/30)-180=122.53 average angle=122.565
c.)122.0+(14/30)=122.46 303.0+(18/30)-180=123.60 average angle=123.030
d.)122.0+(22/30)=122.73 303.0+(20/30)-180=123.67 average angle=123.200
2.)Finding Angle for Each Emission line of Hydrogen angle+(minutes/30)=final angle 1 angle +(minutes/30)-180=final angle 2 (angle 1+angle 2)/2=average angle a.) 116.0+(15/30)=116.50 296.0+(14/30)-180=296.47 average angle=116.485
b.) 120.0+(30/30)=121.0 300.0+(27/30)-180=120.90 average angle=120.950
c.) 125.0+(30/30)=126.0 305.0+(28/30)-180=125.93 average angle=125.965
3.) Finding Wavelength of Hydrogen Angle=slope(wavelength-2) + intercept
a.) 116.485= -2.697x10-8(wavelength-2) + 131.2 wavelength= 4.27E-05 cm
b.) 120.950= -2.697x10-8(wavelength-2) + 131.2 wavelength= 4.91E-05 cm c.) 125.965= -2.697x10-8(wavelength-2) + 131.2 wavelength= 6.59E-05 cm
4.)Finding Ionization Energy
a.) For ground State n1=1
n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/1 +0) lambda=9.25E-6 m
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/9.25-6 m =2.149E-20J
b.) For first excited State n1=2 n2=∞ 1/lambda = R ( 1/n2 + 1/n2) lambda= 1/[R(1/n2 + 1/∞) lambda= 1/[108098.35(1/4 +0) lambda=3.7E-5 m
E=hc/lambda
E=(6.626E-34J*s x 3.00E+08m/s)/3.7E-5 m =5.372E-21J
Appendix II: List of Formulas and Equations Used
5.) 1/lambda = R ( 1/n2 + 1/n2) 6.)E=hc/lambda
Appendix III : Charts and Graphs
Ra 1.)Non-adjusted measured values
Exercise C4 - The Emission Spectrum of Atomic Hydrogen
Colour
Wavelength (cm)
Vernier 1
Final Angle 1
Vernier 2
Mercury
Angle
Minutes
Angle
Minutes
4.36E-05
116.0
26
117.29
296.0
25
5.46E-05
122.0
18
122.60
302.0
16
5.77E-05
122.0
14
122.46
303.0
18
5.79E-05
123.0
22
122.73
303.0
20
Helium
4.47E-05
130.0
29
130.97
310.0
4 5.02E-05
132.0
4
132.13
312.0
28
5.88E-05
134.0
11
134.36
315.0
9
Hydrogen
4.34E-05
116.0
15
116.50
296.0
14 4.86E-05
120.0
30
121.00
300.0
27
6.56E-05
125.0
30
126.00
305.0
28
Final Angle 2
Angle 2 - 180°
Average Angle
1/(wavelength)2
296.83
116.83
117.060
5.265E+08
302.53
122.53
122.565
3.354E+08
303.60
123.60
123.030
3.004E+08
303.67
123.67
123.200
2.983E+08
310.13
130.13
130.550
5.005E+08
312.93
132.93
132.530
3.976E+08
315.30
135.30
134.830
2.897E+08
Wavenumber (cm-1) n2 1/n22
296.47
116.47
116.485
5.309E+08
2.294E+04
3
0.111111111
300.90
120.90
120.950
4.232E+08
2.058E+04
4
0.0625
305.93
125.93
125.965
2.322E+08
1.524E+04
5
0.04
2.) Helium omitted from Data
Exercise C4 - The Emission Spectrum of Atomic Hydrogen
Colour
Wavelength (cm)
Vernier 1
Final Angle 1
Vernier 2
Mercury
Angle
Minutes
Angle
Minutes
4.36E-05
117.0
9
117.30
296.0
25
5.46E-05
122.0
18
122.60
302.0
16
5.77E-05
122.0
14
122.46
303.0
18
5.79E-05
123.0
22
122.73
303.0
20
Hydrogen
4.27E-05
116.0
15
116.50
296.0
14
4.91E-05
120.0
30
121.00
300.0
27
red
6.59E-05
125.0
30
126.00
305.0
28
Final Angle 2
Angle 2 - 180°
Average Angle
1/(wavelength)2
296.83
116.83
117.065
5.265E+08
302.53
122.53
122.565
3.354E+08
303.60
123.60
123.030
3.004E+08
303.67
123.67
123.200
2.983E+08
Wavenumber (cm-1) n2 1/n22
296.47
116.47
116.485
5.493E+08
2.344E+04
3
0.111111111
300.90
120.90
120.950
4.153E+08
2.038E+04
4
0.063
305.93
125.93
125.965
2.304E+08
1.518E+04
5
0.04
3.) Angle vs Wavelength without Helium
4.)Angle Vs. 1/Wavelength2 Without Helium
5.)Wavenumber Vs. 1/n22 Without Helium
References
Pavia, J. Introduction to Spectroscopy, 4th Edition, Nelson. Ottawa, 2009.
Zubrick, J.W. The Chem Lab Survival Manual: A Students Guide to Techniques, Wiley. New York, 1988
Adair, P.L.Molecular Hydrogen Emission. Massachusetts, 1982.