Experiment 12

The Line Spectra

(Rydberg Constant)



The objectives are (1) to use the emission spectrum of hydrogen atom in order to verify the relation between energy levels and photon wavelength, and (2) to calculate Rydberg Constant: R = 1.097x107m-1.




   A computer with Internet connection, a calculator, a few sheets of paper, and a pencil




   When an electron in an atom receives some energy by any means, it moves to a greater radius orbit with an energy level equal to that electron's energy.  Such atom is then said to be in an excited state.  The excited state is unstable however, and the electron returns to lower levels (orbits) by giving off its excess energy in the form of electromagnetic radiation (a photon of light).  Max Planck showed that the frequency f of a particular transition between two energy levels depends on the energy difference between those two levels given by

En - Em = hf


wherEn is the energy of the n-th level, Em the energy of the m-th level (lower than n) and  h = 4.14x10 -15 eV-sec. is the Planck's constant.   f  is the frequency of the emitted photon.



   Possibilities for the occurrence of electron jump from one level to other levels are numerous.  It depends on the amount of energy an electron receives.  An electron can get energized when a photon hits it, or is passed by another more energetic electron that repels it, or by any other means.  The electron return can occur in one step or many steps depending on the amount (s) of energy it loses in different steps.  In the above figure, for each possibility, the red arrow shows an electron going to a higher energy level, and the black arrows show possible return occurrences.


   Hydrogen is the simplest of atoms.  It has one proton and one electron.  This means that although there are so many transitions associated with that single electron of each hydrogen atom, heavier atoms that contain more electrons will have much more possible transitions.


Note: There are three new links that need to be added to the JAVA safe list:

          1) http://www.colorado.edu/,

          2) http://mo-www.harvard.edu/, and

          3) http://online.cctt.org/.


   Click on the following link for a better understanding of the transitions:

http://www.colorado.edu/physics/2000/quantumzone/lines2.html . 


   In the above applet, if you click on a higher orbit than where the electron is orbiting, a wave signal must be received by the electron (from outside) to give it energy to go to that higher level.  If the electron is already in a higher orbit and you click on a lower orbit, then the electron loses excess energy and gives off a wave signal before going to that lower orbit.


   Also click on the following link:   http://www.walter-fendt.de/ph14e/bohrh.htm   and try both options of "Particle Mode" and "Wave Mode".  You can put the mouse on the applet near or exactly on any circle and change the orbit of the electron to anywhere you wish; however, there are only discrete orbits with circumferences that each can accommodate an integer multiple of a certain wavelength.  It is at those special orbits that the applet shows principal quantum numbers for the electron on the right side of the applet.


   In Fig. 1, possible transitions from ground state E1 to other states are shown for hydrogen atom.  This simply means:  E1 to E2,   E1 to E3,   E1 to E4,  and so on.  


   The possibilities for electron return are also shown.  The greater the energy difference between two states, the more energetic the released photon is when an excited electron returns to lower orbits.  If the return is very energetic, the wavelength may be too short to fall in the visible range and cannot be seen in spectrometer.  


   On the other hand, some transitions are weaker and result in larger wavelengths in the infrared region that cannot be seen either.  However, there are some intermediate transitions that fall in the visible range and can be seen.


Grouping of the Transitions:


Transitions made from higher levels to the first orbit form the Lyman Series.




Transitions made from higher levels to the second orbit form the Balmer Series.



Transitions made from higher levels to the third orbit form the Paschen Series.




Transitions made from higher levels to the fourth orbit form the Pfund Series.




Emission and Absorption Spectra


   A hot gas emits light because of the energy it receives by any means to stay hot.   As was mentioned earlier, the received energy by an atom sends its electrons to higher levels, and in their returns, the electrons emit light of different wavelengths.  The emitted wavelengths can be observed in a prism spectrometer in the form of a few lines of different colors.  Each element has its own unique spectral lines that can be used as an ID for that element.  Such spectrum coming from a hot gas is called the "emission spectrum."  For a hot gas spectral lines are discrete.


   For white light entering a spectrometer the spectrum is a continuous band of rainbow colors.  This continuous band of colors in a spectrometer ranges from violet to red and gives the following colors violet, blue, green, yellow, orange, and red.  Light emitted from the Sun contains so many different colors (or electronic transitions) that its spectrum gives variety of colors changing gradually from violet to red It contains so many different violets, blues, greens, yellows, oranges, and reds that it appears continuous. 


   When white light passes through a cold gas, that cold gas absorbs only a certain energy-level photons and lets the remaining photons pass through.  The cold gas absorbs only those photons that it can emit when it is hot.  Now, if you are observing the continuous spectrum of a streak of sunlight into a spectrometer, and then place a cold tube of a certain gas in the way of sunlight before entering the spectrometer, the cold gas will absorb those photons that it can normally emit when hot and therefore some dark lines will appear inside the continuous spectrum you were observing before.  The dark lines are called the "absorption spectrum."  It is very interesting to see that the absorption (dark) lines happen exactly at wavelengths that the emission lines occur when the gas is hot and emissive itself.




Click on http://mo-www.harvard.edu/Java/MiniSpectroscopy.html .  


   The "Mini Spectroscopy" Applet appears.   At the very top it has a dropdown window that allows you chose different hot gases.  First Select the hydrogen gas.  The very top picture is what you see of hot (excited) hydrogen if you view it through a spectrometer.  All a spectrometer does is that it passes the received beam of light through a prism causing different colors to separate.   It has a scale installed in it that allows the viewer to read the wavelengths of each spectral line in either  nano-meter or Angstrom.   1nm = 10 -9m  and  1Angstrom = 10 -10m.  


This applet is calibrated in nm.  For hydrogen, you should see a red band at about 650nm, a light blue at about 490nm , a dark blue at about 440nm, and a violet color at about 410nm.


1) Read the exact values of the wavelengths from the peaks on the second diagram and record them.  If you place the mouse on the second diagram an move it, a vertical line appears and helps you locate the peaks exactly and read the wavelengths exactly.  If  the top figure does not show you the dark blue and the violet colors clearly, click on the following applet and you will see them more clearly:



2) Use the Balmer Series formula (above) and calculate Rydberg Constant R by using each wavelength you obtained for hydrogen atom.

3) Average the 4 values you obtained for R in the previous step.  This is your measured value of Rydberg Constant.

4) Compare it with the accepted value of R = 1.097x107m-1 by calculating a % error.

5) Click on the 2nd link and observe the emission spectra for other atoms.  Why do other atoms have more transitions (spectral lines) in the visible region? Give this an explanation as part of your conclusion.


Given:  Raccepted = 1.097x107m-1.


Measured The hydrogen visible wavelengths are:


   λ62 =       nm,   λ52 =       nm,   λ42 =        nm,  λ32 =       nm.


  { λ62 means the λ corresponding to transition from n = 6 to n = 2 }.




Use the Balmer Series equation to calculate R for each of the measured wavelengths.  Next, find the average of R values to get your measured R.


Comparison of the Results:  Calculate a % error on R.


Conclusion: To be explained by students.  

Also, explain why heavier elements have more transitions visible to us.


Discussion:  To be explained by students.