Experiment 12

Line Spectra

The Rydberg Constant



1) To use the emission spectrum of hydrogen atom in order to verify the relation between energy levels and the photon wavelength, and

2) to calculate Rydberg’s constant, R = 1.097x107m-1.




A computer with the internet connection, a calculator (The built-in calculator of the computer may be used.), a few sheets of paper, and a pencil




When an electron in an atom receives some energy by any means, it moves to a bigger radius orbit whose energy level fits 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 by giving off its excess energy in the form of electromagnetic radiation ( visible light is a small part of the E&M waves spectrum).  Max Planck showed that the frequency of occurrence ( f ) of a particular transition between two energy levels in an atom depends on the energy difference between those two layers.

En - Em = hf


In this formula En is the energy of the n-th level, Em the energy of the m-th level (lower than the n-th level) and  h = 4.14x10-15 eV-sec is the Planck’s constant.  ( f ) is the frequency of the released 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 each possibility, the red arrow shows electron going to a higher energy level, and the black arrows show possible return occurrences.


Hydrogen is the simplest atom.  It has one proton and one electronClick on the following applet for a better understanding of the transitions:   http://www.colorado.edu/physics/2000/quantumzone/lines2.html .  In this 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 whose each circumference is 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.


For hydrogen atom, possible transitions from the ground state (E1) to 2nd state (E2), 3rd state (E3), and 4th state (E4) are shown in Fig. 1.   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 the spectroscope.  Some transitions are weak and result in larger wavelengths in the infrared region that cannot be seen either.  However some intermediate energy transitions 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 at 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 emission spectrum.  For a host 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 it is hot and therefore some dark lines will appear inside the continuous spectrum you were observing before.  The dark lines are called 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 the following link:    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 light emitted from an excited gas through a prism (a triangular piece of glass) causing different colors to separate.  Under the separated colors it show a scale with nm or Angstrom graduations.  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 graph and record them.  If you place the mouse on the second graph an move it, a vertical line appears and helps you locate the peaks exactly and read the wavelength 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 better:               http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/dualnature/discharge/index.html .
  2. Use the formula for Balmer Series above and calculate the Rydberg’s 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 the 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 other atoms have more transitions (spectral lines) in the visible region? Give this an explanation as part of your conclusion.






            Raccepted = 1.097x107m-1




The hydrogen visible wavelengths are:


                        λ62 =   ?   nm,   λ52 =   ?   nm,   λ42 =    ?   nm,  λ32 =   ?    nm.  ( λ62 means the wavelength corresponding to the 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 value of R.  This gives the measured value for R.


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


Conclusion:        To be explained by students.  Also, explain why heavier elements have more transitions visible to us.


Discussion:        To be explained by students