Part 1



When light from a "white" source, such as the sun, is passed through a prism, it is dispersed into the broad spectrum of colours that we know as the rainbow. There are a number of light sources that do not produce a complete rainbow, but rather a series of sharp lines of very specific wavelengths. One example would be the familiar neon tube. This computerized exercise concerns a similar tube filled with hydrogen at low pressure.

An electric discharge is passed through the hydrogen gas. The electrons which constitute the discharge collide with the hydrogen molecules, H2, and supply sufficient energy to break the H-H bonds. The resulting hydrogen atoms are left in an "excited" state and in losing their excess energy, they emit a bluish glow, which, if it is dispersed by a prism, is seen to consist of four lines in the visible region of the spectrum. A number of other lines can be detected at specific wavelengths in the ulta-violet and infra-red regions of the electromagnetic spectrum using appropriate equipment.

One of the earliest challenges to scientists interested in atomic structure was to explain the appearence of these line spectra. A second challenge was to explain how atoms bond together to form molecules. The first of these problems was addressed by the theory proposed by to Niels Bohr, which, while it does not attempt to provide an explanation of bonding, has proved a vital step in the evolution of the modern theory of atomic structure.


The first steps in the development of the theory of atomic structure were:

  1. The discovery of the electron. The famous experiments of Thompson (the cathode ray tube), and Millikan (the oil drop experiment), characterized the electron as being a light, electrically negative particle.

  2. The discovery of the nucleus. One of the key experiments was performed to Rutherford (the alpha- particle scattering experiment). This experiment indicated that most of the mass of the atom was concentrated in a very small portion of the total volume of the atom, and that this part, called the nucleus, had a positive electrical charge. Rutherford envisioned an atom rather like a planetary system, with the light electrons in orbit around a heavy nucleus.


Bohr added a mathematical framework to this picture of the atom which contained some radical suggestions for explaining the line spectra already mentioned. In its simplest form, the theory treats one electron in an circular orbit. The electron is electrostatically attracted to the nucleus, and this force of attraction is balanced by the centrifugal force due to its circular motion. If the charge on the electron is e, and its mass is m, and if its velocity is v, in an orbit of radius r, about a nucleus with a charge Ze, where Z is the atomic number, which is 1 in the case of hydrogen, then:

Electrostatic attraction: Ze2/r2 =
Centrifugal force: mv2/r

Rearranging this gives a formula for the radius of the orbit:

r = Ze2/mv2 . . . . . . . . . . . .(1)

The major new proposal by Bohr was that the electron's momentum could not take any value, but was "quantized" to an integral multiple of h/2p, where h is a universal constant called Planck's constant. Thus:

Electron Momentum: mvr = nh/2p

where n is a whole number between 1 and infinity called the principal quantum number. Rearranging this gives a formula for the velocity:

v = nh/2pmr

If we substitute for v in equation (1) we obtain:

r = n2h2/4p 2mZe2

In trying to rationalize the line spectra of excited hydrogen atoms, we must consider the energy of the atom, or more specifically the electron. Its total energy is the sum of its electrostatic potential energy, and its kinetic energy, which are given by:

- Ze2/r + mv2/2 . . . . . . . . . (2)

(The first term, the potential energy, is negative because, by convention, the energy associated with a stabilization is negative, and the electron has zero potential energy when it is an infinite distance from the nucleus. The electron, which is attracted to the nucleus is, of course, more stable when it moves closer to the nucleus.)

We can eliminate r and v from equation (2) to obtain:

En = - 2p 2me4Z2/n2h2

En is the energy of the electron in an orbit with quantum number n. An atom will emit light when an electron moves from a higher energy orbit to one of lower energy. The frequency, of the light depends on the energy difference between initial and final orbits according to the Planck equation:

E = hn

The frequency of the light and its wavelength, l, are related by the formula:

n = c/l


The formula for the Energy applies to any atom or ion which has only one electron, ie H, He+, Li2+, B3+, etc, provided the appropriate value of Z is used.

In addition, any calculations must be done using consistent units thus if the charge on the electron is given in electrostatic units (esu), the rest of the units must be based in the "centimetre/gram/second" (c.g.s) system. To use quantities in S.I. units (with the electron charge specified in coulombs) divide by (4pe)2. See the list of constants given at the end.


The following section contains two embedded programs (Java applets) for you to study.


They are associated with questions which you must answer as your laboratory report.

The first applet (below) starts by showing you the complete emission spectrum of hydrogen. Because of the limited resolution of the computer screen, many of the lines appear to run together. In a real laboratory experiment, only portions of the spectrum can be observed at one time using spectrometers with appropriate optics, but the resolution is much better, thus more lines can be resolved. The simulation continues by showing how spectroscopists analysed the complex pattern into five separate series (which bear their names). Finally, you can look at these individual series expanded.

Question 1.

What are the wavelengths for the first four or five lines of each of these series.

You can click the buttons to move backward or forwards through the displays, and then, when you have finished, go on to the questions below.

Question 2.

Which end of each series corresponds to the higher energy light: the higher or lower wavelength?

Question 3.

Which end of each series corresponds to the higher frequency light: the higher or lower wavelength?

At this point you should be wondering about the relationship between the five series of lines and Bohr theory outlined above. The next simulation will ask you to identify spectral lines in terms of the initial and final orbits of the electrons whose jumps produce them. These orbits are chosen by specifying the quantum numbers n. You can do this by trial and error, but the sooner you figure out the answers logically, the sooner you will be able to answer the remaining questions which appear after the applet:

Question 4.

Explain, in words, why the programmer, quite correctly, has shown that the electron moves at different speeds in the different orbits.

(If your computer is slow this may not be evident!)

Question 5.

Explain, in words why the orbits which are shown, are not evenly spaced.

When you have succeeded in correctly predicting the electron jumps, or "transitions" as the spectroscopists call them, of several randomly chosen lines without making a mistake, you have probably figured out the system.

Question 6.

The general form of the equation for the energy of a transition is:

DE = (1/n12-1/n22)

Give the formulae corresponding to each of the five named series according to your results in the second part of the "experiment."

Here are a few last questions:

Question 7.

What is the value of (in cm-1) based on the wavelength of the first line in the Balmer series (which you determined at the beginning of this lab)?

(You may be interested to know that the constant, , known as the Rydberg constant, is one of the most accurately measured constants.)

Question 8.

What value of (in cm-1) do you calculate (using the data given below) for hydrogen and for He+?

Question 9.

Will the lines in the spectrum of He+ appear at longer or shorter wavelengths than the equivalent hydrogen lines?

Procede to Part 2

Some useful constants

In c.g.s Units:

m = 9.1094 x 10-28 g e = 4.8032 x 10-10 esu
h = 6.6261 x 10-27 erg s c = 2.9979 x 1010 cm s-1
p = 3.1416 e = 8.854 x 10-12 C2 m-1 J-1
(1 erg = 1 g cm2 s-2) (1 esu2 = 1 g cm3 s-2)

In S.I. Units:

m = 9.1094 x 10-31 kg e = 1.6022 x 10-19 C
h = 6.6261 x 10-34 J s c = 2.9979 x 108 m s-1