THE ELECTRONIC SPECTRUM OF HYDROGEN - THE BOHR ATOM
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.
A BRIEF HISTORY
The first steps in the development of the theory of atomic structure
THE BOHR THEORY
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 =|
Rearranging this gives a formula for the radius of the orbit:
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:
If we substitute for v in equation (1) we obtain:
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:
(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 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
The frequency of the light and its wavelength, l, are related by the formula:
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 EXPERIMENT The following section contains two embedded programs (Java applets) for you
IF THEY ARE NOT DISPLAYED:
They are associated with questions which you must answer as your laboratory
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.
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.
What are the wavelengths for the first four or five lines of
each of these series.
The following section contains two embedded programs (Java applets) for you to study.
IF THEY ARE NOT DISPLAYED:
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.
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.
Which end of each series corresponds to the higher energy light: the higher or lower wavelength?
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:
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!)
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.
The general form of the equation for the energy of a transition is:
DE = Â (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:
What is the value of Â (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, Â, known as the Rydberg constant, is one of the most
accurately measured constants.)
to Part 2
What value of Â (in
cm-1) do you calculate (using the data given below) for hydrogen
and for He+?
Will the lines in the spectrum of He+ appear at longer
or shorter wavelengths than the equivalent hydrogen lines?
Some useful constants
to Part 2
|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
|(1 erg = 1 g cm2 s-2)||(1 esu2 = 1 g cm3 s-2)|
|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|