Photons and Electrons ©98
Experiment 9
Objective: To determine the energy of photons generated in atoms and to find the
charge-to-mass ratio of the electron.
DISCUSSION:
Diffraction of photons (i.e. light):
A simplified model of the modern concept of the atom is that it consists of a
massive positively charged nucleus about which orbit not-so-massive negatively charged
electrons. Only certain orbits are allowed. Since each of these orbits corresponds to a
definite total energy for the electron involved, when an electron jumps from one orbit to
another, energy must be absorbed or emitted by the atom. If an electron jumps from a
"lower" to a "higher" orbit, energy is absorbed by the atom; and if from higher to lower,
energy is emitted from the atom.
The energy absorbed or emitted can come in a variety of forms. For example, an
electric field can be used to force electrons to a higher orbit. Energy is absorbed from
whatever mechanism is used to generate the field. Once in the higher orbit, the electron
very likely will fall to some lower orbit. The energy can be emitted in the form of light.
This light is visible provided the energy lost by the atom lies in the range of about 2.8
10-19 J to 5.0 10-19 J for each electronic transition. (This is approximately the range of
energies per atom which is involved in chemical interactions. For this reason, many
exothermic chemical reactions, such as the rapid combination of oxygen and carbon to
form carbon dioxide, give off visible light.)
The electronic orbits allowed for a given atom are unique to that atom, and so the
energy of the light given off is peculiar to the atom. The human eye is roughly able to
measure the energy of the light by its color. For example, the energy of a photon of red
light is about 3×10-19 J. On the other hand, the energy of a photon of blue light is about
4.5×10-19 J.
A very accurate way to measure the energy of photons is to allow them to pass
through a diffraction grating. The grating bends or diffracts a small part of the light
incident on the grating into distinct rays, the directions of which depend upon the energy
of the photons in the ray.
The equation which governs this diffraction is usually derived from the model in
which light is represented as an electromagnetic wave rather than as a beam of photons.
It can be shown that the angle of diffraction , as depicted in Fig. 1, is given by
sinθ 
 ,
(1)
a
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where  is the wavelength of the light and a is the distance between adjacent grid-lines of
the grating. Most of the light passes through the grating undiffracted.
Figure 1: The diffraction of a beam of light by a diffraction grating.
Eq. (1) gives what is known as first-order diffraction. The equation governing
second-order diffraction is
2
sinθ 
.
(2)
a
Both first- and second-order diffraction patterns will be visible in the experiment, but the
first order will be much brighter and easier to see. The second order diffraction pattern
will be visible at higher diffraction angles.
The connection between the wave model of light and the photon model is found in
the relationship between the wavelength  of the wave and the energy E of the
corresponding photon. This relationship is that
hc
Ε
,
(3)
λ
where h is Planck’s constant equal to 6.625 10-34 J•s, and c is the speed of light equal to
2.9979 108 m/s.
Charge-to-mass ratio of the electron:
The diffraction grating gives one method by which the electronic structure of
atoms can be observed. In the second part of this experiment, the electron itself is
examined. In particular, the charge-to-mass ratio of the electron is measured.
When an electron moves at a velocity v at right angles to a uniform magnetic field
whose field strength is B, the electron experience a centripetal force which causes the
electron to move in a circle of radius r. The connection between these variables can be
shown to be
Bqr  mv ,
(4)
9-2
where m is the mass of the electron and q is its charge. Of these variables, the speed of
the electron cannot be determined directly. However, if the speed is acquired because the
electron is accelerated from rest though a potential difference V, it can be shown that
v
2Vq
.
(5)
m
Eqs. (4) and (5) can be combined to eliminate the unknown velocity. We can then solve
for the ratio q/m:
q
2V

.
(6)
m B2 r 2
In this experiment, electrons are evaporated from the filament of a vacuum tube
and accelerated through a known potential difference V provided by a direct current
power supply. The electrons strike a phosphorescent screen in the tube, causing the
screen to glow in a certain pattern. When the vacuum tube is placed inside a current
carrying solenoid, the solenoid’s magnetic field interacts with the electrons in the field,
causing the pattern to change in a way which enables one to measure the radius of
curvature r of the circles in which the electrons move. The magnetic field strength B is
proportional to the current in the solenoid. (The constant of proportionality between the
solenoid’s magnetic field and current will be given by the instructor.)
EXERCISES FOR DIFFRACTION:
1. Using the Hg gas tube as a reference, find the value of a for your diffraction grating.
(Mercury has a green line at 5461Å)
2. Determine several wavelengths of light emitted by a different gas tube.
3. Calculate the energies for the photons corresponding to these wavelengths.
4. Determine the percent deviation from the expected values in the photon energies.
(See the Appendix for more accurate values.)
EXERCISES FOR CHARGE-TO-MASS RATIO:
1. With the equipment provided, establish that magnetic field in the solenoid which, for
a given accelerating voltage, causes electrons in the vacuum tube to move in circles
whose radii can be determined. Note the current used to produce this magnetic field
and determine the corresponding value of the magnetic field strength.
2. Determine the charge-to-mass ratio for the electron.
3. Calculate the percent deviation of your measured value from the accepted value
(1.7588 1011 C/kg).
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