QUANTUM MECHANICS
OLD QM
13. PLANCK IN 1900
All bodies emit EM radiation (‘light’) because of their temperature.
If the body is hotter than the surroundings, it will cool by radiation. (It will
also cool by conduction and convection, but radiation will work even in a vacuum.)
If the surroundings are hotter, it will warm up by absorbing the ambient radiation,
but that doesn't change the fact that it is emitting radiation.
Spectrum. The shape of the spectrum, as determined by experiment, looked like
this...
As the temperature increases, the peak shifts to higher frequency i.e. lower
wavelength: Wien’s Displacement Law. Mathematically,
EXAMPLE: Where is the peak radiation of our bodies?
SOLUTION: Human body temperature is 37°C, or 310K. (The temperature used
in Wien's Law must be in Kelvin.)
Wien’s Law and the shape of the spectrum were known from experiment before
1900. What was not known was why this is how things worked.
In 1900 Planck was trying to calculate this spectrum theoretically. He was trying
to calculate the blackbody spectrum, an idealization of emission from solid objects,
liquids, and dense gases. Imagine a surface that absorbes all light hitting it,
a so-called blackbody. Most real-life surfaces reflect at least some of the light
hitting them, but in the lab a blackbody can be made by using a cavity: a box
completely closed except for a small hole. Any light that enters the hole bounces
around until absorbed. Any light escaping the hole has been reemitted and absorbed
by the walls many times before finally chancing to escape. The spectrum of the
light escaping from such a cavity is what Planck was trying to calculate.
The problem was, the classical laws of physics led to the result that the intensity
went to infinity for short wavelengths. Clearly that can’t be right. This
problem was called the ultraviolet catastrophe. The allowed energies in a classical
system are continuous. That is, there are no restrictions placed on what . Planck
found that he could calculate the correct spectrum if instead the energy were
discrete, i.e. only particular energies were allowed. That is, If , the correct
spectrum is calculated, i.e. it matches the experimental results. He called the
proportionality constant h, now known as Planck’s constant. By matching
the equations with the experimental data, we find . We don’t notice the
effects of this discretization is our everyday lives because h is really, really
small.
This discretization is like the classical theory of standing waves, where (in
one dimension) if the cavity has length , then the allowed wavelengths are such
that
or in terms of frequency,
In a letter to R.W.Wood, Planck described his quantization assumption “an
act of desperation.” “I knew that the problem (of the equilibrium
of matter and radiation) is of fundamental significance for physics; I knew the
formula that reproduces the energy distribution in the normal spectrum; a theoretical
interpretation had to be found at any cost, no matter how high.”
14. WAVE-PARTICLE DUALITY
Einstein’s Nobel Prize was not awarded in recognition of relativity, but
rather for theory regarding the photoelectric effect: the ejection of electrons
from a surface by light hitting that surface. (This is how solar cells work.)
Here's a diagram showing the photoelectric experiment. Light hits a metal (the
anode) that is part of an electric circuit, ejecting an electron. There is another
electrotrode (the cathode) which, if the electron hits, will carry the electron
away (a current, measured in amps as we saw before.)
The KE of the ejected electrons can be determined by turning up the voltage V,
putting a negative voltage on the cathode. If the stopping potential is Vo, the
maximum KE of the electrons is KEmax=eVo. One electron-volt is a unit of energy.
The conversion is .
The wave theory of light cannot explain 3 facts about the photoelectric effect:
(1) The KE of the photoelectrons is independent of the light intensity. In the
wave theory of light, the greater the amplitude of the wave, the greater the
energy or intensity of the wave. (2) There is a characteristic cutoff frequency
below which no electrons are ejected, no matter how intense the radiation. According
to the wave theory, the electron should occur at all frequencies of the light,
provided that the amplitude (i.e. the energy) of the wave is sufficient to wrest
the electron from its atom. (3) There is no time lag between illumination and
ejection, even at very low illumination levels. Especially in a gas, collective
absorption can be ruled out.
Mostly on the basis of these failures, Einstein proposed that light was composed
of particles. These particles are called photons. If the energy of light is concentrated
into small packets, rather than spread out in a wave, then that little packet
must have an energy . This is consistent with Planck’s hypothesis of 1900,
but the conception of where that energy is is radically different: as a bundle
rather than spread out in a wave. This energy equation explains paradoxes (1)
and (2) from above. If one electron is ejected by one photon, then doubling the
intensity (the total energy) means doubling the number of photons, not their
individual energies, therefore the number of electrons ejected (measured as a
current) doubles. If the energy of the individual photons is less than the binding
energy, B (a.k.a. the work function) of the electron to an atom, it doesn’t
matter how many photons there are, the electron isn’t ejected. The small
size of the photon explains paradox (3).
Note that h can be determined as Planck did, by matching the theoretical blackbody
spectrum to the experimentally observed spectrum, and also by predicting the
kinetic energy of photoelectrons -- a totally different phenomenon. The same
h is gotten for both, a strong hint that we’re on to something.
But wait, haven’t we already settled on light as being a wave, based on
Young’s double-slit and other interference experiments, polarization, and
Maxwell’s Equations? That’s what all the physicists thought when
they first heard of Einstein’s photoelectric paper. We’re used to
thinking of a wave as something spread out and a particle as being in one place
at a each moment. But wait again in the equation , the frequency is specifically
a wave property! How could Einstein call light a particle, and then turn around
and accept this equation. This is the wave-particle duality: looking at light
solely as a wave cannot explain the photoelectric effect; viewing it solely as
a particle cannot explain polarization and interference. Somehow, light is both
a wave and a particle, and our discomfort with that is our problem.
Einstein's assertion that light can be thought of as a particle was resisted
by many physicists. Bohr, for one, didn't like it. Another experiment, in 1922,
helped force the acceptance of the particle aspect of light. Arthur Compton scattered
X-rays off unbound electrons, a phenomenon that the wave theory of light could
not account for at all. Light behaved like colliding pool balls, and this phenomenon
came to be called the Compton Effect.
15. THE EMBARASSMENT OF SPECTROSCOPY
Emission spectra and absorption spectra.
The spectrum of hydrogen looks like this...
The spectrum obviously has some regularity. Hydrogen is the simplest of the atoms;
other elements and molecules have much more complex patterns. Each element has
a distinctive fingerprint. The Hydrogen lines converge to 364.6 nm. Balmer, in
1885, found a formula predicting the location of the lines:
where n=3 for H?, n=4 for H?, etc. This is a purely mathematical match to the
experimental data -- no physics was involved.
The then-current models (guesses) of atomic structure (the plum-pudding model,
for instance) couldn’t explain any of this. The electrons embedded in the
pudding (like the mass on a spring) should be able to absorb or emit any frequency
of light. Why the preference for those particular wavelengths? And there was
another serious problem: electrons being accelerated within the atom do not radiate
energy away, like Maxwell's equations says they should. Maxwell's equations were
known to work — in antennas, for instance.
16. BOHR AND HIS ATOM
In 1913 Bohr solved this embarassing situation with his planetary model of the
hydrogen atom. His description of the atom was as follows:
•
An electron in an atom moves in a circular orbit about the nucleus under the
influence of the electrostatic attraction of the (negative) electron and the
(positive) nucleus, obeying the laws of classical mechanics.
•
Instead of the infinity of possible orbits which would be possible in classical
mechanics, it is only possible for the electron to move in an orbit for which
the orbital angular momentum L=mvr is an integral multiple of Planck’s
constant divided by 2??
We can rewrite the velocity of a particle in a circular orbit as the circumference
divided by the period, or circumference times frequency...
• Despite the fact that the electron is constantly accelerating,
the electron moving in this allowed orbit does not radiate and does not
lose energy.
•
A photon is emitted if an electron, initially moving in an orbit of energy
Ei, discontinuously changes its motion so that it moves in an orbit of
lower total energy Ef. The energy of that photon is equal to the difference
in energy between the two orbits, i.e. . Therefore, the frequency of
the emitted radiation is . For Hydrogen, the energies of the orbits are
(The negative sign indicates that the electron is bound to the atom.
Positive E indicates that the electron is free.)
EXAMPLE: An electron in the n=5 orbit of a hydrogen atom changes to the
n=2 orbit. What is the energy, wavelength, and frequency of the emitted
photon?
SOLUTION: First, let's calculate the energy...
Now, the frequency and wavelength...
This is the H? line!
Similarly, if a photon happens by with exactly the right energy, the electron
will absorb that energy and change orbits accordingly.
These are postulates, but together they form a model, which Balmer’s
simple formula did not pretend to do. These postulates are a mixture of classical
(electrostatic centripetal force) and nonclassical physics (quantization).
This fusion is typical of the Old Quantum Mechanics.
17. APPLICATIONS OF SPECTROSCOPY
Uses of spectroscopy: (1) identify the constituents of something without even
coming near it, (2) determine the temperature of something by using Wien’s
displacement law (e.g. temperatures of stars, and optical pyrometer), (3) determine
the radial velocity of an object, using the Doppler shift.
Doppler shift
redshifts and blueshifts
EXAMPLE: the H? line is observed in a stellar spectrum at a wavelength of 437
nm. (Its rest wavelength is 434 nm.) How fast is the star moving relative to
the telescope?
SOLUTION:
away from the ‘scope.
More detailed analysis of a star’s spectrum can reveal the relative percentage
composition, pressure structure, stellar rotation and atmospheric turbulence.
PROBLEMS
1. Your body temperature is usually around 98.6 °F. This is 37 °C.
What is the peak wavelength that your body is radiating, what is the frequency
of that radiation, and what is the energy (in J and eV) of a single photon
of this light?
2. The burner on your electric stove glows a deep red. Let’s say that
the peak of the radiation is at a wavelength of 0.7 ?m. What is the frequency
of that 0.7 ?m light, and what is the energy (in J and eV) of a single photon
of this light?
3. The surface of the Sun emits a spectrum with peak wavelength around 510
nm, whereas the star Polaris (the North Star) has a peak wavelength of 350
nm. What frequencies do these wavelengths correspond to, and what are the energies
(in eV and J) of those photons? What are the surface temperatures of the Sun
and Polaris?
4. In a thermonuclear explosion, the fireball can be as hot as 107 K. What
is the wavelength at which the radiation emitted is a maximum, and what is
the energy (in J and eV) of one of these photons?
5. The work function for an electron to be ejected from a potassium plate is
2.1 eV. For each of the following light sources, is the light sufficient to
eject an electron from potassium (based on the results of the previous four
problems) and, if it is, with what kinetic energy (in J and eV) is it ejected?
The light sources are: your body, stove burner, Polaris, the Sun, a thermonuclear
fireball.
6. What is the de Broglie wavelength for...
(a) a baseball moving at a speed of 10 m/sec? (A baseball has a mass of 0.15
kg.)
(b) an electron with KE of 50 eV? (The mass of the electron is 9.11x10-31kg.)
7. Calculate the wavelengths (in nm) and frequencies of the photon emitted
by a hydrogen atom whose electron jumps from...
(a) n=2 to n=1 (c) n=4 to n=2
(b) n=3 to n=2 (d) n=7 to n=6
Also, for each photon, tell me whether it is in the UV, visible, IR, or radio
parts of the spectrum.
8. The spectra of two stars are obtained with a spectroscope attached to a
telescope. Ignoring absorption lines, the peak brightness of the spectrum for
star A is at 390 nm, and the peak for star B is at 720 nm. Also, the H? absorption
line (rest wavelength 656.28 nm) is observed in Star A’s spectrum at
a wavelength of 656.99nm; the same line in Star B’s spectrum is at 655.22nm.
Calculate the temperatures of the two stars, and their radial velocities. Make
sure you indicate whether the velocity you calculate is towards or away from
the Earth.
9. A clever lawyer argues that his client was wrongfully accused of running
a red light. Although the stoplight was red ( ), it appeared green ( ) to the
client because of the Doppler shift. Therefore, goes the argument, the client
should not be held responsible. Is this reasonable? How fast would the driver
have to be going for this to work?
10. On the back of this page is (a drawing of) the Sun’s spectrum, plus
the spectra of several gases obtained in a lab, using discharge tubes as we’ve
seen in class.
(a) For each of the lines marked A through S on the solar spectrum, list the
element or elements that might have produced it.
(b) In answering the above question, you should have noticed that some elements
can produce lines at the same wavelengths. If this is so, how can we tell which
element is responsible for a particular line? Using this method, which of the
elements are really present in the Sun?
(c) As we have seen previously, the temperature of the Sun’s surface
is nearly 6000K, which is much too hot for molecular oxygen (O2) to exist there.
(Molecules break into their constituent atoms at high temperatures.) Why do
you think this element showed up in the solar spectrum?
11. Next to each type of spectral feature listed, put the letter corresponding
to the physical information that feature provides about the object (e.g. a
star) emitting the light being analyzed.
? where the brightness is greatest. A. Size
pattern of absorption/emission lines B. Temperature
shift of absorption/emission lines C. Radial velocity
D. Spin
E. Composition
F. Age
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