Chapter Review
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Chapter 30: Quantum Physics Chapter Review |
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Chapter Review
In the previous chapter we discussed the theory of relativity as one of the principle corrections to Newton's laws which become most important for high speeds and strong gravitational fields. In this chapter we discuss the other main correction which becomes important at small size scales; we refer to these results as quantum physics.
30-1 Blackbody Radiation and Planck's Hypothesis of Quantized Energy
One of the first indications of the need for a theory of quantum physics came from the study of the electromagnetic radiation given off by a blackbody. An ideal blackbody is an object that absorbs all of the electromagnetic radiation that is incident upon its surface. This fact means that a blackbody is a perfect absorber of radiation and, in order to maintain thermal equilibrium, a blackbody is also a perfect radiator of electromagnetic energy. One of the most useful properties of blackbody radiation is that the distribution of energy given off only depends on the temperature of the object. In fact, the frequency at which a blackbody gives off the maximum intensity of electromagnetic radiation is directly proportional to the absolute temperature T
fpeak = (1.04 x 10-11s-1 . K-1)T.
This expression is known as Wien's displacement law.
The dependence of the intensity of blackbody radiation on frequency could not be accurately explained by classical physics. In order to reproduce the experimental result, Max Planck hypothesized that the radiant energy in a blackbody must be quantized in integral multiples of the constant h times the frequency
En = nhf, n = 1, 2, 3, ...
where the constant h is called Planck's constant and has the value
h = 6.63 x 10-34 J . s.
Exercise 30.1 Wein's Displacement By how much is the frequency of the most intense light from a blackbody displaced if its temperature increases from 300 K to 380 K?
Solution
Using Wien's displacement law we can see that
f2 - f1 = (1.04 x 1011s-1 . K-1)[T2 - T1] = (1.04 x 1011s-1 . K-1)[80 K] = 8.3 x 1012 Hz
At first sight this may look like an unreasonably large shift, but consider that in this temperature range the frequency peak intensity is already on the order of 1013 Hz.
Physlet Illustration: Blackbody Radiation |
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| This simulation illustrates the fact that the distribution of energy from a blackbody radiator depends only upon its temperature. Consider the circle of light shown above the word "appearance" to be a star seen through a telescope. The three circles above the "star" show the relative energy density observed through red, green, and blue filters. The user may specify the maximum wavelength by clicking on the graph or the temperature by entering in the box above. How can astronomers determine the temperature of stars? | ||
Hints:
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Practice Quiz
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30-2 Photons and the Photoelectric Effect
Planck's idea about electromagnetic radiation being quantized was considered to be a good explanation for blackbody radiation but not a general principle. It was Einstein who proposed that light, in general, comes in bundles of energy, called photons, that obeys Planck's energy quantization result. Einstein argued that the energy of a photon is given by
E = hf.
In addition to providing a more natural explanation of blackbody radiation, Einstein used his photon hypothesis to explain another phenomenon called the photoelectric effect. This effect occurs when light strikes the surface of a metal and ejects electrons. The minimum amount of energy needed to remove an electron from the surface of a metal is called the work function (W0) for that metal. It was observed that no electrons are ejected from a metal unless the frequency of the incident light exceeds a certain cutoff frequency, f0, given by
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It was further found that the maximum kinetic energy of the ejected electrons was independent of the intensity of the incident light and was only a function of its frequency
KEmax = hf - W0..
Neither of these observations could be explained using classical physics. Einstein's photon hypothesis, however, provided a complete and accurate explanation of these results.
Physlet Illustration: Photoelectric Effect |
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In this simulation a photon is incident upon a metal surface. The user may change the metal and vary the energy of the photon between 1.0 and 6.0 eV. What is the work function of the metal? |
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Hints:
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Exercise 30.2 Photons from the Sun: The power output from the sun, called its luminosity, is 3.826 x 1026 W. Estimate the number of photons per second that are leaving the surface of the sun?
Picture the Problem The picture shows the sun with light emanating from its surface.
Strategy Since every photon carries a certain amount of energy, we need to determine how many photons will produce an amount of energy consistent with the sun's luminosity.
Solution
| 1. The amount of energy coming from the sun in one second is: | E = Pt = (3.826 x 1026 J/s)(1.00 s) = 3.826 x 1026 J |
| 2. Most of the sun's energy is radiated near the middle of the electromagnetic spectrum giving an effective frequency of: | feff = 5.45 x 1012 Hz |
| 3. The number n of photons at the above frequency that gives the above energy is: |  |
Insight As you might have expected, a large number of photons are given off in just one second.
Practice Quiz
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