Monday, January 6, 2020
What Is Blackbody Radiation
The wave theory of light, which MaxwellÃ¢â¬â¢s equations captured so well, became the dominant light theory in the 1800s (surpassing NewtonÃ¢â¬â¢s corpuscular theory, which had failed in a number of situations). The first major challenge to the theory came in explaining thermal radiation, which is the type of electromagnetic radiation emitted by objects because of their temperature. Testing Thermal Radiation An apparatus can be set up to detect the radiation from an object maintained at temperature T1. (Since a warm body gives off radiation in all directions, some sort of shielding must be put in place so the radiation being examined is in a narrow beam.) Placing a dispersive medium (i.e. a prism) between the body and the detector, the wavelengths (Ã ») of the radiation disperse at an angle (Ã ¸). The detector, since itÃ¢â¬â¢s not a geometric point, measures a range delta-theta which corresponds to a range delta-Ã », though in an ideal set-up this range is relatively small. If I represents the total intensity of the fraÃ at all wavelengths, then that intensity over an interval Ã ´Ã » (between the limits of Ã » and Ã ´lamba;) is: Ã ´I R(Ã ») Ã ´Ã » R(Ã ») is the radiancy, or intensity per unit wavelength interval. In calculus notation, the Ã ´-values reduce to their limit of zero and the equation becomes: dI R(Ã ») dÃ » The experiment outlined above detects dI, and therefore R(Ã ») can be determined for any desired wavelength. Radiancy, Temperature, and Wavelength Performing the experiment for a number of different temperatures, we obtain a range of radiancy vs. wavelength curves, which yield significant results: The total intensity radiated over all wavelengths (i.e. the area under the R(Ã ») curve) increases as the temperature increases.This is certainly intuitive and, in fact, we find that if we take the integral of the intensity equation above, we obtain a value that is proportional to the fourth power of the temperature. Specifically, the proportionality comes from StefanÃ¢â¬â¢s law and is determined by the Stefan-Boltzmann constant (sigma) in the form:I ÃÆ' T4The value of the wavelength Ã »max at which the radiancy reaches its maximum decreases as the temperature increases.The experiments show that the maximum wavelength is inversely proportional to the temperature. In fact, we have found that if you multiply Ã »max and the temperature, you obtain a constant, in what is known as WeinÃ¢â¬â¢s displacement law:Ã »max T 2.898 x 10-3 mK Blackbody Radiation The above description involved a bit of cheating. Light is reflected off objects, so the experiment described runs into the problem of what is actually being tested. To simplify the situation, scientists looked at a blackbody, which is to say an object that does not reflect any light. Consider a metal box with a small hole in it. If light hits the hole, it will enter the box, and thereÃ¢â¬â¢s little chance of it bouncing back out. Therefore, in this case, the hole, not the box itself, is the blackbody. The radiation detected outside the hole will be a sample of the radiation inside the box, so some analysis is required to understand whatÃ¢â¬â¢s happening inside the box. The box is filled with electromagnetic standing waves. If the walls are metal, the radiation bounces around inside the box with the electric field stopping at each wall, creating a node at each wall.The number of standing waves with wavelengths between Ã » and dÃ » isN(Ã ») dÃ » (8Ãâ¬ V / Ã »4) dÃ »where V is the volume of the box. This can be proven by regular analysis of standing waves and expanding it to three dimensions.Each individual wave contributes an energy kT to the radiation in the box. From classical thermodynamics, we know that the radiation in the box is in thermal equilibrium with the walls at temperature T. Radiation is absorbed and quickly reemitted by the walls, which creates oscillations in the frequency of the radiation. The mean thermal kinetic energy of an oscillating atom is 0.5kT. Since these are simple harmonic oscillators, the mean kinetic energy is equal to the mean potential energy, so the total energy is kT.The radiance is related to the energy d ensity (energy per unit volume) u(Ã ») in the relationshipR(Ã ») (c / 4) u(Ã »)This is obtained by determining the amount of radiation passing through an element of surface area within the cavity. Failure of Classical Physics u(Ã ») (8Ãâ¬ / Ã »4) kTR(Ã ») (8Ãâ¬ / Ã »4) kT (c / 4) (known as the Rayleigh-Jeans formula) The data (the other three curves in the graph) actually show a maximum radiancy, and below the lambdamax at this point, the radiancy falls off, approaching 0 as lambda approaches 0. This failure is called the ultraviolet catastrophe, and by 1900 it had created serious problems for classical physics because it called into question the basic concepts of thermodynamics and electromagnetics that were involved in reaching that equation. (At longer wavelengths, the Rayleigh-Jeans formula is closer to the observed data.) PlanckÃ¢â¬â¢s Theory Max Planck Planck suggested that an atom can absorb or reemit energy only in discrete bundles (quanta). If the energy of these quanta are proportional to the radiation frequency, then at large frequencies the energy would similarly become large. Since no standing wave could have an energy greater than kT, this put an effective cap on the high-frequency radiancy, thus solving the ultraviolet catastrophe. Each oscillator could emit or absorb energy only in quantities that are integer multiples of the quanta of energy (epsilon): E n Ã µ, where the number of quanta, n 1, 2, 3, . . . Ã ½ Ã µ h Ã ½ h (c / 4)(8Ãâ¬ / Ã »4)((hc / Ã »)(1 / (ehc/Ã » kT Ã¢â¬â 1))) kT e Rayleigh-Jeans formula Consequences quantum physics photoelectric effect , by introducing his photon theory. While Planck introduced the idea of quanta to fix problems in one specific experiment, Einstein went further to define it as a fundamental property of the electromagnetic field. Planck, and most physicists, were slow to accept this interpretation until there was overwhelming evidence to do so.