Quanta 2Black-Body RadiationCrisisEmpirical ResultsDerivation by Max PlanckPHY3101 Modern Physics Lecture Notes Quanta 2Quanta 2Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor.Black-Body RadiationHow do night-vision goggles (or cameras) work? You might know that these devices record infrared light, which is not visible to the naked eye. But why is there infrared light at night, when the sun (and even the moon) is not visible? Perhaps you know that infrared light comes from heated objects, but then why do heated objects emit light? This was the question that 19th century physicists tried to solve using the physics known at the time: electromagnetism and thermodynamics, cornerstones of 19th century physics. They set out to explain the spectrum of emitted electromagnetic radiation from heated bodies, but they failed. In the end, a new quantum, the photon, had to be invoked. It was a subtle realization, however, because the spectrum of radiation emitted by hot bodies does not appear quantized at all. A continuous spectrum of wavelengths is emitted.Black body: an idealized object which absorbs all incident radiation and emits none. It makes the calculations simpler because the black-body radiation spectrum will be independent of the material property.One way to imagine a black body is not think of it as a body at all, but rather a cavity with a small hole. Radiation can make it in through the hole, but after internal reflections, very unlikely to escape (hence it is absorbed). When the cavity reaches thermal equilibrium, the emitted radiation from the hole equals the absorbed radiation. This emitted electromagnetic spectrum is what we wish to explain.We will skip the classical calculation of black-body radiation applying only electromagnetism and thermodynamics to our black-body cavity example. The final result is known as the Rayleigh-Jeans Law:D. Acosta Page 1 1/14/2019dRdI TckT ,b g24 Cavity HolePHY3101 Modern Physics Lecture Notes Quanta 2The intensity I T,b g is the total power radiated per unit area, R, per unit wavelength at a given temperature T. The speed of light, c, arises from electromagnetism, and the Boltzmann constant k  138 1023. J / K arises from thermodynamics. This calculation assumes that the average energy of an atomic oscillator (which generates standing waves in the cavity) is kT. CrisisUsing this classical results, we can calculate the total power radiated per unit areafrom the heated object:R I T d  z  ,b g0301The total power is apparently infinite! This cannot be true, however, by empirical observation. This crisis is known as the Ultraviolet Catastrophe, because the integral diverges at the wavelength goes to zero. Classical physics yielded a nonsense result.Empirical ResultsHere is what the radiation spectrum actually looks like. D. Acosta Page 2 1/14/2019T = 6000 KT = 5000 KT = 4000 KWavelength (nm)IntensityPHY3101 Modern Physics Lecture Notes Quanta 2Before discussing the new physics needed, let’s understand what is known empirically about black-body radiation.1. The total power radiated (integral of intensity over wavelength) increases with temperature, as illustrated in the previous figure. This is summarized in the empirical Stefan-Boltzmann Law:2. The black-body spectrum exhibits a peak at a certain wavelength, and this peak shifts to smaller wavelengths as the temperature increases. In fact, it is inversely related to temperature, as expressed in the empirical Wien’s Displacement Law:The Rayleigh-Jeans Law exhibits no peak at all, and diverges as  0For example, if we consider the sun as a black-body, we can derive its surface temperature. The peak emission wavelength is about 500 nm, so this corresponds to a temperature of about 6000 K. For ordinary temperatures of about 300 K, the peak emission is in the infrared region. That is why night-vision devices work.The universe itself has a temperature remnant of the Big Bang. This emission in the microwave region (1 mm), so corresponds to a cold temperature of 3 K.D. Acosta Page 3 1/14/2019peak m - KT  2 898 103.R T I T d Tbg b g  z    ,. emissivity of body ( = 1 for a black body) W m K Stefan - Boltzmann constant-2 -4485 67 10PHY3101 Modern Physics Lecture Notes Quanta 2Derivation by Max PlanckPlanck was able to derive the mathematical form of the black-body radiation spectrum in 1900 by making a rather radical assumption. He assumed that the energy of the atomic oscillators contained in the black-body was quantized:E nhfn. That is, theenergy depends on an integer n=1,2,3,…, the frequency f, and a new constant h. When anoscillator emits radiation, it corresponds to the energy difference between two energy levels:With this assumption, Planck was able to derive the intensity of the black-body radiation (which we will skip):We can look at the limit of very large wavelengths:lim explim ,   FHGIKJ  hckThckTI TckT124bgSo we recover the Rayleigh-Jeans Law in the limit of large wavelengths, but Planck’s derivation solves the ultraviolet catastrophe. By fitting this formula to the empirical data on black-body radiation, Planck derived a new constant of nature:It is small, so atomic oscillators emit nearly continuous energy; but it is not exactly zero. Nature is quantized at a fundamental level. The quantization is necessary to solve the ultraviolet catastrophe.Using Planck’s formula, it is also possible to derive the Stefan-Boltzmann Law and Wien’s displacement Law (which we will skip).D. Acosta Page 4 1/14/2019E E E h fn n   1 I Tc hehc kT,/bg2 1125Planck’s Radiation Lawh  6 626 1034. J - sPlanck’s