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The birth of quantum mechanics Until nearly the close of the 19 th century, classical mechanics and classical electrodynamics had been largely successful.

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Presentation on theme: "The birth of quantum mechanics Until nearly the close of the 19 th century, classical mechanics and classical electrodynamics had been largely successful."— Presentation transcript:

1 The birth of quantum mechanics Until nearly the close of the 19 th century, classical mechanics and classical electrodynamics had been largely successful in describing phenomena in the world. Material particles were determinate objects that obeyed the laws of classical mechanics. Electromagnetic waves were traveling waves of electric and magnetic fields, in which the waves were continuous and exhibited phenomena of interference and refraction that could be explained from their wavelength and frequency.

2 In 1890-1910, there were problems! There were situations where electromagnetic waves exhibited properties that should be associated with particles! Black body radiation Photoelectric effect Frank-Hertz experiment Spectra of emission and absorption by atoms There were situations where material particles exhibited properties that should be associated with waves! Electron diffraction

3 We will study several of these paradoxes, and arrive at the wave-particle duality that spawned quantum mechanics: Light behaves like waves much of the time, but like particles some of the time; Material particles behave like particles much of the time, but like waves some of the time. A successful description of both light and matter must somehow weave together both kinds of properties!

4 First let’s review the statistical mechanics of material particles. (section 1.12 in the text) Suppose we want to calculate the dependence of the density of air as a function of altitude on Earth. We can get it by using the ideal gas law, pV=NRT and the principle of detailed balance

5 The ideal gas law relates pressure p (force per unit area) to the number of moles N, the volume V, and the temperature T. R is the universal gas constant. For describing a gas, it is more convenient to use the number density of particles n=N a N/V of particles: where k = 1.4 x 10 -23 J/ o K = 10 -4 eV/ o K is Boltzmann’s constant.

6 How does the atmosphere vary as a function of altitude? 1)A molecule must have work done upon it to elevate by a distance dz: 2)The density of air decreases with altitude: n(z)

7 Detailed Balance We require that the force on each particle of gas be in balance (otherwise it would rise or fall). Consider a horizontal slice of the atmosphere at altitude z: face area A, thickness dz, mass of each particle m Downward force due to gravity pulling on each particle: Upward force due to the pressure difference between the top and bottom of the slice:

8 Now require F up = F down in each slice of the sky: Now connect n(z) and p(z) through the ideal gas law: So the ideal gas law becomes Solution: The air density decreases exponentially, with scale length

9 This result is easily generalized to any situation where material particles are distributed in a volume of space where the potential energy that varies over the region: This is the Boltzmann distribution function. It governs the distribution of particles in the presence of any interaction potential.

10 Black Body Radiation When a material body is heated, it emits electromagnetic radiation with a broad spectrum. Mystery #1. It is observed experimentally that the total intensity (power per unit area) radiated by a black body is determined solely by its absolute temperature. There is no way to explain this result by treating light as a wave!

11 Stephan-Boltzmann law  = 5.7 x 10 -8 W/m 2 / o K 4 Example: A steel rod is heated red hot (T ~ 700 o C ~1,000 o K). The rod is 1 cm diameter and 1 m long. How much power does it radiate as blackbody radiation? P =  T 4 A = (5.7 x 10 -8 W/m 2 /K 4 )(1,000 K) 4 (  x 10 -2 m)(1 m) = 1,800 W.

12 Mystery #2. The spectrum of light from blackbody radiation cannot be explained by assuming that the light is composed of waves. The spectrum of light is the pattern of intensity as a function of wavelength:

13 The spectrum of black-body radiation can be explained (up to a point!) if we consider the radiation to be produced by oscillations of the atoms in the material. The light emitted should be proportional to the number of modes in which the oscillations of a given wavelength can be excited: Consider the modes that can be excited within a cubic cavity of dimension L. #modes in x: N x = L/  modes in y: N y = L   modes in z: N z = L/

14 The power radiated in wavelength interval d is proportional to the fraction of a wavelength: dP ~ d  So the energy density  (J/m 3 ) is This energy spectrum was derived by Rayleigh and Jeans by assuming that blackbody radiation is emitted from atomic oscillators as a wave process, and that there must be detailed balance between the standing waves that can be supported inside the solid and the emitted radiation that comes out.

15 This Rayleigh-Jeans theory is not too bad in its description of the spectrum of long-wavelength light. Unfortunately, it leads to an ultraviolet catastrophe: The power radiated at short wavelength (high energy) increases without bound! High-frequency (low- ) cutoff requires factor

16 Plank realized that he could (empirically) obtain the observed spectrum IF he assumed that blackbody radiation behaved as if it were emitted by oscillators that could only change energy by integer multiples of some minimum energy step u: E = mu Then he would have from the Boltzmann distribution:

17 The total energy of all the oscillators emitting this particular energy E=mu is then Following the derivation in the book, we calculate the average energy of an oscillator:

18 Put this together with the Rayleigh-Jeans result for the number of oscillators of wavelength : This spectrum matched experimental observation only if the energy u were inversely proportional to wavelength : h = 6.6 x 10 -34 J s = 2,000(2  ) eV Å/c This only makes sense if light is emitted in quantized packets: it behaves like a particle when it is emitted!

19 Example: We can estimate the temperature at the surface of a star by determining the wavelength corresponding to the maximum intensity in its spectrum, and assume that the emission is a blackbody spectrum. This wavelength is 550 nm (red) for the Sun, 430 nm (blue) for the North Star, and 290 nm (ultraviolet) for Sirius. Calculate the surface temperatures.

20 This is a transcendental equation. We must obtain an approximate solution. The solution will be near  = 5. hc = 2000 eV Å = 200(2  ) eV nm

21 Sun pk = 550 nmT = 4,600 o K North Star pk = 430 nmT = 5,800 o K Sirius pk = 290 nmT = 8,700 o K

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