Presentation is loading. Please wait.

Presentation is loading. Please wait.

Photon Statistics. 18.1 Blackbody Radiation 1.The energy loss of a hot body is attributable to the emission of electromagnetic waves from the body. 2.The.

Similar presentations


Presentation on theme: "Photon Statistics. 18.1 Blackbody Radiation 1.The energy loss of a hot body is attributable to the emission of electromagnetic waves from the body. 2.The."— Presentation transcript:

1 Photon Statistics

2 18.1 Blackbody Radiation 1.The energy loss of a hot body is attributable to the emission of electromagnetic waves from the body. 2.The distribution of the energy flux over the wavelength spectrum does not depend on the nature of the body but does depend on its temperature. 3.Electromagnetic radiation can be regarded as a photon gas.

3 The spectrum of blackbody radiation energy for three temperatures: T 1 >T 2 >T 3

4 4.Photons emitted by one energy level can be absorbed at another, so the total number of photons is not constant, i.e. ΣN J = N does not apply. (i.e. For phonon gas, N is not independent!) 5.The Lagrange multiplier α that was determined by ΣN J = N becomes 0, so that e - a = e -0 = 1 6. Photons are bosons of spin 1 and thus obey Bose- Einstein Statistics

5 For a continuous spectrum of energy The energy of a photon is h v So: Recall that f(v)dv is the number of quantum states with frequency v in the range v to v + d v

6 For phonon gas (chapter 16) In a photon gas, there are two states of polarization corresponding to the two independent directions of polarization of an electromagnetic wave. As a result Where c is the speed of electromagnetic wave (i.e. light) g(v)

7 The energy within the frequency range v to v + d v equals the number of photons within such a range times the energy of each photon: u u u u

8 The above is the Planck radiation formula, which gives the energy per unit frequency. When expressed in terms of the wavelength Since u u

9 Hereμ(λ)is the energy per unit wavelength The Stephan-Boltzmann Law states that the total radiation energy is proportional to T 4 u

10 The total energy can be calculated Setting one has The integral has a value of with

11 Particle flux equals, where is the mean speed and n is the number density. In a similar way, the energy flux e can be calculated as Assuming this is the Stephan-Boltzmann Law with the Stephan-Boltzmann constant.

12 The wavelength at which is a maximum can be found by setting the derivative of equal to zero! Or equivalently, set (minimum of ) One has is known as Wien’s Displacement Law

13 For long wavelengths: (Rayleigh- Jeans formula) For short wavelength: Sun’s Surface T≈ 6000K, thus u

14 Sketch of Planck’s law, Wien’s law and the Rayleigh-Jeans law.

15 Using The surface temperature of earth equals 300K, which is in the infrared region The cosmic background microwave radiation is a black body with a temperature of 2.735 ± 0.06 K.

16 18.2 Properties of a Photon Gas The number of photons having frequencies between v and v + d v is The total number of photons in the cavity is determined by integrating over the infinite range of frequencies:

17 Leads to Where T is in Kelvins and V is in m 3 The mean energy of a photon The ratio of is therefore of the order of unity. The heat capacity

18 Substitute Entropy

19 Therefore, both the heat capacity and the entropy increase with the third power of the temperature! For photon gas: It shows that F does not explicitly depends on N. Therefore,

20 18.1 a) calculate the total electromagnetic energy inside an oven of volume 1 m 3 heated to a temperature of 400 F Solution Use equation 18.8

21 18.1b) Show that the thermal energy of the air in the oven is a factor of approximately 10 10 larger than the electromagnetic energy. Solution:


Download ppt "Photon Statistics. 18.1 Blackbody Radiation 1.The energy loss of a hot body is attributable to the emission of electromagnetic waves from the body. 2.The."

Similar presentations


Ads by Google