Chemistry 795T. Lecture 7. Electromagnetic Spectrum Black body Radiation. NC State University

Transcription

1 Chemistry 795T Lecture 7 Electromagnetic Spectrum Black body Radiation NC State University

2 Black body Radiation An ideal emitter of radiation is called a black body. Observation: that peak of the energy of emission shifts to shorter wavelengths as the temperature is increased. Wien displacement law λ max T=c 2 /5. The second radiation constant c 2 = 1.44 cm K. λ max T = 2.88 x 10 6 nm-k

3 The wavelength and the frequency The wavelength λ is the distance between the peaks in a traveling wave. In classical physics light is a wave that travels with velocity c. The frequency is ν = c/λ. The wavenumber ν = ν /c. The ~ wavenumber has units of cm -1.

4 The electromagnetic spectrum λ increasing ν decreasing The wavelength and frequency are inversely related.

5 Maxwell s Equations and the Speed of Light Maxwell s theory of electromagnetic radiation as a traveling wave consisting of both electric and magnetic components predicts a relationship between the speed of light and two fundamental constants µ 0 and ε 0. c 2 = µ 1 0 ε 0 This relationship is derived from the Maxwell equation s E =0 B =0 E = B t B = µ 0 ε 0 E t

6 Maxwell s equations can be used to derive a wave equation for electromagnetic radiation. Identity Electromagnetic wave equation E = B t the curl is E = 2 E + E E =0 Since and The result is a wave equation E = B t B = ε 0 µ 0 E t 2 E = ε 0 µ 0 ( E/ t) t = ε 0 µ 0 2 E t 2

7 Dilemma for Classical Physics The maximum in energy for the black body spectrum is not explained by classical physics. The cavity modes of the black body are predicted to be ρ = 8 π k B T λ 4 Where ρ is the radiant energy density. This function increases without bound as λ 0. This law is known as the Rayleigh-Jeans law.

8 The Sun is a Black Body The sun and stars are black bodies The peak of the emission spectrum depends on temperature as indicated by the Wien displacement law.

9 The Planck Distribution Law Planck assumed quantization of cavity modes: E = nhν. (n=0,1,2..) The constant h determines that only those modes with an energy specified by the precise amounts given can be excited. The population of the levels will favor lower energy (quantum number) modes over higher energies.

10 Planck Assumption implies that average energy is temperature dependent In classical physics the average energy in an energy level is < ε > = kt. Quantized levels imply that the average energy in each oscillator is <ε> = hν/(e hν/kt -1). Since c = λν we can also write this as <ε> = hc/λ(e hc/λkt -1). To obtain the Planck formula simply replace kt by the <ε> = hc/ λ(e hc/λkt -1) expression implied by quantization.

11 Mathematical Form of the Planck Law Assume a ladder of energy levels separated by ε = hν. The energy levels will be populated according to a thermal weighting. The higher levels will be less populated than the lower levels. In the Planck theory the energy density becomes: ρ = 8 π hc λ 5 1 e hc /λ k B T 1

12 Planck Distribution Law

13 Consistency with Experiment The temperature behavior of the Rayleigh- Jeans law is recovered because e hc/λkt 1 hc/λkt as T The integral of the total energy is proportional to T 4 which gives the Stefan-Boltzmann law, W = σt 4. W is the flux or energy/area. The Wien displacement law is recovered from differentiation of ρ. Setting dρ/dλ = 0 gives the maximum in the distribution law.

14 Consequences of Planck Law Classical physics fails to describe blackbody radiation. A model that includes quantized modes of electromagnetic radiation succeeds. The constant h = x Js is a fundamental constant the determines the scale of energy quantization.

15 What is the radiant power of the sun? Use the Stefan-Boltzman law W= σ T 4 (W is the flux or power per unit area) σ = kg s -3 K -4 (Watts/m 2 /K 4 ) Assuming that the temperature at the surface is 5500 K and the diameter is 1.4 x 10 6 km. The way that the Sun's diameter is measured is by taking angular diameter measurements and then translating them to linear diameter measurements. The angular diameter of the Sun can be measured using a telescope during a total solar eclipse or by timing Mercury when it is in transit in front of the Sun. The first series of measurements were taken in the early 1700's by Jean Picard in Paris, France.

16 What is the radiant power of the sun? First calculate the area and then the flux (power). The area is A = 4πR 2 A = 4(3.1416)(7 x 10 8 ) 2 A = 6.16 x m 2 The flux is W= σ T 4 W = (5500) 4 W = 5.19 x 10 7 Watts/m 2 The total power is P = WA= (5.19 x 10 7 Watts/m 2 )(6.16 x m 2 ) P = 3.2 x Watts

17 What is the radiant power at the surface of the earth? We use the distance from the earth to the sun to obtain the flux at the earth. The earth is R e = 1.5 x 10 8 km from the sun. The area irradiated is A e = 4πR e 2 A e = 2.83 x m 2

18 What is the radiant flux at the surface of the earth? The flux in space above the earth is called the insolation. The insolation is the power coming from the sun divided by the total area at the radius of the earth. W e = P/A e = (3.2 x Watts)/(2.83 x m 2 ) W e = 1.13 x 10 3 Watts/m 2 This is very close to the measured value for radiation in space above the earth.

19 How much energy does the earth absorb? The earth has a cross-sectional area of A c = πr earth 2 A c = (3.1416)(1.3 x 10 7 m) 2 A c = 5.3 x m 2 P abs = W e A c P abs = (1.13 x 10 3 Watts/m 2 )(5.3 x m 2 ) P abs = 6 x Watts

20 What is the temperature at the surface of the earth? P abs = P emit = 6 x Watts P emit = σ T earth4 A earth [A earth = 4π R earth 2 = 2.1 x m 2 ] T earth = (P emit /σ A earth ) 1/4 T earth = (6 x / /2.1 x ) 1/4 T earth = 266 K This is close, but it is a little frosty. Why is this?

21 What is the temperature at the surface of the earth? P abs = P emit = 6 x Watts P emit = σ T earth4 A earth [A earth = 4π R earth 2 = 2.1 x m 2 ] T earth = (P emit /σ A earth ) 1/4 T earth = (6 x / /2.1 x ) 1/4 T earth = 266 K This is close, but it is a little frosty. Why is this? We ignored the fact that the earth has an atmosphere! The atmosphere does two things.

22 What is the role of the atmosphere? 1. Some molecules in the atmosphere absorb incident light. Ozone absorbs UV light and prevents harmful radiation from reaching the surface of the earth. 2. Molecules can also absorb emitted or radiated light. What is the wavelength of such light? It can be obtained from the Wien displacement law. λ max T = 2.88 x 10 6 nm-k Thus, for the sun with T = 5500 K, λ max = 523 nm For the earth with T = 266 K, λ max = 10,800 nm = 10.8 µm The sun s emission is peaked in the visible region of the Electromagnetic spectrum and the earth emits in the infrared.

23 Absorption by gases in the atmosphere 5500 o K 266 o K 3.5 Selective absorption and emission by atmospheric gases (source: P&O fig 4.2) Electronic Vibrational Rotational

24 Absorption and emission The intrinsic coefficient for absorption B 12 is related to W 12 = N 1 B 12 ρ, where ρ is the energy density: I = c ρνdν Einstein showed that the rate of absorption and stimulated emission are equal. The spontaneous emission rate has a definite relation to the stimulated emission rate: B21 = B 12 A spontaneous = 8πhν 3 21 c 3 B 21

25 Spontaneous emission is fluorescence Stimulated emission is used for lasers spontaneous stimulated N 1 B 12 ρ N 2 A 21 N 2 B 21 ρ