Chemistry 431. Lecture 1. Introduction Statistical Averaging Electromagnetic Spectrum Black body Radiation. NC State University

Chemistry 431 Lecture 1 Introduction Statistical Averaging Electromagnetic Spectrum Black body Radiation NC State University

Overview Quantum Mechanics Failure of classical physics Wave equation Rotational, vibrational and translational motion Electronic structure Spectroscopy Statistical Mechanics Partition function Average energy Entropy

Macroscopic view of equilibrium For a chemical reaction in which X is the reactant and Y is the product X Y with free energy change ΔG o, the equilibrium constant is K = e ΔG o /RT = [Y] [X]

Microscopic view of equilibrium For a system where there are two energy states, which we can call X and Y Y ΔE X with energy difference ΔE, the ratio of population is e ΔE/kT = N Y N X where N Y and N X are the numbers of particles in each state.

Boltzmann s constant Note that we k instead of R in the exponent. K is called Boltzmann s constant and it is related to R by: k = R N A where N A is Avagadro s number. kt has units of Joules and is a measure of the thermal energy of a molecule RT has units of Joules/mole and is a measure of the thermal energy of a mole

Microscopic probability When we look at the world in terms of energy levels we can see that the probability of being in a given level depends on temperature. We define the probability of being in level j as number of molecules in state j P l = total number of molecules or P l = N j N X + N Y

Microscopic probability We can calculate the probability of levels A and B N P X = X = 1 = 1 N X + N Y 1+N Y /N X 1+e ΔE/kT P Y = N Y N X + N Y = N Y /N X 1+N Y /N X = e ΔE/kT 1+e ΔE/kT Of course, P X + P Y = 1

Average energy for a two level system We defined the relative energy of the two states as ΔE = E Y E X. We can (arbitrarily) set the energy of state X to zero, E X = 0. Then the energy of state Y is E Y = ΔE. The average energy of the system is: < E >= Σ P j E j i =1 = e ΔE/kT 1+e ΔE/kTE Y = = P X E X + P Y E Y e ΔE/kT 1+e ΔE/kTΔE

Extension to many levels If we have a system with an infinite number of equally spaced energy levels we can calculate the probability and energy in the same way. P j = e jδe/kt 1+e ΔE/kT ++e 2ΔE/kT ++e 3ΔE/kT +... < E >= Σ P j E j i =1 ΔE = e ΔE/kT ΔE +2e 2ΔE/kT ΔE +2e 3ΔE/kT ΔE +... 1+e ΔE/kT ++e 2ΔE/kT ++e 3ΔE/kT +...

Partition function for ladder of energy levels The denominator gives the average number of levels that are accessible at a given temperature. We can simplify it as follows: Q =1+e ΔE/kT + e 2ΔE/kT + e 3ΔE/kT +... Let x = e ΔE/kT, then Q =1+x + x 2 + x 3 +... Q 1=x + x 2 + x 3 + x 3 +... and xq = x + x 2 + x 3 + x 3 +... since Q 1=xQ Q = 1 1 x = 1 1 e ΔE/kT

Calculating the energy It is not obvious how to obtain the energy for this system. We can consider the general formula and make the substitution β = 1/kT. < E >= Σ P j E j i =1 = e βδe ΔE +2e 2βΔE ΔE +2e 3βΔE ΔE +... 1+e βδe ++e 2βΔE ++e 3βΔE +... = Q/ β Q For the energy ladder we have < E >= 1 e βδe e βδe ΔE 1 e βδe 2 = ΔE e βδe 1

High temperature limit As the temperature approaches infinity all of the levels become equally populated. The average energy can be calculated assuming β = 1/kT << 1. < E >= ΔE e βδe 1 = ΔE 1 βδe+... 1 = 1 β = kt Thus, kt is a classical energy for an averaged system at high temperature. In units of joules/mole < E >=RT The energy of a gas is related to its PV m product (V m is the molar volume) thus, PV m = RT

The wavelength and the frequency of electromagnetic radiation 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.

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

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 = 2.88 x 10 6 nm-k.

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.

UV Catastrophe ρ = 8πk BT λ 4

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.

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

Planck s Innovation Classical radiation All frequencies are possible Quantized radiation Only frequencies nhν are allowed

Mathematical Form of the Planck Law 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

UV Catastrophe ρ = 8πk BT λ 4

Resolution of UV Catastrophe ρ = 8 π hc λ 5 1 e hc /λ k B T 1 ρ = 8πk BT λ 4

Planck Distribution Law

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.

Wien displacement law 0 = ρ λ = λ 8 π hc λ 5 1 e hc /λ k B T 1

W = 0 ρ d λ = 0 Stefan-Boltzmann law 8 π hc λ 5 1 e hc /λ k B T 1 d λ = σ T 4

Consequences of Planck Law Classical physics fails to describe blackbody radiation. A model that includes quantized modes of electromagnetic radiation succeeds. Planck s constant h = 6.626 x 10-34 Js is a fundamental constant the determines the scale of energy quantization.

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.

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 = 5.6704 10-8 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.

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 10 18 m 2 The flux is W= σ T 4 W = 5.6704 10-8 (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 10 18 m 2 ) P = 3.2 x 10 26 Watts

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 10 23 m 2

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 10 26 Watts)/(2.83 x 10 23 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.

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 10 14 m 2 P abs =W e A c P abs = (1.13 x 10 3 Watts/m 2 )(5.3 x 10 14 m 2 ) P abs = 6 x 10 17 Watts

What is the temperature at the surface of the earth? P abs =P emit = 6 x 10 17 Watts P emit = σ T earth4 A earth [A earth = 4π R earth 2 = 2.1 x 10 15 m 2 ] T earth = (P emit /σ A earth ) 1/4 T earth = (6 x 10 17 /5.6704 10-8 /2.1 x 10 15 ) 1/4 T earth = 266 K This is close, but it is a little frosty. Why is this? P.S. What is 266 K in o C? What is 266 K in o F?

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.

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