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Application of quantum in chemistry The Particle in A Box The ‘Classical’ Case The ‘Quantum’ Case The absolutely small particle in the nanometer size box is a quantum particle, and it must obey the Uncertainty Principle, that is, ΔxΔp= h/4π. If V=0 and x= L/2, we know both x and p. The result would be ΔxΔp 0, the same as the classical racquetball. This is impossible for a quantum system. Therefore, V cannot be zero. The particle cannot be standing still at a specific point. If V cannot be zero, then Ek can never be zero. The Uncertainty Principle tells us that the lowest energy that a quantum racquetball can have cannot be zero. Our quantum racquetball can never stand still. Energies of a Quantum Particle in a Box Wave functions must be zero at the walls Nodes are the points where the wave function crosses zero Energies are quantized A Discreet set of energy levels Why are Cherries Red and Blueberries blue ? The Colour of Fruit This energy corresponds to Deep Red Colour If L=0.7 nm, =540 nm If L=0.6 nm, =397 nm Green Colour Blue Colour Particle in a box Step 1: Define the potential energy Step 2: Solve the Schrodinger equation Step 3: Define the wave function Step 4: Determine the allowed energies Step 5: Interpret its meaning Particle in 1-dimensional box • Infinite walls Time Independent Schrödinger Equation 2 d 2 ( x) V ( x) E 2 2m dx Region II Region I Region III PE KE TE Applying boundary conditions: V(x)=0 V(x)=∞ V(x)=∞ Region I and III: 0 L x 2 d 2 ( x) * E 2 2m dx 2 0 d 2 ( x ) 2mE ( x) 2 2 dx Second derivative of a function equals a negative constant times the same function. Functions with this property sin and cos. d 2 sin(ax ) a 2 sin(ax ) 2 dx d 2 cos(ax ) a 2 cos(ax ) dx Copyright – Michael D. Fayer, 2007 b) x=L ψ=0 0 Asin kL Region II: 2 d 2 ( x) E 2 2m dx d 2 ( x) 2m E 2 2 dx This is similar to the general differential equation: d 2 ( x) k 2 2 dx A sin kx B cos kx Applying boundary conditions: a) x=0 ψ=0 0 A sin 0k B cos 0k 0 0 B *1 B 0 ButA 0 kL n Thus, wave function: II A sin nx L But what is ‘A’? Normalizing wave function: Calculating Energy Levels: L 2 ( A sin kx ) dx 1 0 k2 2mE 2 L x sin 2kx A 1 2 4 k 0 2 n sin 2 L 2L L A 1 n 2 4 L 2 L A 1 2 A k 2h2 E 2m4 2 ( n 2 2 h2 E L2 2m4 2 2 L Thus normalized wave function is: II k 2 2 E 2m 2 nx sin L L Thus Energy is: n2h2 E 8mL2 h ) 2 Particle in a 1-Dimensional Box 2 nx sin L L II II 2 2 nx sin L L 2 1) Difference b/w adjacent energy levels: 2) Non-zero zero point energy + + + + + E * 3) Probability density is structured with regions of space demon-strating enhanced probability. At very high n values, spectrum becomes continousconvergence with CM (Bohr’s correspondance principle) Particle in a 3-D box Question: An electron is in 1D box of 1nm length. What is the probability of locating the electron between x=0 and x=0.2nm in its lowest energy state? Question: An electron is in 1D box of 1nm length. What is the probability of locating the electron between x=0 and x=0.2nm in its lowest energy state? Solution: Question: What are the most likely locations of a particle in a box of length L in the state n=3 Example: What are the most likely locations of a particle in a box of length L in the state n=3 Expectation value of position and its uncertainty Expectation values Position Uncertainity Expectation value of Momentum And square of momentum Momentum Estimating pigment length Assumptions: Wavelength of transition for Anthracene Particle in a Box Simple model of molecular energy levels. Anthracene L L6 A electrons – consider “free” in box of length L. Ignore all coulomb interactions. Pigments and Quantum mechanics High degree of conjugation!! • Electrons have wave properties and they don’t jump off the pigments when they reach its ends. • These electrons resonances determine which frequencies of light and thus which colors, are absorbed & emitted from pigments Electron resonances in a cyclic conjugated molecule A crude quantum model for such molecules assumes that electrons move freely in a ring. Resonance condition: R: radius of molecule, λ wavelength of electron Energy is once again quantized. It depends on variable n which posseses discrete values only