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Application of Quantum Theory 1- Particle in 1-D box. The particle is free to move inside the box Consider the ethylene molecule as an example of the free particle move in 1-D box, in which the two electrons, only, are free to move, while the electrons are frozen in bonds with atoms. The length of the box is twice the C=C bond length. V=0 V=0 H H H H x=0 1- Schrodinger Equation is L = 2*1.4 =2.8 A ( ) 2- Using V=0 3- The solution of the 2nd order ordinary differential equation is ( ) ( ) x=L 4- For particle in box =0 at x =0 and =0 at x =L (boundary condition) ( ) ( ) ( ) 5- The last equation are mitts only if the argument of sin function is an integer multiplier of π (0, 180, 360… ) ( ) 6- Note that 7- The energies, wave functions and the probability densities are Energy Level Diagram 15 𝐸𝑛 𝑛 𝑚𝐿 E ψ ψ2 10 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 8- Comments a. The energy increase as the quantum number increase b. The energy separation between energy levels increase as the quantum number increase c. The energy and the energy separation increases as the size of the box decreases d. ψ2 < ψ when ψ have small values while ψ2 = ψ when ψ has the maximum value. (why) e. Node = the point where wave function passes through zero, or the position where probability of finding particle = 0 (No. of nodes = n-1) f. The probability of finding the particle between two points x 1 and x2 are different when n have small values, while as n increases (n >100) these 2 probabilities become the same. Generally the probability density become uniform as n increase, that is to say quantum mechanics results and classical mechanics results tend to agree in the limit of the large quantum numbers 9- The particle in a box model can be applied to electrons moving freely (π electrons) in a molecule a. For example butadiene has an absorption band at 217 nm for the 1st π→ π* transition. As a simple approximation, consider butadiene as being a 1-D box of length 4*1.4 = 5.6 Å. and consider the 4 π electrons to occupy the levels calculated using H2C C C CH2 the particle in box model LUMO H H b. The calculated excitation energy is E3 HOMO HOMO : Highest Occupied Molecular Orbital LUMO : Lowest Unoccupied Molecular Orbital E2 E1 E=ELUMO – EHOMO ( ) h (Plank Constant)= 6.626 x 10-34 JS, m (mass of electron)=9 x 10-31 Kg, L (length of bonds = no. of carbons *1.4 Å , 1 Å =1 x10-10 m c (speed of light)=3 x 108 m/S , 1 m = 109 nm What are the causes of difference between Experimental value (217 nm) and calculated value (204 nm) That is due to the approximations done 1. Box length C=C equal C-C =1.4 Å 2. Assume it linear Calculate (nm) for the 1st π→ π* transition for the following compounds Compound Ethylene Butadiene Hexatriene Octatetraene -carotene No. of C's 2 4 6 8 22 Exp. 162 217 274 304 425 Calculated