Particle-on-a-Ring

The motion of the particle is not separable between the x and y-axes. The problem can be made separable by transforming the coordinates from Cartesian to polar coordinates. In polar coordinates, the variables become the radius of gyration, r, and the angle, , of the particle from the origin. 

The second derivatives with respect to x and y in Equation 3-1 are transformed into polar coordinates using the chain rule. 

Sinee the radius r is constant in this problem, aU of the terms that involve derivatives with respect to r will be zero, reducing the Hamiltonian tojust one variable, (J). 

The moment of inertia, I, of the particle, is equal to its mass times the square of the radius of gyration r. The Hamiltonian in liquation 3-2 is very similar to the Hamiltonian for the one-dimensional Particle-in-a-Box problem (see 

Section 2.4). This will result in the same functional forms for the wavefunctions in terms of the variable (]). The following wavefunction will be used  

The last thing to do is to normalize the wave functions so once again we set the integrated probability to 1. 

Let us take this opportunity to show the wave functions are orthogonal for any m n. 

Since any integer multiple of the fiill 2it range will be the same at the upper and lower limits, they 

The equal value with opposite sign for the n quantum number implies a pair of degenerate energy orbitals but with the particle travehng in opposite directions. 

Let us use the POR model to estimate the HOMO LUMO transition in benzene. 

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I 1 11 Schrodinger equation can be solved exactly for only a few problems, such as the particle in a box, the harmonic oscillator, the particle on a ring, the particle on a sphere and the hydrogen atom, all of which are dealt with in introductory textbooks. A common feature of these problems is that it is necessary to impose certain requirements (often called boundary... 

The 7i-orbitals of benzene, C6H6, may be modeled very crudely using the wavefunctions and energies of a particle on a ring. Lets first treat the particle on a ring problem and then extend it to the benzene system. 

The Schrodinger equation for a particle on a ring then becomes H P = E P... 

The particle on a ring is unduly restrictive to correspond immediately to a real situation. We now consider an obvious extension—the particle on a sphere—... 

Fig. 3.13 Particle-on-a-ring wavefunctions. (a) One that satisfies the boundary condition (eqn 3.49) ...

This gives the magnitude of the angular momentum vector. The 0 dependence of the functions in Table 3.1 is exactly the same as for the particle on a ring, and m indeed has the same significance it gives the component of angular momentum about the z axis ... 

Kinetic energy of a particle on a ring in terms of the moment of inertia... 

Fig. 9. 32 Two solutions of the Schrodinger equation for a particle on a ring. The circumference has been opened out into a straight line the points at =0 and 2tc are identical.

Just as the particle in a box gives us some imderstanding of the distribution and energies of n electrons in linear conjugated systems, the particle on a ring is a useful model for the distribution of k electrons around a cyclic conjugated system. 

The previous particle-in-a-box (PIB) and particle-on-a-ring (POR) problems both had V=0 and only dealt with the kinetic energy operator. The essence of the harmonic oscillator is a parabolic kjp"... 

Derive the formula for the energy of a particle-on-a-ring, normalize the wave function and apply the Perimeter Model to anthracene (C14H10) to estimate the wavelength X of the first IT —> IT transition (453 nm). 

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