Quantum Chemistry-4th Lecture Particles in Boxes PDF

Quantum Chemistry 4th lecture Particles in Boxes (SOLUTIONS TO THE SCHRÖDINGER EQUATION ) # Free particle and the particle in a box Learning objectives By the end of this section you should be able: - Obtain the wavefunctions and energy levels for a particle in a box...

Quantum Chemistry 4th lecture Particles in Boxes (SOLUTIONS TO THE SCHRÖDINGER EQUATION ) # Free particle and the particle in a box Learning objectives By the end of this section you should be able: - Obtain the wavefunctions and energy levels for a particle in a box. - Apply the particle in a box model to electrons in one dimensional and to π electrons in conjugated electrons. - Schrödinger equation is a 2nd-order diff. eq. We can find two independent solutions Φ1 (x) and Φ2 (x) The general solution is a linear combination A and B are then determined by boundary conditions on Ψ(x) and Ψ’(x). Additionally, for physically reasonable solutions we require that Ψ(x) and Ψ’(x) be continuous function. 1 (I) Free particle V(x) = 0 Then and The wave equation becomes:- with solutions 2 Free particle ⇒ no boundary conditions ⇒ any A and B values are possible, any So any wavelike solution (traveling wave or standing wave) with any wavelength, wavevector, momentum, and energy is possible. (II) Particle in a box The wavefunctions are zero outside the box where xa. Wavefunctions are everywhere continuous. Therefore, the wavefunctions must be zero at the walls at x= 0 and x= a. The boundary conditions are therefore Ψ(0)=0 and Ψ(a)=0. We now apply each condition in turn to a general solution of the form Particle can’t be anywhere with V(x) = ∞ 3 ⇒Ψ(x < 0, x > a) = 0 For 0 ≤ x ≤ a , Schrödinger equation is like that for free particle. With same definition Again with solutions But this time there are boundary conditions! First, at x=0, Continuity of Ψ(x)⇒ Ψ(0) = Ψ(a) = 0 because cos 0=1 and sin 0=0. Therefore, to satisfy the condition Ψ(0)=0 we require sumation=0. 4 Next, at x=a, after setting sumation=0, Can’t take B = 0 (no particle anywhere) ⇒ must have sin(ka) = 0 ⇒ ka = nᴫ n = 1, 2, 3,... ⇒ k is not continuous but takes on discrete values k = nᴫ/a Thus integer evolves naturally!! Note/ the value n=0 is excluded because it would give sin ka=0 for all x, and the particle would not be found anywhere. The integer n is an example of a quantum number, a number that labels a state of the system and that, by the use of an appropriate formula, can be used to calculate the value of an observable of the system. So solutions to the Schrödinger equation are These solutions describe different stable (time-independent or “stationary”) states with energies 5 Energy is quantized!! And the states are labelled by a quantum number n which is an integer. There now remains only the constant B to determine before the solution is complete. The probability of finding the particle somewhere within the box must be 1, so the integral of Ψ2 over the region between x=0 and x=a must be equal to 1. The wave function is normalized when Therefore, for this problem the integration is over the interval in which x can vary, which is from 0 to a. If we let B be the normalization constant, Therefore, the solution of above eq. is , B=(2/a)1/2. The complete solution is Properties of the stationary states (a) The energy spacing between successive states gets progressively larger as n increases 6 (b) The wavefunction Ψ(x) is sinusoidal, with the number of nodes increased by one for each successive state (c) The energy spacings increase as the box size decreases. We’ve solved some simple quantum mechanics problems! The P-I-B model is a good approximation for some important cases, e.g. pi- bonding electrons on aromatics. Electronic transitions shift to lower energies as molecular size increases. 7 8 Example/ π Electrons in Conjugated Molecules Butadiene, CH,=CH-CH=CH,, will be used to illustrate the way in which the particle-in-a-box wavefunctions can be applied to conjugated molecules, a term applied to hydrocarbon molecules with alternate single and double carbon-carbon bonds. To simplify the problem further we will assume that the carbon framework is linear and that the potential energy of the π electrons remains constant as they move along the molecule. The second assumption is suppose the actual value of V to be used is arbitrary, provided it remains constant, and we shall put V = 0. 9 Two electrons can occupy each state, with their spins paired. Since butadiene has four π electrons, the two lowest energy states will be fully occupied. Experimentally it is found that butadiene absorbs electromagnetic radiation with a wavelength equal to 217 nm. The energy of the photon is used to excite a π electron from the n = 2 to the n = 3 state 10 11 -The Particle in a Three-Dimensional Box We have a particle in a box of dimensions a, b, and c in the x, y, and z directions, respectively. Inside the box, the Hamiltonian can now be written as The Hamiltonian operator can be written using the kinetic energy expressed in terms of the momentum :- Therefore, the equation Becomes 12 Or This is a partial differential equation in three variables (x, y, and z) and the simplest method to solve this eq. is the method of separation of variables. To separate the variables, it is assumed that the desired solution, ψ, can be factored into three functions, each of which is a function of one variable only. This product of three functions is now written in place of ψ, but we will use the simplified notation X = X(x), etc.: Since YZ is not a function of x, XZ is not a function of y, and XY is not a function of z, we can remove them from the derivatives to give divided by XYZ, we obtain Since each term on the left-hand side is a function of only one variable, each will be independent of any change in the other two variables. Each term must be equal to a constant, which we will call −k2. Since there must be three such constants, we will call the constants −k2x, −k2y, and −k2z for the x, y, and z directions, respectively. We thus have three equations 13 The sum of the three constants must be equal to the right-hand side of above Eq. Therefore: The energy is the sum of the contributions from each degree of freedom in the x, y, and z coordinates therefore Since the energy associated with the degree of freedom in the x direction is not dependent on the y and z coordinates, we can separate Eq. to give By substitution in eq. Then The solution of this equation is 14 Similarly The general solution can be written as the product of the three partial solutions where nx, ny, and nz , are the quantum numbers for the x, y, and z components of energy, respectively. The general solution can be simplified somewhat to give we will find expressions for the energies We can now write the expression for the energy levels based on the x component as Also 15 The total energy, E, is where nx = 1, 2, 3,... ; ny = 1, 2, 3,... ; and nz = 1, 2, 3.... 16