#### Particle in One Dimensional Box Problem, Chemistry tutorial

Introduction

The Schrodinger wave equation is of particular interest in chemistry since it is appropriate to behaviour of object down to molecular and atomic dimensions. In a learn of molecules and atoms, the quantities that must be of prime interest are the energies and the positions of the atoms.

The Schrodinger technique yields the energy straight. It doesn't provide precise information about the positions of the atoms in the system. Only the probability that the particle is at a specified position is attained from wave mechanics. The lack of information about the precise position of a particle at several specific times seems to be a trait of the difficulty rather than a defect in the Schrodinger process.

Definition of the Particle in a Box

The easiest system of physical significance is that of a particle constrained to shift in a box of given dimension. The particle can be detained to assured region in space via permitting the potential energy to be unlimited in any region outside that of interest. If the potential energy is zero inside the described region (that is the box), we have an easy system that still has physical application: translational motion of molecules. The levels can be utilized in statistical mechanics to compute energies of molecules. The element in a box model can as well be utilized to clarify the electronic properties of several molecules. We suppose that the particle can shift in a line only (one-dimensional). The potential outside the box of length 'a' is and obviously = 0 in any region where V =. This is since no normal particle will have sufficient energy to exist in the region where V =.

The Schrödinger wave equation

In this chapter, we will, for the sake of together brevity and simplicity, only consider the time-independent Schrodinger wave equation in one dimension. We won't think the full equation in its complete gruesome splendor (Redwoods, 2009). The time-independent Schrodinger wave equation that could validly be termed Schrodinger's law is specified via the differential equation.

d2φ/dx2 = -2m/x2[B-U(x)] φ(x)

where φ (x) is the is the wave function, m is mass, is Planck's constant separated via  2π,  E  is the total energy of the particle, and U(x) is the potential energy  function of the particle. As whenever one ingests something disagreeable and the natural reaction is nausea, so as well is the natural reaction to this equation. Though, comfort might be taken if we think that acceleration is the 2nd derivative of the location function and, consequently, could be written in the equation.

α=d2x/dt2

As certainly as acceleration simplifies to incredible more palatable, The Schrodinger wave equation must simplify (a little, at least). To discover common solutions to this equation, boundary situations must be established. The principle conditions that it must adhere to are

1.   φ (x) -> 0 as  π -> +- ∞

2.    φ (x) = 0 if x is in someplace it is physically impossible to be

3.  (x) is a continuous function

4.  (x) is a normalized function

In such pages, we will, again for brevity and simplicity, consider the case of a particle in a one-dimensional box of ideal stiffness, these that its walls are impenetrable. Let the box have length L.  As might be seen in the illustration, the potential energy function has 2 states:

1. U(x) =0 for 0 <=x<=L

2. U(x) = for x<0 or x>L Fig: Potential Energy of a Particle in a Box

Since it is physically impossible for the particle to be outside of the box, t is the 1st state that is of interest.  Indeed, this simplifies the wave equation considerably, by the term U(x) dropping out. Consequently, the wave equation equivalent to the particle in the box is given via equation.

d2φ/dx2 = -2m/h2 Eφ(X)

Before assailing this equation through a display of mathematical acumen, let us ask ourselves  what functions 2nd derivative is merely several negative steady all of the terms on the right-hand side save (x) multiple of itself? To simplify, let

B2 = 2mE/h2

d2φ/dx2 = -E2φ(x)

It becomes clear; a trigonometric function like sine or cosine would be a good candidate for (x). As a result, our guess for the solutions to the wave equations is equation.

Φ(x) = sinBx

By the 1st above boundary condition, it is recognized that

Φ(π - L) = sin BL=0

Therefore,

BL-nx => B- M π/L

Where n=1, 2, 3,...  When the smoke clears, we have the equation

Φ(x) - Asin nπx/L

Where A is the function's amplitude. To find out the amplitude, recall the 4th boundary condition, (x) is a normalized function. Mathematically, this means

In  words,  this  states  that  the  probability  of  discovering  the  particle somewhere on the x-axis is one or 100%.  Waving hands a bit to omit the gory details, this provides equation.

A= √Z/L

Gasping for breath, we at last have unearthed the solution to the wave equation for the particle of the nth quantum state in the rigid box we have equation.

Φx(x) = √2/Lsin nπx/L     for 0<=x<=L

Φx(x)= 0                           for x<0 or x>L

The utility of this solution lies primarily in that the probability of discovering the particle at some position x is given by the square of (x) which provides an equation.

Pn(x) = | Φx(x)|2  =  2/L sin2  nπx/L

The importance of this association is best demonstrated graphically. Consider a particle in the third quantum state. Fig: The Third Quantum State in a 1-D Box

It can be seen that there are regions where the probability of discovering the particle is zero so termed nodes.  This isn't unique to merely put the particle in a stiff box model.  It is examined in more sophisticated ones these as the model of an electron orbiting a nucleus (Redwoods, 2009). This equation can be utilized to compute the probability of discovering the particle at any point, x. The probability is [(n,x)]2. If the values of n, x and a are recognized, the probability can be evaluated.

Orthogonal and Orthonormal Wave Functions

Two wave functions are said to be normalized when

ΦL**  jd  =1  if  i = j.

However,  ∫ ΦLjd  = 0 if i ≠ j

Such wave functions are said to be orthogonal to each other.

The last 2 expressions can be joined to attained j d = σij

Where σij = 0 for i ≠ j

σij = 1 for i = j

Wave functions that satisfy ∫ ΦL* jd = σij are said to be orthonormal. Hence wave functions that are normalized and are orthogonal to each other are sometimes called orthonormal functions

Free electron model (femo)

The easy computation done for the particle in a one-dimensional box can be applied to approximation the absorption frequency of several organic molecules presumably conjugated dienes. The technique frequently employed is the free electron model (FEMO) for the π electrons of such molecules. The energy for the lowest electronic transition is that needed to elevate an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). For molecules through conjugated dienes, it has been originate that the electronic absorption bands shift to longer wavelengths as the number of conjugated dienes is amplified. Each carbon atom gives one electron to the π-system but such electrons are free to shift the whole length of the series of π orbitals and aren't localised on a specified carbon atom. In FEMO, it is assumed that the π-system is a region of uniform potential and that the potential energy rises sharply to infinity at the ends of the system. Therefore the energy level, E, available to the electrons would be supposed to be analogous to that of a particle restricted to move in one dimension. The π electrons are allocated to orbitals so that there are 2 in each level (with opposite spin) starting through the lowest. For a totally conjugated system (hydrocarbon), the number of the HOMO is n =N/2 where N is the number π electrons included and parallels the number of carbon atoms in theπ π system. For absorption, an electron from the HOMO is excited to the LUMO with quantum number n' = (N/2 + 1). The difference in energy between these levels is

ΔE = h2(n'2-n2) / 8ma2= h2/8ma2{[(N/2) + 1]2 - (N/2)2}

The absorption frequency in wave numbers is ?E/hc = h(N+1)/8ma2c

To estimate 'a' two techniques are utilized. Firstly, when end consequences aren't neglected, 'a' equals the sum of bond lengths and half the bond length of extensions at both extremes. Secondly, when end consequences are neglected, 'a' = 1.54(N + 1) where N is number of carbon atoms in the π-system.

Instance

Compute the lowest absorption wave numbers for octatetraene neglecting end effects. Octatetrene has the carbon skeleton below.

C=C-C=C-C=C-C=C

Because there are 8 ? electrons in the system (one from each carbon atom) there will be four levels occupied via electrons (in pairs). This means the HOMO is 4 while the LUMO is 5.

Neglecting end results means that 'a' = 1.54(N + 1) = 1.54(8-1) = 10.78Å

ΔE = [52h2/8ma2]- [42h2/8ma2] = 9h2/8ma2

But ΔE = hν = 9h2/8ma2

Therefore, wave number = [9x6.626x10-34]/[8x9.11x10-31x(10.78x10- 10)2x3x108]

=2.347x106 m-1

= 2.347 x104 cm-1

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