--%>

Schrodinger equation with particle in a box problem.

Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.

So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functions for a particle that is free to move in three dimensions are deduced. A molecule of a gas enclosed in a cubic container provides a specific example that is dealt with in the section after the general procedure has been developed.

For any three-dimensional problem, the potential energy is, general, a function of three coordinates. For a cubic potential box, the Cartesian coordinates are convenient. The differential equation that must be solved is now the Schrodinger equation in three dimensions.

1310_Particle in a box.png 

For a "cubic box," the potential function can be expressed in terms of separate x, y, and z components,
98_Particle in a box1.png 

Each of the potential function components for a "particle-in-a-box" is like the one-dimensional potential for a "particle-on-a-line".

For three-dimensional systems, the solution function ψ depends on the three coordinates necessary to locate a point in space. It is often profitable to try to separate such systems into parts, with each part involving only one coordinate. On the basis we try the substitution

ψ (x, y, z) = Ø(x)Ø(y)Ø(z)

Substitution of (2) from (1) gives

1350_Particle in a box2.png 

Division by Ø(x)Ø(y)Ø(z) gives

1180_Particle in a box3.png 

For the equation to be satisfied for all values of x, y and z, each term on the left must equal a component of ε, and we can write

ε = εx + εy + εz

The Schrodinger equation can then be broken down into three identical equations of the type

1394_Particle in a box4.png 

Or

578_Particle in a box6.png 

These equations are identical to that written for one-dimensional problem. The solution to the three-dimensional cubic-box problem is therefore

ψ =  Ø(x)Ø(y)Ø(z)

With

1809_Particle in a box7.png

   Related Questions in Chemistry

  • Q : Coordination compounds discuss the

    discuss the practical uses of coordination compounds, give reactions involves and explain whats happening in the process

  • Q : Quastion of finding vapour pressure

    Vapour pressure of CCl425Degree C at is 143mm of Hg0.5gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4 .Find the vapour pressure of the solution (Density of CCl4 = = 1.58g /cm2): (a)141.43mm (b)

  • Q : Molarity based question Help me to

    Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M

  • Q : Laws of Chemical Combination Laws of

    Laws of Chemical Combination- In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chem

  • Q : Effect of addition of mercuric iodide

    Give me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the:(a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

  • Q : Molality of Sulfuric acid Choose the

    Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml]  (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6

  • Q : What are biodegradable polymers?

      These are polymers that can be broken into small segments by enzyme-catalysed reactions. The required enzymes are produced by microorganism. It is a known fact that the carbon-carbon bonds of chain growth polymers are inert to enzyme-catalysed reactions, and hence they are non biod

  • Q : Problem on mol fraction of naphthalene

    At 20°C the solubility of solid naphthalene in hexane is 0.09 mol/mol of solution. Use this information and the data below to estimate the following for this system: a) The mol fraction of naphthalene in the vapour phase in equ

  • Q : Finding Molarity of final mixture Can

    Can someone help me in finding out the right answer. 25ml of 3.0 MHNO3 are mixed with 75ml of 4.0 MHNO3. If the volumes are adding up the molarnity of the final mixture would be: (a) 3.25M (b) 4.0M (c) 3.75M (d) 3.50M

  • Q : Relative lowering of the vapour pressure

    Choose the right answer from following.The relative lowering of the vapour pressure is equal to the ratio between the number of: (a) Solute moleules and solvent molecules (b) Solute molecules and the total molecules in the solution (c) Solvent molecules and the tota