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.

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

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

Division by Ø(x)Ø(y)Ø(z) gives

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

Or

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

#### Related Questions in Chemistry

• ##### Q :Amount of glucose in blood What is the

What is the normal amount of glucose in 100ml of blood (8–12 hrs after meal) is: (i) 8mg (ii) 80mg (iii) 200mg (iv) 800mg Choose the right answer from above.

• ##### Q :Explain Phase Rule The relation between

The relation between the number of phases, components and the degrees of freedom is known as the phase rule. One constituent systems: the identification of an area on a P-versus-T with one phase of a component system illustrates the two degrees of freedom that

• ##### Q :Molarity of the final mixture Can

Can someone please help me in getting through this problem. Two solutions of a substance (that is, non electrolyte) are mixed in the given manner 480 ml of 1.5M first solution + 520 ml of 1.2M second solution. Determine the molarity of the final mixture

• ##### Q :Explosions produce carbon dioxide

Illustrate all the explosions produce carbon dioxide?

• ##### Q :Problem on making solutions The weight

The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g  (b) 1g  (c) 2g  (d) 10g Choose the right answer from above.

• ##### Q :Define alum Illustrate alum?

Illustrate alum?

• ##### Q :Electrons present in a benzene How

How many electrons are present in a benzene?

• ##### Q :Relative lowering in vapour pressure of

Give me answer of this question. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is: (a) Henry law (b) Dulong and Petit law (c) Raoult's law (d) Le-Chatelier's pri

• ##### Q :P block why BiF3 is ionic whereas other

why BiF3 is ionic whereas other trihalides are covalent in nature?

• ##### Q :Explain Second Order Rate Equations.

Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o