--%>

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 : What is laser and explain its working?

    Laser action relies on a non-Boltzmann population inversion formed by the absorption of radiation and vibrational deactivation that forms a long lived excited electronic state. An excited state molecule can move to a lower energy state or return to the

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Dipole attractions-London dispersion

    Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Infrared Adsorption The adsorption of

    The adsorption of infrared radiation by diatomic molecules increases the vibrational energy fo molecules and gives information about the force constant for the "spring" of the molecule.;The molecular motion that has the next larger energy level spacing aft

  • Q : How to calculate solutions ionic

    Transference numbers and molar conductors can be used to calculate ionic mobilities. This tables under is giving the transference numbers for positive ions at 25 degree C and the values obtained by extrapolation to infinite dilution:

    Q : Unit of molality Select the right

    Select the right answer of the question. The unit of molality is: (a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre

  • Q : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2

  • Q : What are diazonium salts? The diazonium

    The diazonium salts are represented by the general formula ArN2 +X where X- ion may be anion such as (Cl) ¨, B ¨r, HSO

  • Q : Explain equilibrium and molecular

    The equilibrium constant can be treated as a particular type of molecular distribution. Consider the simplest gas-phase reaction, one in which molecules of A are converted to molecules of B. the reaction, described by the equation

    Discover Q & A

    Leading Solution Library
    Avail More Than 1444354 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads
    No hassle, Instant Access
    Start Discovering

    18,76,764

    1942371
    Asked

    3,689

    Active Tutors

    1444354

    Questions
    Answered

    Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

    Submit Assignment

    ©TutorsGlobe All rights reserved 2022-2023.