--%>

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 : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998

  • Q : Film Mass Transport Sulfur trioxide

    Sulfur trioxide (SO3) is manufactured by the gas-phase oxidation of SO2 over a platinum catalyst: SO2 + ½ O2 à SO3 The catalyst is a non-porous ext

  • Q : Pressure and power for adiabatic

    a) Air flowing at 1 m3/s enters an adiabatic compressor at 20°C and 1 bar. It exits at 200°C. The isentropic efficiency of the compressor is 80%. Calculate the exit pressure and the power required. b) Steam enter

  • Q : Problem on bubble point The following

    The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery.

    Q : Number of mlecules in methane Can

    Can someone please help me in getting through this problem. The total number of molecules in 16 gm of methane will be: (i) 3.1 x 1023 (ii) 6.02 x 1023 (iii) 16/6.02 x 1023 (iv) 16/3.0 x 1023

  • Q : Problem on melting of ice A) It has

    A) It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. Suppose that the width of the skate in contact with the ice has been reduced by sh

  • Q : Hydrocarbons list and identify

    list and identify differences between the major classes of hydrocarbons

  • Q : Normality of solution containing

    Can someone please help me in getting through this problem. Determine the normality of a solution having 4.9 gm H3PO4 dissolved in 500 ml water: (a) 0.3  (b) 1.0  (c) 3.0   (d) 0.1

  • Q : Equimolar solutions Select the right

    Select the right answer of the question. Equimolar solutions in the same solvent have : (a)Same boiling point but different freezing point (b) Same freezing point but different boiling poin (c)Same boiling and same freezing points (d) Different boiling and differe

  • Q : Question based on vapour pressure and

    Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)