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# Application of Schrodinger wave equation: Particle in a box

Consider one dimensional closed box of width L. A particle of mass ‘m’ is moving in a one-dimensional region along X-axis specified by the limits x=0 and x=L as shown in fig. The potential energy of particle inside the box is zero and infinity elsewhere.

I.e Potential energy V(x) is of the form

V(x) = {o; if o<x<L

∞: elsewhere

The one-dimensional time independent Schrodinger wave equation is given by

d^{2}ψ/dx^{2}+ 2m/Ћ^{2}[E-V] ψ=0 (1)

Here we have changed partial derivatives in to exact because equation now contains only one variable i.e x-Co-ordinate. Inside the box V(x) =0

Therefore the Schrodinger equation in this region becomes

d^{2}/ψ/dx^{2}+ 2m/Ћ^{2}Eψ=0

Or d^{2}ψ/dx^{2}+ K^{2}ψ=0 (2)

Where k= 2mE/Ћ^{2 }(3)

K is called the Propagation constant of the wave associated with particle and it has dimensions reciprocal of length.

The general solution of eq (2) is^{ }

^{ }Ψ=A sin Kx + B cos K x (4)

Where A and B are arbitrary conditions and these will be determined by the boundary conditions.

(ii) **Boundary Conditions**

The particle will always remain inside the box because of infinite potential barrier at the walls. So the probability of finding the particle outside the box is zero i.e.ψx=0 outside the box.

We know that the wave function must be continuous at the boundaries of potential well at x=0 and x=L, i.e.

Ψ(x)=0 at x=0 (5)

Ψ(x)=0 at x= L (6)

These equations are known as Boundary conditions.

**(iii) Determination of Energy of Particle**

Apply Boundary condition of eq.(5) to eq.(4)

0=A sin (X*0) +B cos (K*0)

0= 0+B*1

B=0 (7)

Therefore eq.(4) becomes

Ψ(x) = A sin Kx (8)

Applying the boundary condition of eq.(6) to eq.(8) ,we have

0=A sin KL

Sin KL=0

KL=nπ

K=nπ/L (9)

Where n= 1, 2, 3 – - -

A Cannot be zero in eq. (9) because then both A and B would be zero. This will give a zero wave function every where which means particle is not inside the box.

**Wave functions.** Substitute the value** **of K from eq. (9) in eq. (8) to get

Ψ(x)=A sin(nπ/Lx)

As the wave function depends on quantum number π so we write it ψ_{n}. Thus

Ψ_{n}=A sin (nπx/L)0<x<L

This is the wave function or eigen function of the particle in a box.

Ψ_{n}=0 outside the box

**Energy value or Eigen value of particle in a box:** Put this value of K from equation (9) in eq. (3)

nπ/L = 2m E/Ћ^{2}

Squaring both sides

n^{2}π^{2}/L^{2}=2mE/Ћ^{2}

E=n^{2}π^{2}Ћ^{2}/2mL^{2}

Where n= 1, 2, 3… Is called the Quantum number

As E depends on n, we shall denote the energy of particle ar E_{n}. Thus

E_{n}= n^{2}π^{2}Ћ^{2}/2mL^{2} (10)

This is the eigen value or energy value of the particle in a box.

This entry was posted by shiva on November 16, 2011 at 5:18 pm, and is filed under Quantum Physics. Follow any responses to this post through RSS 2.0. You can leave a response or trackback from your own site. |