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Engineering-Physics-06PHY12--BE-I-Semester--VTU-Belagaum-Unit-1-Modern-physics-->View question

Asked On2017-06-12 05:08:27 by:milan-ransingh

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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

d2ψ/dx2+ 2m/Ћ2[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

d2/ψ/dx2+ 2m/Ћ2Eψ=0

Or                 d2ψ/dx2+ K2ψ=0                                          (2)

Where                       k=    2mE/Ћ(3)

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

he 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


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.

Answerd on:2017-06-13 Answerd By:prajwalamv

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