Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
Question
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Chapter 34, Problem 53P
To determine

The value of xandx2 .

Expert Solution & Answer
Check Mark

Answer to Problem 53P

The value of x is 0 and value of x2 is L2[11212π2] .

Explanation of Solution

Given:

The one-dimensional box region is L2xL2 .

The one-dimensional box length is L .

Centre at origin.

The particle mass is m

The wave function for n=1,3,5,... is ψ(x)=2LcosnπxL .

The wave function for n=2,4,6,... is ψ(x)=2LsinnπxL .

State is ground (n=1) .

Formula used:

The expression for x is given by,

  x=xψ2(x)dx

The expression for x2 is given by,

  x2=x2ψ2(x)dx

The integral formula,

  θ2sin2θdθ=θ36(θ2418)sin2θθcos2θ4+c

Calculation:

The x is calculated as,

  x=xψ2(x)dx=L/2L/2x( 2 L cos πx L )2dx

The function x( 2 L cos πxL)2 is odd function.

Solving further as,

  x= L/2 L/2 x( 2 L cos πx L )2dx=0

The x2 is calculated as,

  x2=x2ψ2(x)dx=L/2L/2x2( 2 L cos πx L )2dx

The function x2( 2 L cos πxL)2 is even function.

Solving further as,

  x2= L/2 L/2 x 2( 2 L cos πx L )2dx=20L/2x2( 2 L cos πx L )2dx

Let, πxL=θ .So,

  πdxL=dθ

Solving further as,

  x2=20 L/2 x 2( 2 L cos πx L )2dx=20π/2 ( Lθ π )2( 2 L cosθ)2(Ldθπ)=4L2π30π/2θ2cos2θdθ=4L2π3[0π/2θ2dθ0π/2θ2sin2θdθ]

Solving further as,

  x2=4L2π3[0 π/2 θ 2dθ0π/2θ2 sin2θdθ]=4L2π3[{ θ 3 3}0π/2{ θ 3 6( θ 2 4 1 8 )sin2θ θcos2θ4}0π/2]=4L2π3[{π324}{π3480π8}]=L2[11212π2]

Conclusion:

Therefore, the value of x is 0 and value of x2 is L2[11212π2] .

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