Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
Question
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Chapter 34, Problem 51P

(a)

To determine

The value of xandx2 for a particle in nth state.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The value of x is 12L and value of x2 is 13L2L22n2π2 for a particle in nth state.

Explanation of Solution

Given:

The one-dimensional box region is 0xL .

The particle is in the nth state.

The wave functionis ψn(x)=2LsinnπxLfor,n=1,2,3,... .

Formula used:

The expression for x in nth state is given by,

  x=x( ψ n ( x ))2dx

The expression for x2 in nth state is given by,

  x2=x2( ψ n ( x ))2dx

The integral formula,

  θsin2θdθ=θ24θsin2θ4cos2θ8+c

The integral formula,

  θ2sin2θdθ=θ36(θ2418)sin2θθcos2θ4+c

Calculation:

Let, nπxL=θ .

By differentiating both sides,

  nπdxL=dθdx=Ldθnπ

The limit is 0toL changes to 0tonπ .

The x is calculated as,

  x=x ( ψ n ( x ) )2dx=0Lx( 2 L sin nπx L )2dx=0nπ( Lθ nπ)2Lsin2θ(Ldθnπ)=(2Ln2π2)0nπθsin2θdθ

Solving further as,

  x=( 2L n 2 π 2 )0nπθ sin2θdθ=(2L n 2 π 2)[ θ 24θsin2θ4cos2θ8]0nπ=(2L n 2 π 2)[n2π2400]=L2

The x2 is calculated as,

  x2=x2 ( ψ n ( x ) )2dx=0Lx2( 2 L sin nπx L )2dx=0nπ( Lθ nπ )22Lsin2θ(Ldθnπ)=(2L2n3π3)0nπθsin2θdθ

Solving further as,

  x2=( 2 L 2 n 3 π 3 )0nπθ2 sin2θdθ=(2 L 2 n 3 π 3)[ θ 36( θ 2 4 1 8)sin2θθcos2θ4]0nπ=(2 L 2 n 3 π 3)[n3π360nπ4]=13L2L22n2π2

Conclusion:

Therefore, the value of x is 12L and value of x2 is 13L2L22n2π2 for a particle in nth state.

(b)

To determine

The comparison of xandx2 for n1 with the value of problem 50 .

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The value of xandx2 for the classical probability distribution function, P(x)=1L and for the wave function ψn(x)=2LsinnπxL in the nth state while n1 is same.

Explanation of Solution

Calculation:

The value of x in both the cases is same and equalto L2 .

The value of x2 for n1 is calculated as,

  x2=13L2L22n2π2=13L20=13L2

Which is same as in case of problem 50 .

Conclusion:

Therefore, the value of xandx2 for the classical probability distribution function, P(x)=1L and for the wave function ψn(x)=2LsinnπxL in the nth state while n1issame.

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