Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 34, Problem 48P

(a)

To determine

The probability that the particle will be found in the region.

(a)

Expert Solution
Check Mark

Answer to Problem 48P

The probability that the particle will be found in the region is 0.50 .

Explanation of Solution

Given:

The one-dimensional box regionis 0xL .

The particle is in the ground state.

The given region is 0<x<L2 .

Formula used:

The expression for probability for finding the particle in ground stateis given by,

  P=2Lsin2πxLdx

From integral formula,

  sin2dθ=θ2sin2θ4+c

Calculation:

Let, πxL=θ .

By differentiating both sides,

  πdxL=dθdx=Ldθπ

The limit is 0toL2 changes to 0toπ2 .

The probabilityis calculated as,

  P=2L sin2 πxLdx=0L/22Lsin2πxLdx=0π/22Lsin2θ(Ldθπ)=2π0π/2sin2θdθ

Solving further as,

  P=2π0π/2 sin2θdθ=2π[θ2sin2θ4]0π/2=2π[π40]=0.5

Conclusion:

Therefore, the probability that the particle will be found in the region is 0.50 .

(b)

To determine

The probability that the particle will be found in the region.

(b)

Expert Solution
Check Mark

Answer to Problem 48P

The probability that the particle will be found in the region is 0.196 .

Explanation of Solution

Given:

The given region is 0<x<L3 .

Formula used:

The expression for probability for finding the particle in ground state is given by,

  P=2Lsin2πxLdx

From integral formula,

  sin2dθ=θ2sin2θ4+c

Calculation:

Let, πxL=θ .

By differentiating both sides,

  πdxL=dθdx=Ldθπ

The limit is 0toL3 changes to 0toπ3 .

The probability is calculated as,

  P=2L sin2 πxLdx=0L/32Lsin2πxLdx=0π/32Lsin2θ(Ldθπ)=2π0π/3sin2θdθ

Solving further as,

  P=2π0π/3 sin2θdθ=2π[θ2sin2θ4]0π/3=2π[π6sin2π/34]=0.196

Conclusion:

Therefore, the probability that the particle will be found in the region is 0.196 .

(c)

To determine

The probability that the particle will be found in the region.

(c)

Expert Solution
Check Mark

Answer to Problem 48P

The probability that the particle will be found in the region is 0.909 .

Explanation of Solution

Given:

The given region is 0<x<3L4 .

Formula used:

The expression for probability for finding the particle in ground state is given by,

  P=2Lsin2πxLdx

From integral formula,

  sin2dθ=θ2sin2θ4+c

Calculation:

Let, πxL=θ .

By differentiating both sides,

  πdxL=dθdx=Ldθπ

The limit is 0to3L4 changes to 0to3π4 .

The probability is calculated as,

  P=2L sin2 πxLdx=03L/42Lsin2πxLdx=03π/42Lsin2θ(Ldθπ)=2π03π/4sin2θdθ

Solving further as,

  P=2π0 3π/4 sin2θdθ=2π[θ2sin2θ4]03π/4=2π[3π8sin6π/44]=0.909

Conclusion:

Therefore, the probability that the particle will be found in the region is 0.909 .

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