Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 41, Problem 63CP

(a)

To determine

The energy of the state.

(a)

Expert Solution
Check Mark

Answer to Problem 63CP

The energy of the state will be 3ω2.

Explanation of Solution

Write the Schrodinger equation for a simple harmonic oscillator.

  22md2ψdx2+12mω2x2ψ=Eψ

Here, is the reduced Plank’s constant, m is the mass of article, ω is the angular frequency of oscillator, and E is the energy of oscillator.

Write the given wave function.

  ψ=Bxe(mω/2)x2

Differentiate the above equation with respect to x.

  dψdx=d(Bxe(mω/2)x2)dx=Be(mω/2)x2+Bx(mω2)2xe(mω/2)x2=Be(mω/2)x2B(mω)x2e(mω/2)x2

Differentiate again the above equation.

  d2ψdx2=d2(Be(mω/2)x2B(mω)x2e(mω/2)x2)dx2=Bx(mω)xe(mω/2)x2B(mω)2xe(mω/2)x2B(mω)x2(mω)xe(mω/2)x2=3B(mω)xe(mω/2)x2+B(mω)2x3e(mω/2)x2

Conclusion:

Rewrite the Schrodinger equation by substituting the equations for dψdx and d2ψdx2.

  22m[3B(mω)xe(mω/2)x2+B(mω)2x3e(mω/2)x2]+12mω2x2[Bxe(mω/2)x2]=E[Bxe(mω/2)x2](3ω2)[Bxe(mω/2)x2]+(12mω2x2)[Bxe(mω/2)x2] +(12mω2x2)[Bxe(mω/2)x2]=E[Bxe(mω/2)x2](3ω2)(Bxe(mω/2)x2)=E(Bxe(mω/2)x2)

Compare the coefficients of Bxe(mω/2)x2.

    E=3ω2

Therefore, the energy of the state will be 3ω2.

(b)

To determine

The position at which there is no any possibility to find a particle.

(b)

Expert Solution
Check Mark

Answer to Problem 63CP

Particle cannot be found at x=0.

Explanation of Solution

The possibility of finding a particle depends on the value of wave function. A particle cannot be found at a position where wave function vanishes. The wave function vanishes at x=0. That means, there will be no any probability foe finding a particle at x=0.

Therefore, the particle cannot be found at x=0.

(c)

To determine

The most probable position for finding the particle.

(c)

Expert Solution
Check Mark

Answer to Problem 63CP

The maximum probability for particles are at x=±mω.

Explanation of Solution

Point of maximum probability is a point of maxima.

Write the condition to be satisfied at the point of maximum probability of finding the particle.

  dψdx=0

Conclusion:

Substitute Bxe(mω/2)x2 for ψ in the above equation.

  dψdx=d(Bxe(mω/2)x2)dx=Be(mω/2)x2B(mω)x2e(mω/2)x2=01(mω)x2=0

Rewrite the above relation in terms of x.

    x=±mω

Therefore, the maximum probabilities for particles are at x=±mω.

(d)

To determine

The value of B required for the normalization of wave function.

(d)

Expert Solution
Check Mark

Answer to Problem 63CP

B is (4m3ω3π3)1/4.

Explanation of Solution

Write the normalization condition.

  |ψ|2dx=1

Conclusion:

Substitute Bxe(mω/2)x2 for ψ in the above equation.

  B2x2e(mω/)x2dx=12B214π(mω/)3=1B2π1/22(mω)3/2=1

Rewrite the above relation in terms of B.

  B=21/2π1/4(mω)3/4=(4m3ω3π3)1/4

Therefore, B is (4m3ω3π3)1/4.

(e)

To determine

The classical probability of finding the particle in the width δ centred at x=2(mω)12.

(e)

Expert Solution
Check Mark

Answer to Problem 63CP

Classical probability is zero in the width δ.

Explanation of Solution

Write the equation for potential energy.

  U=12mω2x2

Here, U is the potential energy.

Conclusion:

Substitute 2(mω)12 for x in the above equation.

  U=12mω2(2(mω)12)2=2ω

The total energy of oscillator is 32ω. The calculated result, 2ω is greater than 32ω. That means the probability of finding the particle in width δ is zero.

Therefore, the classical probability is zero in the width δ.

(f)

To determine

The real probability of finding the particle in the width δ centred at x=2(mω)12.

(f)

Expert Solution
Check Mark

Answer to Problem 63CP

The real probability in width δ is 8δ(mωπ)1/2e4.

Explanation of Solution

Write the equation to calculate the real probability.

  P=|ψ|2dx

Here, P is the probability.

Conclusion:

Substitute Bxe(mω/2)x2 for |ψ| and δ for x in the above equation to find P.

  P=(Bxe(mω/2)x2)2δ=B2x2e(mω/2)x2δ

Rewrite the above relation by substituting (4m3ω3π3)1/2 for B and 2(mω)12 for x in the above equation to find P.

    P=(4m3ω3π3)1/2(2(mω)12)2e(mω/2)x2δ=δ2π1/2(m3/2ω3/23/2)(4mω)e(mω/)4(/mω)=8δ(mωπ)1/2e4

Therefore, the real probability in width δ is 8δ(mωπ)1/2e4.

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Chapter 41 Solutions

Physics for Scientists and Engineers With Modern Physics

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