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Chapter 28, Problem 55P

(a)

To determine

The energy of the particle in terms of h, m and L.

(a)

Expert Solution
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Answer to Problem 55P

The energy of the particle in terms of h, m and L is E=h24π2mL2.

Explanation of Solution

Write the Schrodinger’s equation.

  22md2Ψdx2+UΨ=EΨ        (I)

Here, is the reduced Planck’s constant, m is the mass of the particle, Ψ is the given wave function, U is the potential energy and E is the total energy of the particle.

Write the equation for the potential energy.

  U(x)=2x2mL2(L2x2)        (II)

Here, L is the half of the length of the potential well.

Write the expression of the given wavefunction.

  Ψ(x)=A(1x2L2)        (III)

Take the derivative of the above wave function with respect to x.

  dΨdx=A(02xL2)=2AxL2

Take the derivative of the above equation with respect to x.

  d2Ψd2x=ddx(2AxL2)=2AL2        (IV)

Put equations (II), (III) and (IV) in equation (I) and rearrange it.

  22m(2AL2)+(h2x2mL2(L2x2))A(1x2L2)=EA(1x2L2)2AmL22x2AmL2(L2x2)(L2x2)L2=EA(1x2L2)2mL22x2mL4=E(1x2L2)2mL2(1x2L2)=E(1x2L2)        (V)

Equation (V) will be true for all x if

  E=2mL2        (VI)

Write the equation for the reduced Planck’s constant.

  =h2π        (VII)

Here, h is the Planck’s constant.

Conclusion:

Put equation (VII) in (VI).

  E=(h2π)2mL2=h24π2mL2

Therefore, the energy of the particle in terms of h, m and L is E=h24π2mL2.

(b)

To determine

The normalization constant A.

(b)

Expert Solution
Check Mark

Answer to Problem 55P

The normalization constant A is 1516L.

Explanation of Solution

The given wavefunction is an even wavefunction.

Write the normalization condition for the given even wave function.

  20L|Ψ2|dx=1        (VIII)

Put equation (III) in equation (VIII) and rearrange it.

  20L|[A(1x2L2)]2|dx=120LA2(1x2L2)2dx=12A20L(12x2L2+x4L4)dx=1

Integrate the above equation.

  2A2[x2x33L2+x55L4]0L=12A2(L2L33L2+L55L40)=12A2(L2L3+L5)=1A2(16L15)=1

Rearrange the above equation for A.

  A2=1516LA=1516L

Conclusion:

Therefore, the normalization constant A is 1516L.

(c)

To determine

The probability that the particle is located in the range L/3xL/3.

(c)

Expert Solution
Check Mark

Answer to Problem 55P

The probability that the particle is located in the range L/3xL/3 is 0.580.

Explanation of Solution

Write the equation for the probability that the particle lies in the range L/3xL/3 for the given even wavefunction.

  P=20L/3Ψ2dx        (IX)

Here, P is the probability.

Put equation (III) in equation (IX) and rearrange it.

  P=20L/3|[A(1x2L2)]2|dx=20L/3A2(1x2L2)2dxP=2A20L/3(12x2L2+x4L4)dx

Integrate the above equation.

  P=2A2[x2x33L2+x55L4]0L/3=2A2(L32(L/3)33L2+(L/3)55L40)=2A2(L32L381L2+L51215L4)=2A2(L32L81+L1215)        (X)

Conclusion:

Substitute 1516L for A in equation (X) to find P.

P=2(1516L)2(L32L81+L1215)=21516L(L32L81+L1215)=4781=0.580

Therefore, the probability that the particle is located in the range L/3xL/3 is 0.580.

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Chapter 28 Solutions

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