Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 7, Problem 7.6IA

(a)

Interpretation Introduction

Interpretation:

The value of Δx and Δpx for the nth state of a particle in a box of length L have to be evaluated.

Concept Introduction:

The wave function represents the exact position of the electron in an atom.  The wave function is represented by ψ in the quantum mechanics.  Expectation value of an operator is known as the average value of the given operator.

(a)

Expert Solution
Check Mark

Answer to Problem 7.6IA

The value of Δx for the nth state of a particle in a box of length L is (L212L22n2π2)1/2.

The value of Δpx for the nth state of a particle in a box of length L is nπL.

Explanation of Solution

The value of Δx is calculated by the following formula.

    Δx=(x2x2)1/2        (1)

The normalized nth state wavefunction of a particle in a box of length L is shown below.

    ψ=(2L)1/2sin(nπxL)

The expectation value of x2 of a particle in a box of length x=0 to x=L is calculated by the following formula.

    x2=L0ψx2ψdx        (2)

Where,

ψ is the normalized wave function.

ψ is the conjugated complex of the normalized wave function.

The wave function is real.  So, ψ is equal to ψ.

Substitute the value of ψ and ψ in equation (1).

    x2=L0(2/L)1/2sin(nπx/L)x2(2/L)1/2sin(nπx/L)dx=L0(2/L)x2sin2(nπx/L)dx=(2/L)L0x2sin2(nπx/L)dx        (3)

The value of a0x2sin2(kx)dx is a36(a24k18k3)sin(2ka)a4k2cos2ka.

Compare the integral a0x2sin2(kx)dx with L0x2sin2(nπx/L)dx the value of a is L and the value of k is (nπL).  Therefore, the value of L0xsin2(nπx/L)dx, using the above integral identity is calculated as shown below.

    L0x2sin2(nπx/L)dx=(L36(L24(nπ/L)18(nπ/L)3)sin(2(nπ/L)L)L4(nπ/L)2cos2(nπ/L)L)=(L36(L24(nπ/L)18(nπ/L)3)sin(2nπ)L34n2π2cos2nπ)=(L36(L24(nπ/L)18(nπ/L)3)(0)L34n2π2(1))(

Substitute the value of 0Lx2sin2(nπx/L)dx in equation (3).

    x2=(2/L)(L36L34n2π2)=(2/L)L2(L23L22n2π2)=(L23L22n2π2)

Therefore, the value of x2 is (L23L22n2π2).

The expectation value of x2 of a particle in a box of length in range of x=0 to x=L is calculated by the following formula.

    x2=(0Lψxψdx)2        (4)

Substitute the value of ψ and ψ in equation (1).

    x2=(0L(2/L)1/2sin(nπx/L)x(2/L)1/2sin(nπx/L)dx)2=(0L(2/L)xsin2(nπx/L)dx)2=((2/L)0Lxsin2(nπx/L)dx)2        (5)

The value of 0axsin2(kx)dx is a2414k(asin(2ka))18k2(cos2ka1).

Compare the integral 0axsin2(kx)dx with 0Lxsin2(nπx/L)dx the value of a is L and the value of k is (nπL).  Therefore, the value of 0Lxsin2(πx/L)dx, using the above integral identity is calculated as shown below.

    0Lxsin2(nπx/L)dx=(L2414×nπL(Lsin(2×nπL×L))18×(nπL)2(cos(2×nπL×L)1))=L24L4nπ(Lsin(2nπ))L28n2π2(cos(2nπ)1)=L24L4nπ(0)L232π2(11)(sin(2nπ)=0andcos(2nπ)=1)=L24

Substitute the value of 0Lxsin2(nπx/L)dx in the expectation value of x in equation (5).

    x2=((2/L)(L24))2=(L2)2=L24

Therefore, the value of x2 is L24.

Substitute the value of x2 and x2 in equation (1).

    Δx=((L23L22n2π2)L24)1/2=(L23L22n2π2L24)1/2=(L212L22n2π2)1/2

Hence, the value of Δx for the ground state of a particle in a box of length L is (L212L22n2π2)1/2.

The value of Δpx is calculated by the following formula.

    Δpx=(px2px2)1/2        (6)

The expectation value of the momentum of the electron is calculated by the following expression.

    px2=0L(2/L)1/2sin(nπx/L)px2(2/L)1/2sin(nπx/L)dx=(2/L)0Lsin(nπx/L)(iddx)2sin(nπx/L)dx(px=iddx)=22i2L0Lsin(nπx/L)ddx(ddxsin(nπx/L))dx(ddxsinx=cosx)=22L0Lsin(nπx/L)(nπ/L)ddxcos(nπx/L)dx(i2=1)

On further solving the above integration,

    px2=22L0Lsin(nπx/L)(nπ/L)(sin(nπx/L))ddx(nπx/L)dx=22L0Lsin(nπx/L)(nπ/L)sin(nπx/L)ddx(nπx/L)dx=22L0Lsin(nπx/L)(nπ/L)2sin(nπx/L)dx=22L(nπ/L)20Lsin2(nπx/L)dx        (7)

The value of 0asin2kxdx is 12a14ksin2ka.

Compare the integral 0Lsin2(nπx/L)dx with 0asin2kxdx the value of k is (nπL).  Therefore, the value of 0Lsin2(nπx/L)dx, using the above integral identity is calculated as shown below.

    0Lsin2(nπx/L)dx=12L14(nπL)sin2(nπL)L=L2L4nπsin2nπ(sin2nπ=0)=L2

Substitute the value of 0Lsin2(nπx/L)dx in equation (7).

    px2=22L(nπ/L)2L2=2n2π2L2

Therefore, the value of px2 for the normalized wavefunction (2/L)1/2sin(nπx/L) is 2n2π2L2.

The expectation value of the square of the momentum of the particle is calculated by the following expression.

    px2=(0L(2/L)1/2sin(nπx/L)px(2/L)1/2sin(nπx/L)dx)2=(2iL0Lsin(nπx/L)ddxsin(nπx/L)dx)2(p^x=iddx)=(2iL0Lsin(nπx/L)cos(nπx/L)×ddx(nπx/L)dx)2(ddxsinx=cosx)=(2iL0Lsin(nπx/L)cos(nπx/L)×(nπ/L)dx)2

The value of 0asinAxcosBxdx is cos(BA)a12(BA)cos(A+B)a12(A+B).

Compare the integral 0Lsin(nπx/L)cos(nπx/L)dx with 0asinAxcosBxdx the value of a is L and the values of A and B are (nπL).  Therefore, the value of 0Lsin(nπx/L)cos(nπx/L)dx, using the above integral identity is calculated as shown below.

0Lsin(nπx/L)cos(nπx/L)dx=cos(((nπ/L)(nπ/L))L)12((nπ/L)(nπ/L))cos(((nπ/L)+(nπ/L))L)12((nπ/L)+(nπ/L))=cos(0)12((nπ/L)(nπ/L))cos(2nπ)12((nπ/L)+(nπ/L))=112((nπ/L)(nπ/L))112(2nπ/L)(cos0=1andcos2nπ=1)=0

Substitute the value of 0Lsin(nπx/L)cos(nπx/L)dx in equation (2).

    px2=(2iL×0×(π/L))2=0

Therefore, the value of px2 for the normalized wavefunction (2/L)1/2sin(nπx/L) is 0.

Substitute the value of px2 and px2 in equation (7).

    Δpx=(2n2π2L20)1/2=nπL

Hence, the value of Δpx is nπL.

(b)

Interpretation Introduction

Interpretation:

The value of Δx and Δpx for the νth state of a harmonic oscillator have to be evaluated.

Concept Introduction:

A system that experiences a restoring force when displaced from its equilibrium position is known as the harmonic oscillator.  The restoring force experienced by the system is directly proportional to the displacement.  Masses connected to the spring and the pendulums are the examples of harmonic oscillator.

(b)

Expert Solution
Check Mark

Answer to Problem 7.6IA

The value of Δx for the νth state of a harmonic oscillator is (ν+12)1/2((mkf)1/2)1/2.

The value of Δpx for νth state of a simple harmonic oscillator is (12(kfm)1/2)1/2.

Explanation of Solution

The value of x for the νth state of a harmonic oscillator is zero.  Therefore, the value of x2 for the νth state of a harmonic oscillator is zero.

The value of x2 for the νth state of a harmonic oscillator is (ν+12)(mkf)1/2.  The value of kf represents the force constant and m represents the mass of particle.

Substitute the value of x2 and x2 in equation (1).

    Δx=((ν+12)(mkf)1/20)1/2=((ν+12)(mkf)1/2)1/2=(ν+12)1/2((mkf)1/2)1/2

Hence, the value of Δx for the νth state of a harmonic oscillator is (ν+12)1/2((mkf)1/2)1/2.

The value of px for the νth state of a harmonic oscillator is zero.  Therefore, the value of px2 for the νth state of a harmonic oscillator is zero.

The value of px2 for the νth state of a harmonic oscillator is 12(kfm)1/2.  The value of kf represents the force constant and m represents the mass of particle.

Substitute the value of px2 and px2 in equation (6).

    Δpx=(12(kfm)1/20)1/2=(12(kfm)1/2)1/2

Hence, the value of Δpx for νth state of a simple harmonic oscillator is (12(kfm)1/2)1/2.

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Chapter 7 Solutions

Atkins' Physical Chemistry

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