Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 28, Problem 71P

(a)

To determine

The sketches of the wave function of a particle confined in a finite box.

(a)

Expert Solution
Check Mark

Answer to Problem 71P

The wave function is given below.

Explanation of Solution

For a particle in a box, the wave function in the nth state has n1 nodes.

For n=1 state, the wave function is given below:

Physics, Chapter 28, Problem 71P , additional homework tip  1

Since the wave function is a sinusoidal curve decaying outside the box, the box contains less than half the wavelength. Thus,

λ2>L and λ<

i.e.λ>2L1 and λ<2L11

Here, λ is the wavelength and L is the length of the box.

For n=2 state, the wave function is given below:

Physics, Chapter 28, Problem 71P , additional homework tip  2

Since the wave function is a sinusoidal curve decaying outside the box, the box contains less than one wavelength and more than half the wavelength.

λ>L and λ<2L

i.e.λ>2L2 and λ<2L21

For n=3 state, the wave function is given below:

Physics, Chapter 28, Problem 71P , additional homework tip  3

Since the wave function is a sinusoidal curve decaying outside the box, the box contains less than 1.5 times the wavelength but more than one wavelength.

3λ2>L and λ<L

i.e.λ>3L2 and λ<2L31

From all the above given wave functions, it can be generalised that

2Ln<λn<2Ln1                                                             (I)

Here, λn is the wave function of the nth state.

(b)

To determine

Check whether the condition (n1)2E1<En<n2E1  is satisfied for a particle confined in a finite box.

(b)

Expert Solution
Check Mark

Answer to Problem 71P

Yes. The given condition is valid for a particle confined in a finite box.

Explanation of Solution

The energy of 3d level is 1.6 eV and the energy of 3p level is 3.0 eV.

Write the expression for energy of a particle in a infinite box in terms of wavelength

En=h22mλn2                                                                    (II)

Here, En is the energy of the nth state, h is the Planck’s constant and m is the mass of the particle.

Rearranging (II) for λn2

λn2=h22mEn                                                           (III)

Write the expression for ground state energy

E1=h28mL2                                                           (IV)

Write expression (I)

2Ln<λn<2Ln1                                                            (I)

Taking reciprocal and squaring (I)

(n1)2(2L)2<1λn2<n2(2L)2                                                             (V)

While taking reciprocal, the direction of inequality changes.

Multiplying (III) by E1(2L)2

(n1)2E1<(2L)2E1λn2<n2E1                                                         (VI)

Substituting (III) in (VI)

(n1)2E1<(2L)2E12mEnh2<n2E1                                                          (VII)

Substituting for E1 from (IV) in (VII)

(n1)2E1<(h28mL2)8mL2Enh2<n2E1

Cancelling out the common terms in numerator and denominator

(n1)2E1<En<n2E1

Thus, the given condition is valid.

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Chapter 28 Solutions

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