Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

An electron is in a certain energy state in a one-dimensional, infinite potential well from x = 0 to x = L =200PM electron’s probability density is zero at x = 0.300 L , and x = 0.400 L ; it is not zero at intermediate values of x. The electron then jumps to the next lower energy level by emitting light. What is the change in the electron’s energy?

The change in electrons energy or the energy of the photon is

$2.86 \times 10^{- 17} J$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Introduction

In quantum mechanics, the particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle free to move in a small space surrounded by impenetrable barriers.

Step 2: Concept

Use the concept of probability density of a practice in an infinite one dimensional box.

The probability density of a wave function of a practice in an infinite one dimensional box of length is given as follows;

${|ψ_{n} (x)|}^{2} = \frac{2}{L} \sin^{2} (\frac{n π}{L} x)$

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

The probability density is zero if the angle of sin is integral multiple of $π$ .

$\sin^{2} (\frac{n πx}{L}) = 0 \frac{n πx}{L} = m π$

This is possible only if is integer that is m = 1,2,3,4...

The probability density becomes is not zero between x = 0.300 to x = 0.400L .

Thus is only possible if the state of the electron is n = 10.

The next lower energy level is n' = 9 if the electron jumps from the state n = 10 to the state n' = 9 a photon with energy equal the energy difference between the states is emitted.

Step 3: The energy of the photon emitted is given as follows:

$∆ E = \frac{n^{2} h^{2}}{8 m L^{2}} - \frac{n^{' 2} h^{2}}{8 m L^{2}} = \frac{h^{2}}{8 m L^{2}} (n^{2} - n^{' 2})$

Here, K is the Planck’s constant, m is the mass of electron, and is the length of the box.

Substitute $6.63 \times 10^{- 34} j . s for h, 9.11 \times 10^{31} k g for m for L, 10 for n, and for 9 for n'$ .

$∆ E = \frac{{(6.63 \times 10^{- 34} j . s)}^{2}}{(8) (9.11 \times 10^{31} kg) {(200 \times 10^{- 12} m)}^{2}} ({(10)}^{2} - {(9)}^{2}) = 2.86 \times 10^{- 17} J$

Therefore, the change in electrons energy or the energy of the photon is $2.86 \times 10^{- 17} J$ .

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Step 1: Introduction

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Step 3: The energy of the photon emitted is given as follows:

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