Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be $\frac{3}{2} k_{B} T$ . Show that the wavelength of a thermal particle is given by
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

As a result, were able to demonstrate that :

$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Given and unknowns.

Particle's average kinetic energy is $KE = \frac{3}{2} k_{B} T$

$λ = \frac{h}{\sqrt{{3mk}_{B} T}}$

Step 2: Concept Introduction

The following equation can be used to describe the de Broglie wavelength:

$p= \frac{h}{λ} . ........ (1)$

When it comes to momentum, it's best to think of it this way:

$p = mv . ...... (2)$

The kinetic energy, on the other hand, can be expressed by the following equation:

$KE = \frac{1}{2} {mv}^{2} . ....... (3)$

Step 3:Expression for wavelength.

Get the expression for using Equations (1) and (2) $λ$ :

$\frac{h}{λ} =mv$

$λ= \frac{h}{mv}$

Step 4: Expression for v.

Now use the average kinetic energy and Equation (3) to get the expression for the particle's speed because don't have one

$KE= \frac{1}{2} {mv}^{2}$

$KE= \frac{3}{2} k_{B} T$

$\frac{1}{2} {mv}^{2} = \frac{3}{2} k_{B} T$

$v^{2} = \frac{2× \frac{3}{2} k_{B} T}{m}$

$v= \sqrt{\frac{{3k}_{B} T}{m}}$

Step 5: Derived expression for v.

The generated expression is then plugged into the expression of $λ$ :

$\begin{matrix} λ= \frac{h}{mv} \\ = \frac{h}{m× (\sqrt{\frac{{3k}_{B} T}{m}})} \\ = \frac{h}{\sqrt{{3mk}_{B} T}} \end{matrix}$

As a result, were able to demonstrate that $λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

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Modern Physics

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Short Answer

A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be $\frac{3}{2} k_{B} T$ . Show that the wavelength of a thermal particle is given by
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

Step by Step Solution

Step 1:Given and unknowns.

Step 2: Concept Introduction

Step 3:Expression for wavelength.

Step 4: Expression for v.

Step 5: Derived expression for v.

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Answers without the blur.

Short Answer

A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be 32kBT . Show that the wavelength of a thermal particle is given by λ=h3mkBT

Step by Step Solution

Step 1:Given and unknowns.

Step 2: Concept Introduction

Step 3:Expression for wavelength.

Step 4: Expression for v.

Step 5: Derived expression for v.

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A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be $\frac{3}{2} k_{B} T$ . Show that the wavelength of a thermal particle is given by
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$