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Q18E

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Found in: Page 134

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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

As a result, were able to demonstrate that :

$\text{λ=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}$

See the step by step solution

## Step 1:Given and unknowns.

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

$\lambda \text{=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}$

## Step 2: Concept Introduction

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

${\mathbf{\text{p=}}}\frac{\mathbf{\text{h}}}{\mathbf{\text{λ}}}{\mathbf{.}}{\mathbf{........}}{\mathbf{\text{(1)}}}$

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

${\mathbf{\text{p}}}{\mathbf{=}}{\mathbf{\text{mv}}}{\mathbf{.}}{\mathbf{......}}{\mathbf{\text{(2)}}}$

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

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

## Step 3:Expression for wavelength.

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

$\frac{\text{h}}{\text{λ}}\text{=mv}$

$\text{λ=}\frac{\text{h}}{\text{mv}}$

## 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

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

$\text{KE=}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

${\text{v}}^{\text{2}}\text{=}\frac{\text{2×}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}}{\text{m}}$

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

## Step 5: Derived expression for v.

The generated expression is then plugged into the expression of $\text{λ}$:

$\begin{array}{c}\text{λ=}\frac{\text{h}}{\text{mv}}\\ \text{=}\frac{\text{h}}{\text{m×}\left(\sqrt{\frac{{\text{3k}}_{\text{B}}\text{T}}{\text{m}}}\right)}\\ \text{=}\frac{\text{h}}{\sqrt{{\text{3mk}}_{\text{B}}\text{T}}}\end{array}$

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