Chapter 4: Q18E (page 134)

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
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

Short Answer

Expert verified

As a result, were able to demonstrate that :

$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

Step by step solution

Step 1:Given and unknowns.

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

$λ = \frac{h}{\sqrt{{3mk}_{B} T}}$

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) $λ$ :

$\frac{h}{λ} =mv$

$λ= \frac{h}{mv}$

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

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

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

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

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

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

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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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
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$

Short Answer

Step by step solution

Step 1:Given and unknowns.

Concept Introduction

Step 3:Expression for wavelength.

Expression for v.

Derived expression for v.

One App. One Place for Learning.

Most popular questions from this chapter

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A particle is “thermal” if it is in equilibrium with its surroundings – its average kinetic energy would be32kBT. Show that the wavelength of a thermal particle is given byλ=h3mkBT

Short Answer

Step by step solution

Step 1:Given and unknowns.

Concept Introduction

Step 3:Expression for wavelength.

Expression for v.

Derived expression for v.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Engineering Physics

Magnetism and Electromagnetic Induction

Particle Model of Matter

Physics of Motion

Atoms and Radioactivity

Translational Dynamics

Study anywhere. Anytime. Across all devices.

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
$λ= \frac{h}{\sqrt{{3mk}_{B} T}}$