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 by

λ=h3mkBT

Short Answer

Expert verified

As a result, were able to demonstrate that :

λ=h3mkBT

Step by step solution

01

Step 1:Given and unknowns.

Particle's average kinetic energy isKE=32kBT

λ=h3mkBT

02

Concept Introduction

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

p=hλ.........(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=12mv2........(3)

03

Step 3:Expression for wavelength.

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

hλ=mv

λ=hmv

04

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=12mv2

KE=32kBT

12mv2=32kBT

v2=32kBTm

v=3kBTm

05

Derived expression for v.

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

λ=hmv=h(3kBTm)=h3mkBT

As a result, were able to demonstrate thatλ=h3mkBT

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