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Expert-verified Found in: Page 134 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # The average kinetic energy of a particle at temperature ${\mathbf{\text{T}}}$ is$\frac{\mathbf{\text{3}}}{\mathbf{\text{2}}}{{\mathbf{\text{k}}}}_{{\mathbf{\text{B}}}}{\mathbf{\text{T}}}$ . (a) What is the wavelength of a room-temperature (${\mathbf{\text{22°C}}}$)electron? (b) Of a room-temperature proton? (c) In what circumstances should each behave as a wave?

Relation between Wavelength and temperature

a)$\text{6.289 nm}$

b)$\text{0.146 nm}$

c) For the electron to express its wave nature, only a few nanometers are required; nevertheless, for the proton to manifest its wave nature at the same temperature, these dimensions must be lowered by at least forty times.

See the step by step solution

## Step 1: Concept Introduction

For the average kinetic energy of the particle$\frac{\mathbf{\text{3}}}{\mathbf{\text{2}}}{{\mathbf{\text{k}}}}_{{\mathbf{\text{B}}}}{\mathbf{\text{T}}}$ , the wavelength of the particle of mass m can be given as,

${\mathbit{\lambda }}{\mathbf{\text{=}}}\frac{\mathbf{\text{h}}}{\sqrt{{\mathbf{\text{3mk}}}_{\mathbf{\text{B}}}\mathbf{\text{T}}}}$…………………..(1)

## Step 2:Calculation of wavelength

(a)

Use equation (1) to calculate the wavelength of the particle, such that,

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{e}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{\text{6.63}×{\text{10}}^{\text{-34}}\text{\hspace{0.17em}}J\cdot s}{\sqrt{\text{3}×\left(\text{9.1}×{\text{10}}^{\text{-31}}\text{kg}\right)×\left(\text{1.38}×{\text{10}}^{\text{-23}}\text{\hspace{0.17em}}J/K\right)×\text{(295\hspace{0.17em}K)}}}\\ \text{}=\text{\hspace{0.17em}6.289\hspace{0.17em}nm}\end{array}$

## Step 3:Analysis of Mass and Proton.

(b)

We'll do the same analysis as part (a), but this time we'll use the proton's mass.

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{p}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{\text{6.63}×{\text{10}}^{\text{-34}}\text{\hspace{0.17em}}J\cdot s}{\sqrt{\text{3}×\left(\text{1.6726}×{\text{10}}^{\text{-27}}\text{kg}\right)×\left(\text{1.38}×{\text{10}}^{\text{-23}}\text{\hspace{0.17em}}J/K\right)×\text{(295\hspace{0.17em}K)}}}\\ =\text{0.146\hspace{0.17em}nm}\end{array}$

## Step 4: Circumstance.

(c)

When the dimension of the slit is approximately near to their wavelength, each particle will behave as a wave, e.g. diffracted from a single slit. As a result, while a few nanometers will suffice for the electron to express its wave nature, the proton will require these dimensions to be decreased forty times before it can manifest its wave nature at the same temperature.

So,

Therefore

a)$\text{6.289 nm}$

b)$\text{0.146 nm}$

c) For the electron to express its wave nature, only a few nanometers are required; nevertheless, for the proton to manifest its wave nature at the same temperature, these dimensions must be lowered by at least forty times. ### Want to see more solutions like these? 