Question

Determine the Compton wavelength of the electron, defined to be the wavelength it would have if its momentum were${\mathbf{\text{m}}}_{\mathbf{\text{e}}}\mathbf{\text{c}}$.

Short answer

Compton Wavelength of the electron

$\lambda =\text{2.43}\times {\text{10}}^{\text{-12}}\text{m}$

Step-by-step solution

Step 1:Given and unknowns.

${\text{m}}_{\text{e}}{\text{=9.1×10}}^{\text{-31}}\text{kg--}$the electron's mass

${\text{p=m}}_{\text{e}}\text{c--}$the electron's momentum

Concept Introduction

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

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

Expression of wavelength.

Know that the electron's momentum is${\text{p=m}}_{\text{e}}\text{c}$ .

Get the wavelength expression as follows

$\text{p=}\frac{\text{h}}{\text{λ}}$

${\text{p=m}}_{\text{e}}\text{c}$

${\text{m}}_{\text{e}}\text{c=}\frac{\text{h}}{\text{λ}}$

$\text{λ=}\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}$

Compton Wavelength of Electron.

Using the wavelength's derived expression, Obtain$\lambda$as:

$\begin{array}{c}\lambda =\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}\\ =\frac{\text{6.626}\times {\text{10}}^{\text{-34}}J\cdot s}{(\text{9.1}\times {\text{10}}^{\text{-31}}kg)\times (\text{3.0}\times {\text{10}}^{\text{8}}m/s)}\\ =\text{2.43}\times {\text{10}}^{\text{-12}}\text{m}\end{array}$ $\begin{array}{c}\lambda =\frac{\text{h}}{{\text{m}}_{\text{e}}\text{c}}\\ =\frac{\text{6.626}\times {\text{10}}^{\text{-34}}J\cdot s}{(\text{9.1}\times {\text{10}}^{\text{-31}}kg)\times (\text{3.0}\times {\text{10}}^{\text{8}}m/s)}\\ =\text{2.43}\times {\text{10}}^{\text{-12}}\text{m}\end{array}$

The Compton Wavelength of the electron is$\lambda =\text{2.43}\times {\text{10}}^{\text{-12}}\text{m}$ .