Question

Roughly speaking for what range of wavelengths would we need to treat an electron relativistically, and what would be the corresponding range of accelerating potentials? Explain your assumptions.

Short answer

Rane of Wavelength and Range of Accelerating potentials:

$\text{λ}\le \text{2.43}\times {\text{10}}^{\text{-11}}\text{\hspace{0.17em}m,\hspace{0.17em}and\hspace{0.17em}}V\ge \text{2.556\hspace{0.17em}}k\text{eV}$

Step-by-step solution

Step 1:Concept Introduction

The wavelength of any moving charged particle is given by the,

$\lambda =\frac{\text{h}}{\sqrt{\text{2mqV}}}$…………………..(1)

Where m, q, and V are the mass, charge, and applied potential difference.

Step 2:Formula.

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{{\text{m}}_{\text{e}}\text{v}}\\ \text{λ}=\frac{\text{6.63}\times {\text{10}}^{\text{-34}}J\cdot s}{(\text{9.1}\times {\text{10}}^{\text{-31}}\text{\hspace{0.17em}kg})\times (\text{3}\times {\text{10}}^{\text{7}}\text{\hspace{0.17em}m/s})}\\ \text{λ}=\text{2.43}\times {\text{10}}^{\text{-11}}\text{m}\end{array}$

Relativistic Treatment.

Relativistic treatment is required for wavelengths equal to or smaller than this value.

$\begin{array}{c}\text{V}=\frac{{\text{h}}^{\text{2}}}{{\text{2m}}_{\text{e}}{\text{qλ}}^{\text{2}}}\\ \text{V}=\frac{{(\text{6.63}\times {\text{10}}^{\text{-34}}\text{J}\cdot \text{s})}^{\text{2}}}{\text{2}\times (\text{9.1}\times {\text{10}}^{\text{-31}}\text{kg})\times (\text{1.6}\times {\text{10}}^{\text{-19}}\text{\hspace{0.17em}C})\times {(\text{2.43}\times {\text{10}}^{\text{-11}}\text{\hspace{0.17em}m})}^{\text{2}}}\\ \text{V}=\text{2.556\hspace{0.17em}keV}\end{array}$

Relativistic therapy will be required if the potential is equal to or greater than this value.

$\text{λ}\le \text{2.43}\times {\text{10}}^{\text{-11}}\text{\hspace{0.17em}m,\hspace{0.17em}and\hspace{0.17em}}V\ge \text{2.556\hspace{0.17em}}k\text{eV}$