Introduction
Wavelength
DiffractionOrder Graphite Grating Tasks
Introduction
Wavelength
DiffractionIn 1924 Louis de Broglie theorized that not only light posesses both wave and particle properties, but rather particles with mass - such as electrons - do as well. The wavelength of these 'material waves' - also known as the de Broglie wavelength - can be calculated from Planks constant \(h\) divided by the momentum \(p\) of the particle. In the case of electrons that is $$\lambda_{\text{de Broglie}} = \frac {h}{p_\text e}=\frac {h}{m_\text e\cdot v_\text e}$$ The acceleration of electrons in an electron beam gun with the acceleration voltage \(V_{\rm{a}}\) results in the corresponding de Broglie wavelength $$\bbox[8px,border:2px solid red]{\lambda_{\text{de Broglie}} = \frac {h}{m_\text e\cdot \sqrt{2\cdot \frac{e}{m_\text e}\cdot V_{\text a}}}=\frac {h}{\sqrt{2\cdot m_\text e \cdot e\cdot V_{\text a}}}}$$ Proof of the de Broglie hypothesis will be experimentally demonstrated with the help of an electron diffraction tube on the following pages.
For electons with acceleration voltage \(V_{\rm{a}}\), the following constansts will be used to calculate the de Broglie wavelengths:
$h=6{.}6\cdot 10^{-34}\,\text J\cdot \text s$, $ m_{\text e}=9{.}1\cdot 10^{-31}\,\text{kg}$, $e=1{.}6\cdot 10^{-19}\,{\text C}$
| acceleartion voltage Va | de Broglie wavelength $\lambda_{\text {de Broglie}}$ |
|---|---|
| 10 V | $3.87\cdot 10^{-10}\,\text m $ |
| 100 V | $1.22\cdot 10^{-10}\,\text m $ |
| 1000 V | $3.87\cdot 10^{-11}\,\text m $ |
| 10000 V | $1.22\cdot 10^{-11}\,\text m $ |
| V | ${}$ |
Note: The input only allows an acceleration voltage up to \(V_{\rm{a}}=100000\,\rm{V}\). For higher acceleration voltage relativistic calculation should be used.
