Use the measurement table to compare the wavelength expected by de Broglie $\lambda_{\text {de Broglie}}=\frac{h}{\sqrt{2\cdot m_{\text e}\cdot e\cdot U_{\text b} }}$ with the wavelength $\lambda_{\text {Experiment}}=2\cdot d\cdot \sin\left(\frac{1}{2}\cdot \tan^{-1}\left(\frac{r}{L-R+\sqrt{R^2-r^2}}\right)\right)$ provided by the inner interference ring in the experiment.
Given: $m_\text e=9.1\cdot 10^{-31}\,\text{kg}$; $e=1.6\cdot 10^{-19}\,\text{C}$; $h=6.6\cdot 10^{-34}\, \text J \cdot \text s$; $d=2.13\cdot 10^{-10}\, \text m$; $L=12.7\,\text {cm}$; $R=6.35\,\text {cm}$
Voltage \(V_{\rm{a}}\) | \(\lambda_{\rm{de-Broglie}}\) | Radius \(r_{\rm{inner}}\) | \(\lambda_{\rm{Experiment}}\) |
---|---|---|---|
kV | \( \) | cm | \( \) |
kV | \( \) | cm | \( \) |
kV | \( \) | cm | \( \) |