(Solved): a. The de Broglie Wavelength vs. Voltage In the cathode ray tube the electron ...

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a. The de Broglie Wavelength vs. Voltage In the cathode ray tube the electron is accelerated through high voltage \( \mathrm{V} \). Its energy and momentum are then given by \[ E=\frac{p^{2}}{2 m}=e V \] Solving for the momentum, and substituting into Eq. 1 gives: \[ \lambda=\frac{h}{\sqrt{2 e V m}} \] You should verify for yourself that this can be re-written in the practical form \[ \lambda(\text { Angstroms })=\sqrt{\frac{151.3}{V(\text { volts })}} \]