# Find how to the ratio of de Broglie wavelength of fast moving electron and cutoff wavelength of X-Ray if electron is accelerated by potential difference V in the coolidge tube?

Eshan
4 years ago
Dear student,

The cutoff wavelenth corresponds to the energy of electron incident on the atom.

Hence$\dpi{80} \dfrac{hc}{\lambda}=eV\implies \lambda=\dfrac{hc}{eV}$

Fast moving electron will also have the same energy, correspoding to the same case.
Hence$\dpi{80} p=\sqrt{2meV}$
Therefore de-broglie wavelength is$\dpi{80} \lambda_B=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}$

Hence required ratio=$\dpi{80} \dfrac{\lambda_B}{\lambda}=\sqrt{\dfrac{eV}{2mc^2}}$