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Grade: 12th pass

                        

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?

2 years ago

Answers : (1)

Eshan
askIITians Faculty
2095 Points
							Dear student,

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

Hence\dfrac{hc}{\lambda}=eV\implies \lambda=\dfrac{hc}{eV}

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

Hence required ratio=\dfrac{\lambda_B}{\lambda}=\sqrt{\dfrac{eV}{2mc^2}}
2 years ago
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