Grade 12th passModern Physics

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?

Tapajeet Roy

8 Years agoGrade 12th pass

1 Answer

Eshan

8 Years ago

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.
Hence $p=\sqrt{2meV}$
Therefore de-broglie wavelength is $\lambda_B=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}$

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

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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?

Tapajeet Roy

1 Answer

Eshan

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