Electrons with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

Options

A. `lambda_(0)=(2mclambda^(2))/(h)`
B. `lambda_(0)=(2h)/(mc)`
C. `lambda_(0)=(2m^(2)c^(2)lambda^(2))/(h^(2))`
D. `lambda_(0)=lambda`

Answer

1

Explanation

`lambda=(h)/(p) = (h)/(sqrt(2mE_(k)))=E_(k)=(h^2)/(2mlambda^(2))`
also `(hc)/(lambda_(0)) =E_(k)=(h^(2))/(2mlambda^(2))(or)lambda_(0)(2mclamda^(2))/(h)`