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. `lamda_(0)=(2mclamda^(2))/(h)`
B. `lamda_(0)=(2h)/(mc)`
C. `lamda_(0)=(2m^(2)c^(2)lamda^(3))/(h^(2))`
D. `lamda_(0)=lamda`

Answer

A

Explanation

`{:(lamda_(0)=(hc)/(KE_(e))),(lamda_(0)=(hc)/(h^(2)//2mlamda^(2))):|{:(lamda=(h)/(sqrt(2mKE_(e)))),(KE_(e)=(h^(2))/(2mlamda^(2))):}`
`lamda_(0)=(2mc)/(h)lamda^(2)`