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^(2))/(h^(2))`
D. `lamda_(0)=lamda`

Answer

A

Explanation

(a) : Let K be the kinetic energy of the incident electron. Its linear momentum, `p=sqrt(2mK)`
The Broglie wavelength is related to the linear momentu as
`(hc)/(lamda_(0))=K=K(h^(2))/(2mlamda^(2))orlamda_(0)=(2mclamda^(2))/(h)`