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

Options

A. `lambda_(0)=(2mc lambda^(2))/h`
B. `lambda_(0)=(2h)/(mc)`
C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`
D. `lambda_(0)=lambda`

Answer

A

Explanation

`lambda=h/sqrt(2meV) implies V=h^(2)/(2melambda^(2))`
`(hc)/(lambda_(0))=eV=eh^(2)/(2melambda^(2))`
`lambda_(0)=(2mclambda^(2))/h`