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

Answer

C

Explanation

`lambda = (h)/(p) rArr p = (h)/(lambda)`
`E = (p^(2))/(2 m) = (h^(2))/(2m lambda^(2))`
Also in `X` - ray `E = (hc)/(lambda_(0))`
`:. (hc)/(lambda_(0)) = (h^(2))/(2 m lambda^(2)) rArr lambda_(0) = ( 2 mc lambda^(2))/(h)`