Electrons of mass m with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength `(lambda_(0))` 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^(3))/(h^(2))`
D. `lambda_(0)=lambda`

Answer

A

Explanation

`p=(h)/(lambda)`
`K.E=(p^(2))/(2m)=(h^(2))/(2mlambda^(2))`
If entire `K.E` of electron is converted into photon then
`(h^(2))/(2mlambda^(2))=(hc)/(lambda_(0))`
`lambda_(0)=(2mclambda^(2))/(h)`