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^2c^2lamda^2)/h^2`
D. `lamda_0=lamda`

Answer

A

Explanation

`lamda=h/(mv)impliesv=h/(mlamda)`
Kinetic energy of electron can be written as follows :
`K.E.=1/2mv^2=1/2xxmxx(h/(mlamda))^2`
`impliesK.E.=h^2/(2mlamda^2)`
For cut - off wavelength, we can write as follows :
`K.E.=h^2/(2mlamda^2)=(hc)/lamda_c`
`implieslamda_c=(2mclamda^2)/h`