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. `lamda_0=(2mclamda^2)/h`
B. `lamda_0=(2h)/(mc)`
C. `lamda_0=(2m^2c^2lamda^3)/h^2`
D. `lamda_0=lamda`

Answer

A

Explanation

Kinetic energy of electrons
`K=P_2/(2m) =(h//lamda^2)/(2m) =h^2/(2mlamda^2)`
So, maximum energy of photon = K
`(hc)/lamda_0 =h^2/(2mlamda^2):. lamda_0=(2mclamda^2)/h`