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)=(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

Cut-off wavelength occures when incoming electron losses its complete energy is collison . This energy appears in the from X-rays .
Given , mass of electrons =m
de-Broglie wavelength =`lambda`
so, kinetic energy of electron `=(p^(2))/(2m)=((h/(lambda)))/(2m)=(h^(2))/(2mlambda^(2))`
Now, maximum energy of photon cab be given by
`" " E=(hx)/(lambda_(0))=(h^(2))/(2mlambda^(2))rArrlambda_(0)=(hcxx2lamda^(2)*m)/(h^(2))=(2mclambda^(2))/(h)`