Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `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`

Electrons with de- Broglie wavelength `lambda` fall on the target in a

1.	Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `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» Correct Answer - A 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)`