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

`(hc)/lambda_(c)=KE_(e)=p_(e)^(2)/(2m_(e))=(h/lambda)^(2)/(2m)=h^(2)/(2mlambda^(2))Rightarrow lambda_(c)=(2hcmlambda^(2))/h^(2)Rightarrow lambda_(c)=(2cmlambda^(2))/h`