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. `lambda_(@) = (2mclambda^(2))/h`
B. `lambda_(@) = (2h)/(mc)`
C. `lambda_(0) = (2m^(2)c^(2)lambda^(2))/(h^(2))`
D. `lambda_(@) = lambda`

Answer

A