Electrons of mass m with de-Broglie wavelength `lambda` fall on the target in an X-ray tube. The cut-off wavelength `(lambda_(0))` of the emitted X-ray is

Options

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

Answer

A

Explanation

Momentum of striking electrons `p=h/lambda`
`:.` Kinetic energy of striking electrons
`K=p^(2)/(2m)=h^(2)/(2mlambda^(2))`
This also, maximum energy of X-ray photons.
Therefore, `(hc)/lambda_(0)=h^(2)/(2m lambda^(2)) implies lambda_(0)=(2m lambda^(2)c)/(h)`