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

Answer

C

Explanation

The cut-offwavelength is given by
`lambda_(0) = (hc)/(eV)`
According to de Broglie equation.
`lambda = (h)/(P) = (h)/(sqrt(2 meV)) implies lambda^(2) = (h^(2))/(2 meV)`
`implies V = (h^(2))/(2 me lambda^(2))`
From (i) and (ii) `lambda_(0) = (hc xx 2 me lambda^(2))/(e h^(2)) = (2 me lambda^(2))/(h)`