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) = (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» Correct Answer - C
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)`


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