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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)=(2mc lambda^(2))/h`B. `lambda_(0)=(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`D. `lambda_(0)=lambda` |
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Answer» Correct Answer - A `lambda=h/sqrt(2meV) implies V=h^(2)/(2melambda^(2))` `(hc)/(lambda_(0))=eV=eh^(2)/(2melambda^(2))` `lambda_(0)=(2mclambda^(2))/h` |
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