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)=(2mclambda^(2))/h`B. `lambda_(0)=(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`D. `lambda_(0)=lambda`

Electrons with de- Broglie wavelength `lambda` fall on the target in a

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)=(2mclambda^(2))/h`B. `lambda_(0)=(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`D. `lambda_(0)=lambda`
Answer» Correct Answer - A `(hc)/lambda_(c)=KE_(e)=p_(e)^(2)/(2m_(e))=(h/lambda)^(2)/(2m)=h^(2)/(2mlambda^(2))Rightarrow lambda_(c)=(2hcmlambda^(2))/h^(2)Rightarrow lambda_(c)=(2cmlambda^(2))/h`

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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)=(2mclambda^(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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