Electrons with de-Broglie wavelength `lamda` fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is:

Options

A. `lamda_(0)=(2mclamda^(2))/(h)`
B. `lamda_(0)=(2h)/(mc)`
C. `lamda_(0)=(2m^(2)c^(2)lamda^(3))/(h^(2))`
D. `lamda_(0)=lamda`

Answer

A

Explanation

`p=sqrt(2mk)`
`lambda=(h)/(p)=(h)/(sqrt(2mK))`
`or K=(h^(2))/(2mlambda^(2))`
Cut of wavlength of X-ray is related to K.E. as
`(hc)/(lambda_(0))=K=(h^(2))/(2mlambda^(2))`
`rArr lambda_(0)=(2mclambda^(2))/(h)`