1.

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:

Answer»

`lamda_(0)=(2mclamda^(2))/(h)`
`lamda_(0)=(2h)/(mc)`
`lamda_(0)=(2m^(2)c^(2)lamda^(3))/(h^(2))`
`lamda_(0)=lamda`

Solution :`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)`


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