1.

A beam of 35.0 keV electrons strikes a molybdenum target, generating the X-rays. What is the cutoff wavelength ?

Answer»

`35.5 pm`
`40.0 pm`
`15.95 pm`
`18.2 pm`

Solution :The cut-off wavelength `lambda_(MIN)` corresponds to an electron TRANSFERRING (approximately) all of its energy to an `X-ray` photon, thus producting a photon with the greatest POSSIBLE frequency and least possible wavelength.
From relation
`lambda_(min) = (HC)/(K_(0)) = ((4.14 xx 10^(-5))(3xx10^(8)))/(35.0 xx 10^(3))`
`= 3.55 xx 10^(-11) = 35.5 pm`.


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