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A beam of 35.0 keV electrons strikes a molybdenum target, generating the X-rays. What is the cutoff wavelength ? |
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Answer» `35.5 pm` 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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