The K α X-ray of molybdenum has wavelength 0 . 071 nm . If the energy of a molybdenum atom with a K electron knocked out is 27 . 5 keV , the energy of this atom when an L electron is knocked out will be keV . (Round off to the nearest integer ) h = 4 . 14 × 10 - 15 eV s , c = 3 × 10 8 m s - 1

Question

The K α X-ray of molybdenum has wavelength 0 . 071 nm . If the energy of a molybdenum atom with a K electron knocked out is 27 . 5 keV , the energy of this atom when an L electron is knocked out will be keV . (Round off to the nearest integer ) h = 4 . 14 × 10 - 15 eV s , c = 3 × 10 8 m s - 1,

MARKS App · Accepted Answer

E k α = E k - E L h c λ k α = E k - E L E L = E k - h c λ k α = 27 . 5 KeV - 12 . 42 × 10 - 7 eV m 0 . 071 × 10 - 9 m E L = ( 27 . 5 - 17 . 5 ) keV = 10 keV

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