JEE Main202127 Jul 2021Evening ShiftPhysicsAtomic PhysicsActual
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
Correct answer
0
Step-by-step solution
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