The `Ka` X-ray of molybdenum has wavelength 71 pm. If the energy of a molybdenum atom with a K electron knocked out is 23:32 keV, what will be the energy of this atom when an L electron is knocked out?

Explanation

`E_(K)-E_(L)=(hc)/lambda_(K_(alpha))`
`lambda_(K_(alpha))=71 p m=0.071 nm`
`E(eV)=1242/(lambda(nm))=1242/0.071=17.5xx10^(3) eV=17.5 keV`
`E_(K)-E_(L)=E`
`23.32-E_(L)=17.5`
`E_(L)=5.82 ke V`