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?

Answer

`5.86 keV`

Explanation

Given `lambda_(K alpha)=71 p m=0.71 Å`
`E_(K)-E_(L)=(hc)/(lambda_(K alpha))=(12400 eV-Å)/(0.71 Å)=17.46 keV`
Thus `E_(L)=E_(k)-17.46 keV`
`=23.32 keV-17.46 keV=5.86 keV`