Determine the energy of the characteristic X-ray `(K_beta)` emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1).

Answer

66090

Explanation

Energy associated with the electron with the electron in the K - shell is approximately
`E_(K) = - (74-1)^(2)(13.6 eV)=-724 eV `
An electron in the M-shell is subjected to an effective nuclear charge that depends on the number of electrons in the n = 1 and n = 2 states because these electrons shield the M electrons from the nucleus . Because there are eight electrons from the nucleus. Because there are eight electrons in the n=2 state and one remaining in the n=1 ,roughly nine electrons shield M electrons from the nucleus .
So `Z_(eff) =Z-9`
Hence , the energy associated with an electron in the M-shell is
`E_(M) = (-13.6Z_("eff)^(2))/(3^(2))eV=(-13.6(Z-9)^(2))/(3^(2))eV`
`=-((13.6)(74-9)^(2))/9eV = - 6384 eV `
Therefore ,emitted X - ray has an energy equal to
`E_(M)-K_(K)={-6384-(-72479)}eV`
= 66090 eV