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

B, C, D

Explanation

Energy associated with the electron in the K-shell is approximately `E_K =- (74-1)^2 (13.6 eV) =- 72474 eV`
An electron in the M-shell is subjected to an effective nuclear that depends on the number of electrons in the n=1 and n =2 states because these electrons shield the M electrons form the nucleus. Because there are eight electrons in the n=2 state and one remaining in the n=1 state, 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/(9) eV =- 6384 eV`
Therefore, emitted X-ray has an energy equal to
`E_M - E_K = {-6384 - (-72474)}eV =- 66090eV`