Question

he K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. (a) Determine the ionization energies of the L, M, and N shells.

L shell	___keV
M shell	___keV
N shell	___keV

Answer 1

For tunsten Z = 74,  Pulling the last electron off a tungsten atom having a single electron is calculable from the Bohr model as (74)²(13.6 eV) = 74 keV

The electron being -74 keV deep in the potential of the ionized tungsten atom. That it is actually 69.5 keV comes from the presence of those other electrons around the nucleus.

delta E(KL) = h c / lambda(KL)

hc = 1240 eV

delta E(KL) = 1240 / 0.0185 nm

= 67.03 KeV

delta E(KM) = 1240 / 0.0209 nm

= 59.33 KeV

delta E(KN) = 1240 / 0.0215 nm

= 57.67 KeV

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The ionization energies are, therefore, the differences between the photon energies and the K-shell ionization energy:

69.5 keV - 67.03 keV = 2.47 keV

69.5 keV - 59.33 keV = 10.17 keV

69.5 keV - 57.67 keV = 11.83 keV

As the L-shell is the next deepest after the K- shell, its ionization energy = 11.83 keV

for the M-shell, the ionization energy must = 10.17 keV

for the N-shell, the ionization energy must = 2.47 keV