Answer
See explanation.
Work Step by Step
Equation 41.45 gives the energy of an atomic level, taking screening into account.
It is given in terms of n and $Z_{eff}$ by $E_n=-\frac{Z_{eff}^2}{n^2}(13.6 eV)$.
Find the effective atomic number seen by some electrons of potassium.
4s state, energy is -4.339 eV.
$E_{4s}=-\frac{Z_{eff}^2}{4^2}(13.6 eV)= -4.339 eV$
$Z_{eff}=4\sqrt{\frac{4.339eV}{13.6 eV }}= 2.26$
4p state, energy is -2.73 eV.
$E_{4p}=-\frac{Z_{eff}^2}{4^2}(13.6 eV)= -2.73 eV$
$Z_{eff}=4\sqrt{\frac{2.73eV}{13.6 eV }}= 1.79$
4d state, energy is -0.94 eV.
$E_{4d}=-\frac{Z_{eff}^2}{4^2}(13.6 eV)= -0.94 eV$
$Z_{eff}=4\sqrt{\frac{0.94eV}{13.6 eV }}= 1.05$
The higher $l$ states, with more angular momentum, spend more time away from the nucleus. This places them outside the electron charge clouds of the inner filled shells, bringing the effective atomic number closer to 1.0.
