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.
a. 2s state of lithium, energy is -5.391 eV.
$E_{4s}=-\frac{Z_{eff}^2}{2^2}(13.6 eV)= -5.391 eV$
$Z_{eff}=2\sqrt{\frac{5.391eV}{13.6 eV }}= 1.26$
b. 4s state of potassium, energy is -4.339 eV.
$E_{4s}=-\frac{Z_{eff}^2}{4^2}(13.6 eV)= -4.339 eV$
$Z_{eff}=4\sqrt{\frac{4.339eV}{13.6 eV }}= 2.26$
c. We see that $Z_{eff}$ increases as we move down a column in the periodic table. The screening is less complete as n increases.
