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 for an outer electron.
a. Z=4 for Be, so singly-ionized Be has 3 electrons. The electron configuration for $Be^{+}$ is $1s^2 2s$.
In this ion, there is one electron outside a set of filled subshells. We assume full screening by the inner-subshell electrons, so there are 4 protons and 2 electrons interior to the 2s electron. $Z_{eff}=4-2=2$ for the 2s electron.
$E_{2s}=-\frac{2^2}{2^2}(13.6 eV)= -13.6 eV$
b. Similarly, $Z_{eff}=2$ for the N shell of calcium, where n = 4 for the outermost electron.
$E=-\frac{2^2}{4^2}(13.6 eV)= -3.4 eV$