Answer
a. -5.3 eV.
b. 4.4 eV.
Work Step by Step
a. Estimate the electrostatic potential energy for the point charges.
$$PE=k\frac{q_1q_2}{r}=-\frac{(8.99\times10^9 Nm^2/C^2)(1.60\times10^{-19}C)^2}{0.27\times10^{-9}m}$$
$$=-8.52\times10^{-19}J\approx-5.3eV$$
b. As we just calculated, the potential energy of the 2 ions is negative. This means that 5.32 eV is released when the ions are brought together from a large separation.
When the electron is transferred to the F atom, 3.41 eV is released. We are also told that 4.34 eV is absorbed in the process if taking an electron from K.
Calculate the binding energy of KF relative to free K and free F atoms.
$$E=5.32eV+3.41eV-4.34eV\approx 4.4 eV$$
