Final answer:
The element with the second smallest ionization energy among Strontium, Potassium, Cesium, and Barium is Potassium (K), based on its higher position in group 1 (alkali metals) compared to Cesium (Cs) but lower than the heavier alkaline earth metals.
Step-by-step explanation:
To determine which of the elements Strontium (Sr), Potassium (K), Cesium (Cs), and Barium (Ba) has the second smallest ionization energy, we can refer to the periodic table trends. Ionization energy increases from left to right across a period and from bottom to top within a group (column). In this case, we are considering elements from the same group (alkaline earth metals for Sr and Ba, and alkali metals for K and Cs), which means we should focus on the vertical trend.
Among the given elements, as we move down a group, the ionization energy decreases because of the increasing atomic size and the shielding effect of additional electron shells, which makes electrons easier to remove. Therefore, Cesium (Cs), being at the bottom of group 1 (alkali metals), would have the smallest ionization energy, followed by Barium (Ba), then Strontium (Sr) since they are heavier alkali earth metals, while Potassium (K), being higher up in the same group as Cesium, would have a higher ionization energy than Cesium but typically lower than that of the heavier alkaline earth metals. Thus, Potassium (K) would have the second smallest ionization energy among the choices provided.