There are three factors which decide how easily an electron can be removed, in this order of priority:
- Number of shells (distance from the nucleus in effect)
- Effect of shielding
- Nuclear attractive force from protons
First ionisation
Both K (potassium) and Ca (calcium) are in Period 4.
To ionise once, we must remove one electron from these two metals.
K is in Group 1. It, therefore, has one electron in its outer shell (in the s orbital). Ca is in Group 2, and has two electrons in its outer shell (again both in the s orbital).
Ca has the same number of shells as K, a similar amount of shielding, but more protons in the nucleus. This means there is a stronger attraction between the nucleus and the electron to be removed, meaning more energy is required to remove it (moving it to a potential of 0 eV).
Ca has a higher first ionisation energy than K.
In the case of K, this leaves it with a full third shell, as a K+ ion. For Ca, it becomes a Ca+ ion, with one electron still remaining in the 4th shell.
Second ionisation
In this case, we are ionising the K+ ion and the Ca+ ion. These are formed after the first ionisation.
K+ has the electron configuration [Ar]. This means it has three full shells. Ca+ has the same electron configuration as an atom of K, i.e. [Ar]4s1.
Ca+ has more shells and shielding than K+. This outweighs the fact that Ca+ has more protons in the nucleus, meaning means there is a weaker attraction between the nucleus and the electron to be removed, meaning less energy is required to remove it.
K+ has a higher ionisation energy than Ca+, so K has a higher second ionisation energy than Ca.
Hope this helped :)
