Question

How do you calculate calcium and potassium ionization energy?

Answer 1

As we know that the Ionization energy is abbreviated as I.E. and it is also known as ionization potential (older term). If you take any element in a gaseous state or in an isolated state or neutral state, to ionize the elements, i.e. to remove an electron to get positive charge, it will take some energy. For the ionization of an element, some amount of energy is required, this is called ionization energy.

Complete step by step answer:
We need to remember that in the periodic table, $118$ elements are arranged in a systematic manner. There are some properties which are related to these elements, which defines the way in which these elements are reacted. Since the elements are arranged in the periodic table in a very well organized manner even the properties follow some particular trends. In that, one of the properties is ionization energy.
${M_{(g)}} + IE \to {M^ + }_{(g)} + {e^ - }$
$M$ is the atom, we provide ionization energy (IE) to remove an electron from the atom. When the electron is removed from the atom, it is left with an atom that is positively charged.
We know that the ionization energy usually measured by the tactic is to possess the vapors of the element during a discharge tube and connect it to a source of current. The voltage applied is gradually increased. At a particular voltage there'll be a sudden rise within the current passing through the tube. The energy like this voltage is understood as first ionization energy. The sudden rise within the current is thanks to the liberation of an electron from each neutral atom. If the applied voltage is further increased, there may again be a stage when current shows a sudden rise. This is because the elimination of another electron from each charged ion, produced earlier.
Ionization energy is measured in $eV$, as well as $kilojoules$ or $joules$.
Calcium $(Ca)$ and potassium $(K)$ first ionization energies as follows:
$Ca \to C{a^ + } + {e^ - }$; ${(IE)_1}$ = $589.8kJmo{l^{ - 1}}$
$K \to {K^ + } + {e^ - }$; ${(IE)_1}$ = 418.7$$$kJmo{l^{ - 1}}$$ How tightly the atom associates holds onto its electron is measured by the ionization energy. The tighter an electron is held, the upper is its ionization energy. The developments in ionization energy are just the reverse of these for atomic radii. Usually, because the atomic radii get bigger, ionization energies get lesser and the other way around. The first ionization energy of potassium is 418.7kJmo{l^{ - 1}}$$ and calcium is $589.8kJmo{l^{ - 1}}$$
. That the amount of energy needed to get rid of an outer electron from the atom is known as primary ionization energy. For calcium, larger atoms because we've more electrons and therefore the electrons are at energy levels farther from the nucleus.

Note:
We need to know that in the periodic table, go from left to right across a row, the first ionization increases.
Each accomplished ionization energy is greater than the preceding energy. i.e. $I{E_1} < I{E_2} < ..... < I{E_n}$.
Ionization energy is also measured by spectroscopic technique.
Understanding ionization energy is vital in a chemical reaction so as to know the behavior of whether various atoms make covalent or ionic bonds with one another.