ionization energy

[Clarissa Nava 3H]

is this energy the energy that required in order to remove an electron from an atom ?

[Chem_Mod]

Correct

[904901860]

yes, it also decreases across the period and down a group

[505106414]

When would I use this in a calculation, like what would the problem be asking?

[Nare Arakelian Dis 3E]

The problem could just be asking to compare the ionization energies of two elements and state which is larger. The energy of a photon = ionization energy plus the kinetic energy so the value of the ionization energy could also be used here for a photoelectric effect problem.

[AmyL_3L]

Yes; the ionization energy increases from left to right and bottom to top on the periodic table of elements.

[Rishika Yadav 3D]

You are right, and just as a tip to help you remember, I like to think of it as the energy required to create an ion, thus ionization energy

[Jennifer Yang 3F]

That is correct, and they might ask you to compare the ionization energy of two atoms.

[Tahlia Mullins]

A good way to remember what this is would be to relate its trends to the trends in atomic radius since the larger the atomic radius, the greater the ionization energy, so their trends line up since they are in direct relationship with one another.

[Elizabeth Harty 1A]

Yes and for the periodic trends you can always determine the trend based on stuff you already know (for instance atomic radius). As atomic radius increases as you go down a group, the nuclear pull on the outer electrons is less and thus the ionization energy is lower. Moreover, as you go across a period, the atomic radius decreases and as you adding electrons to the same energy shell, and therefore the ionization energy increases.

[Abraham De Luna]

Ionization energy is indeed the energy required to remove an electron from an atom.

[Sean Tran 2K]

The first ionization energy of an electron is the energy required to remove the first electron from a ground state atom. The trend for this is seen on the periodic table as you go to the right and up.