Question

Question #16f36

Answer 1

Here's what I got.

Explanation:

The trick here is the fact that you can approximate the ionization energy for an atom of potassium with the negative orbital energy of the electron that's being removed from the atom.

This is actually known as Koopman's Theorem

`color(blue)("I.E." = -E_"orbital")" "`, where

`"I.E."` - the ionization enerfy of the atom
`E_"orbital"` - the energy of the orbital that holds the electron that's being removed

The problem provides you with the first ionization energy for a potassium atom. The first ionization energy represents the amount of energy needed to remove one mole of the highest-energy electrons from one mole of gaseous potassium atoms.

Before doing anything else, make sure that you understand what's going on here. Potassium, `"K"`, is located in period 4, group 1 of the periodic table. In a neutral potassium atom, the highest-energy electron is located in the 4s-orbital.

Notice that the 4p-orbitals are higher in energy than the 4s-orbital.

Now, when an electron falls from a higher energy orbital to a lower energy orbital, a photon is emitted. The energy of this photon, which as you know is proportional to its frequency, will be equal to the difference in energy between the two levels of the transition.

So, the problem provides you with the wavelength of the emitted photon, which is equivalent to the difference in energy between the 4p-orbital and the 4s-orbital.

Mathematically, this relationship is expressed as

`color(blue)(E_"photon" = h * nu)" "`, where

`E_"photon"` - the energy of the emitted photon
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`
`nu` - the frequency of the photon

A more useful form of the equation will be

`color(blue)(overbrace(DeltaE)^(color(red)(=E_"photon")) = (h * c)/(lamda)" "`, where

`c` - the speed of light in a vacuum, usually given as `3 * 10^8"m s"^(-1)`
`lamda` - the wavelength of the photon

Plug in your values and solve for `DeltaE` - do not forget to convert the wavelength of the photon from nanometers to meters

`DeltaE = (6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 3 * 10^8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(769 * 10^(-9)color(red)(cancel(color(black)("m"))))`

`DeltaE = 2.585 * 10^(-19)"J"`

The difference in energy levels is defined as

`DeltaE = E_"4p" - E_"4s"" " " "color(red)("(*)")`

Now, use Avogadro's number, which tells you how many atoms you get per mole of an element, to find the first ionization energy for one atom of potassium. Whilke we're at it, let's also convert the energy from kilojoules to joules

`419color(black)(cancel(color(black)("kJ")))/color(red)(cancel(color(black)("mol"))) * overbrace((1color(red)(cancel(color(black)("mole"))))/(6.022 * 10^(23)"atoms"))^(color(purple)("Avogadro's number")) * (10^3"J")/(1color(black)(cancel(color(black)("kJ")))) = 6.958 * 10^(-19)"J"`

According to Koopman's Theorem, this ionization energy will be equal to the negative of the orbital energy of the electron located in the 4s-orbital in a neutral potassium atom.

This means that you have

`E_"4s" = - "I.E."`

`E_"4s" = - 6.958 * 10^(-19)"J"`

Do not get confused by the negative sign! The energies of the orbital levels are always negative (and ionization energy is always positive)!

Moreover, the lower in energy an orbital is, the more negative its energy will be.

Plug this into equation `color(red)("(*)")` to get

`2.585 * 10^(-19)"J" = E_"4p" - E_"4s"`

`E_"4p" = 2.585 * 10^(-19)"J" + E_"4s"`

`E_"4p" = 2.585 * 10^(-19)"J" - 6.958 * 10^(-19)"J"`

`E_"4p" = -4.373 * 10^(-19)"J"`

Therefore, you can say that

`E_"4s" = color(green)(-6.96 * 10^(-19)"J")`

`E_"4p" = color(green)(-4.37 * 10^(-19)"J")`

The answers are rounded to three sig figs.