Question

The ionization energy of aluminum is 577.6kJ/mol. What is the longest wavelength of light to which a single photon could ionize an aluminum atom?

Answer 1

`2.072 * 10^(-7)"m"`

Explanation:

The first thing to do here is to calculate the ionization energy for one atom of aluminium, `"Al"`, by suing the fact that you know the energy needed to ionize one mole of atoms of aluminium.

As you know, one mole of any element contains exactly `6.022 * 10^(23)` atoms of that element, as given by Avogadro's number.

You can thus say that the energy needed to ionize one atomof aluminium is equal to

`577.6 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole Al"))))/(6.022 * 10^(23)"atoms of Al") = 9.5915 * 10^(-22)"kJ / atom"`

Now, the energy of a photon is directly proportional to its frequency as given by the Planck - Einstein equation

`color(blue)(|bar(ul(color(white)(a/a)E = h * nu color(white)(a/a)|)))`

Here

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

Use the above equation to calculate the frequency of a photon that carries enough energy to ionize a single atom of aluminium

`E = h * nu implies nu = E/h`

Notice that Planck's constant is expressed in joules second, `"J s"`, which means that you must convert the ionization energy from kilojoules per atom to joules per atom

`E = (9.5915 * 10^(-22) * 10^(3)color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J")))"s") = 1.4476 * 10^(15)"s"^(-1)`

Now, the frequency of a photon has an inverse relationship with its wavelength as given by the equation

`color(blue)(|bar(ul(color(white)(a/a)lamda * nu = c color(white)(a/a)|)))`

Here

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

Rearrange this equation to solve for `lamda`

`lamda * nu = c implies lamda = c/ (nu)`

Plug in your values to find

`lamda = (3 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))) )/(1.4476 * 10^(15)color(red)(cancel(color(black)("s"^(-1))))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.072 * 10^(-7)"m")color(white)(a/a)|)))`

The answer is rounded to four sig figs.