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?

Aug 18, 2016

$2.072 \cdot {10}^{- 7} \text{m}$

Explanation:

The first thing to do here is to calculate the ionization energy for one atom of aluminium, $\text{Al}$, 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 $6.022 \cdot {10}^{23}$ 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

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

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} E = h \cdot \nu \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$E$ - the energy of the photon
$\nu$ - its frequency
$h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$

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

$E = h \cdot \nu \implies \nu = \frac{E}{h}$

Notice that Planck's constant is expressed in joules second, $\text{J s}$, 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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} l a m \mathrm{da} \cdot \nu = c \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$l a m \mathrm{da}$ - the wavelength of the photon
$c$ - the speed of light in a vacuum, usually given as $3 \cdot {10}^{8} {\text{m s}}^{- 1}$

Rearrange this equation to solve for $l a m \mathrm{da}$

$l a m \mathrm{da} \cdot \nu = c \implies l a m \mathrm{da} = \frac{c}{\nu}$

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.