# 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?

##### 1 Answer

#### Explanation:

The first thing to do here is to calculate the *ionization energy* for **one atom** of aluminium, **one mole** of atoms of aluminium.

As you know, **one mole** of any element contains exactly **atoms** of that element, as given by *Avogadro's number*.

You can thus say that the energy needed to ionize **one atom**of 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

*frequency*

**Planck's constant**, equal to

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*, *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

*wavelength* of the photon

*speed of light* in a vacuum, usually given as

Rearrange this equation to solve for

#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**.