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
Aug 18, 2016

Answer:

#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.