How do you calculate the ionization energy of a hydrogen atom in its ground state?

!! VERY LONG ANSWER !!

Start by calculating the wavelength of the emission line that corresponds to an electron that undergoes a #n=1 -> n = oo# transition in a hydrogen atom.

This transition is part of the Lyman series and takes place in the ultraviolet part of the electromagnetic spectrum.

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Your tool of choice here will be the Rydberg equation for the hydrogen atom, which looks like this

#1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)#

Here

• #lamda_"e"# is the wavelength of the emitted photon (in a vacuum)
• #R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)#
• #n_1# represents the principal quantum number of the orbital that is lower in energy
• #n_2# represents the principal quantum number of the orbital that is higher in energy

In your case, you have

#{(n_1 = 1), (n_2 = oo) :}#

Now, you know that as the value of #n_2# increases, the value of #1/n_2^2# decreases. When #n=oo#, you can say that

#1/n_2^2 -> 0#

This implies that the Rydberg equation will take the form

#1/(lamda) = R * (1/n_1^2 - 0)#

#1/(lamda) = R * 1/n_1^2#

which, in your case, will get you

#1/(lamda) = R * 1/1^2#

#1/(lamda) = R#

Rearrange to solve for the wavelength

#lamda = 1/R#

Plug in the value you have for #R# to get

#lamda = 1/(1.097 * 10^(7)color(white)(.)"m") = 9.116 * 10^(-8)# #"m"#

Now, in order to find the energy that corresponds to this transition, calculate the frequency, #nu#, of a photon that is emitted when this transition takes place by using the fact that wavelength and frequency have an inverse relationship described by this equation

#color(blue)(ul(color(black)(nu * lamda = c)))#

Here

• #nu# is the frequency of the photon
• #c# is the speed of light in a vacuum, usually given as #3 * 10^8# #"m s"^(-1)#

Rearrange to solve for the frequency and plug in your value to find

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

#nu = (3 * 10^(8) color(red)(cancel(color(black)("m"))) "s"^(-1))/(9.116 * 10^(-8)color(red)(cancel(color(black)("m")))) = 3.291 * 10^(15)# #"s"^(-1)#

Finally, the energy of this photon is directly proportional to its frequency as described by the Planck - Einstein relation

#color(blue)(ul(color(black)(E = h * nu)))#

Here

• #E# is the energy of the photon
• #h# is Planck's constant, equal to #6.626 * 10^(-34)"J s"#

Plug in your value to find

#E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 3.291 * 10^(15) color(red)(cancel(color(black)("s"^(-1))))#

#E = 2.181 * 10^(-18)# #"J"#

This means that in order to remove the electron from the ground state of a hydrogen atom in the gaseous state and create a hydrogen ion, you need to supply #2.181 * 10^(-18)# #"J"# of energy.

This means that for #1# atom of hydrogen in the gaseous state, you have

#"H"_ ((g)) + 2.181 * 10^(-18)color(white)(.)"J" -> "H"_ ((g))^(+) + "e"^(-)#

Now, the ionization energy of hydrogen represents the energy required to remove #1# mole of electrons from #1# mole of hydrogen atoms in the gaseous state.

To convert the energy to kilojoules per mole, use the fact that #1# mole of photons contains #6.022 * 10^(23)# photons as given by Avogadro's constant.

You will end up with

#6.022 * 10^(23) color(red)(cancel(color(black)("photons")))/"1 mole photons" * (2.181 * 10^(-18)color(white)(.)color(red)(cancel(color(black)("J"))))/(1color(red)(cancel(color(black)("photon")))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J"))))#

# = color(darkgreen)(ul(color(black)("1313 kJ mol"^(-1))))#

You can thus say that for #1# mole of hydrogen atoms in the gaseous state, you have

#"H"_ ((g)) + "1313 kJ" -> "H"_((g))^(+) + "e"^(-)#

The cited value for the ionization energy of hydrogen is actually #"1312 kJ mol"^(-1)#.

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My guess would be that the difference between the two results was caused by the value I used for Avogadro's constant and by rounding.

#6.02 * 10^(23) -> "1312 kJ mol"^(-1)" vs "6.022 * 10^(23) -> "1313 kJ mol"^(-1)#