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

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

Answer:
\(1313 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Explanation:
!! VERY LONG ANSWER !!

Start by calculating the wavelength of the emission line that corresponds to an electron that undergoes a \(n=1 \rightarrow n=\infty\) transition in a hydrogen atom.

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

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

\(\frac{1}{\lambda_{\mathrm{e}}}=R \cdot\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)

Here

• \(\lambda_{e}\) is the wavelength of the emitted photon (in a vacuum)
• \(R\) is the Rydberg constant, equal to \(1.097 \cdot 10^{7} \mathrm{~m}^{-1}\)
• \(n_{1}\) 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

Now, you know that as the value of \(n_{2}\) increases, the value of \(\frac{1}{n_{2}^{2}}\) decreases. When \(n=\infty\) you can say that

\(\frac{1}{n_{2}^{2}} \rightarrow 0\)

This implies that the Rydberg equation will take the form

\(\frac{1}{\lambda}=R \cdot\left(\frac{1}{n_{1}^{2}}-0\right)\)
\(\frac{1}{\lambda}=R \cdot \frac{1}{n_{1}^{2}}\)

which, in your case, will get you

\(\frac{1}{\lambda}=R \cdot \frac{1}{1^{2}}\)
\(\frac{1}{\lambda}=R\)

Rearrange to solve for the wavelength

\(\lambda=\frac{1}{R}\)

Plug in the value you have for \(R\) to get

\(\lambda=\frac{1}{1.097 \cdot 10^{7} \mathrm{~m}}=9.116 \cdot 10^{-8} \mathrm{~m}\)

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

\(\nu \cdot \lambda=c\)

Here

\(ν\) is the frequency of the photon
\(c\) is the speed of light in a vacuum, usually given as \(3 \cdot 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)

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

\(\nu \cdot \lambda=c \Rightarrow \nu=\frac{c}{\lambda}\)

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

\(E=h \cdot \nu\)

Here

\(E\) is the energy of the photon
\(h\) is Planck’s constant, equal to \(6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\)

Plug in your value to find

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 \(2.181 \cdot 10^{-18} \mathrm{~J}\) of energy.

This means that for \(1\) atom of hydrogen in the gaseous state, you have

\(\mathrm{H}_{(g)}+2.181 \cdot 10^{-18} \mathrm{~J} \rightarrow \mathrm{H}_{(g)}^{+}+\mathrm{e}^{-}\)

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 \(6.022 \cdot 10^{23}\) photons as given by Avogadro’s constant.

You will end up with

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

\(\mathrm{H}_{(g)}+1313 \mathrm{~kJ} \rightarrow \mathrm{H}_{(g)}^{+}+\mathrm{e}^{-}\)

The cited value for the ionization energy of hydrogen is actually \(1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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.

\(6.02 \cdot 10^{23} \rightarrow 1312 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { vs } 6.022 \cdot 10^{23} \rightarrow 1313 \mathrm{~kJ} \mathrm{~mol}^{-1}\)