!! 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. Your tool of choice here will be the Rydberg equation for the hydrogen atom, which looks like this
Here
In your case, you have
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
This implies that the Rydberg equation will take the form
which, in your case, will get you
Rearrange to solve for the wavelength
Plug in the value you have for #R# to get
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
Here
Rearrange to solve for the frequency and plug in your value to find
Finally, the energy of this photon is directly proportional to its frequency as described by the Planck - Einstein relation
Here
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 * 10^(-18)# #"J"# of energy. This means that for #1# atom of hydrogen in the gaseous state, you have
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
You can thus say that for #1# mole of hydrogen atoms in the gaseous state, you have
The cited value for the ionization energy of hydrogen is actually #"1312 kJ 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.
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I assume that the way to do this is to start from the Rydberg formula, $$ E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right), $$ set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to $$ E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}. $$ $\endgroup$ |