What is the third-longest wavelength in the absorption spectrum of hydrogen?

The absorption line for hydrogen gas containing atoms in the ground state has the longest wavelength of 121.6nm.

Our goal is to discover the hydrogen atom's absorption spectrum's third longest wavelength.

Assume we have a quantized system, such as the hydrogen atom, which has discrete energy levels rather than a continuum. When an electron absorbs a photon of a specific energy, it makes a quantum jump from a lower energy level ${\mathrm{E}}_{\mathrm{i}}$to a higher energy level ${\mathrm{E}}_{\mathrm{j}}$. The amount of energy required to excite an electron from its ${\mathrm{i}}^{\mathrm{th}}$to its ${\mathrm{j}}^{\mathrm{th}}$state is,

$\begin{array}{l}{\mathrm{E}}_{\mathrm{i}\to \mathrm{j}}=\text{Δ}{\mathrm{E}}_{\mathrm{ij}}\\ ={\mathrm{E}}_{\mathrm{j}}-{\mathrm{E}}_{\mathrm{i}}\left(\text{where}{\mathrm{E}}_{\mathrm{j}}>{\mathrm{E}}_{\mathrm{i}}\right)\\ =\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{i}\to \mathrm{j}}}\Rightarrow {\mathrm{\lambda }}_{\mathrm{i}\to \mathrm{j}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{j}}-{\mathrm{E}}_{\mathrm{i}}}\dots \left(1\right)\end{array}$

${\mathrm{E}}_{\mathrm{i}\to \mathrm{j}}$denotes that the quantized system excites from the ${\mathrm{E}}_{\mathrm{i}}$state to the ${\mathrm{E}}_{\mathrm{j}}$state by absorbing an amount of energy equal to role="math" localid="1650736975140" $∆{\mathrm{E}}_{\mathrm{ij}}$. We may deduce from (1) that ${\mathrm{\lambda }}_{\mathrm{i}\to \mathrm{j}}$is inversely proportional to ${\mathrm{\Delta E}}_{\mathrm{ij}}$, and since ${\mathrm{\Delta E}}_{21}<{\mathrm{\Delta E}}_{31}<{\mathrm{\Delta E}}_{41}\Rightarrow {\mathrm{\lambda }}_{1\to 2}>{\mathrm{\lambda }}_{1\to 3}>{\mathrm{\lambda }}_{1\to 4}$

As a result, the third longest wavelength is ${\mathrm{\lambda }}_{1\to 4}$. The energy of a particular state n for the hydrogen atom where Z=1 is calculated using Eq. (38.38).

${\mathrm{E}}_{\mathrm{n}}=\frac{-13.6\mathrm{eV}}{{\mathrm{n}}^2}\dots \left(2\right)$

We get by swapping (2) for (1),

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{i}\to \mathrm{j}}=\frac{\mathrm{hc}}{\frac{-13.6\mathrm{eV}}{{\mathrm{j}}^2}+\frac{13.6\mathrm{eV}}{{\mathrm{i}}^2}}\text{where}\mathrm{j}>\mathrm{i} \\ {\mathrm{\lambda }}_{1\to 4}=\frac{4.1357\times {10}^{-15}\cancel{\mathrm{eV}}.\cancel{\mathrm{s}}\times 3\times {10}^8\cdot \cancel{{\mathrm{m}.\mathrm{s}}^{-1}}}{\left(\frac{-13.6\cancel{\mathrm{eV}}}{4^2}+\frac{13.6\cancel{\mathrm{eV}}}{1^2}\right)} \\ =9.73\times {10}^{-8}\mathrm{m} \\ =97.3\mathrm{nm} \end{gathered}$$

The wavelength of the third longest wave is 97.3 nm