in order to find the energy differences you need to find the rotational quantity:

\(\frac{hbar^2}{m\cdot Req^2}=\frac{c^2hbar^2}{Req^2mc^2}=\frac{hc^2}{Req^24\pi^2mc^2}\)

\(\displaystyle{\frac{{{\left({1240}{e}{v}\cdot{n}{m}\right)}}}{{{\left({0.5}\cdot{1.008}{u}\cdot{931.5}{M}{e}\frac{{v}}{{u}}\right)}{\left({0.072}{n}{m}\right)}^{{24}}\pi^{{2}}}}}={0.0152}{e}{V}\)

Now, the corresponding wavelengths can be found via the following equation:

\(\displaystyle\lambda={\frac{{{h}{e}}}{{\triangle{E}}}}\)

\(\displaystyle{L}={1}\to{L}={0}\triangle{E}={0.0152}{e}{V}\lambda={81.6}\mu{m}\)

\(\displaystyle{L}={2}\to{L}={1}\triangle{E}={0.0304}{e}{V}\lambda={40.8}\mu{m}\)

\(\displaystyle{L}={3}\to{L}={2}\triangle{E}={0.0456}{e}{V}\lambda={27.2}\mu{m}\)

You can also note that the wavelength is halved each time atransition occurs. Accordingly, the energy is \(\displaystyle\triangle{E},{2}\triangle{E},{3}\triangle{E},{e}{t}{c}\)