Emission Spectrum of Hydrogen and Helium?
We have solved the Schrödinger equation for hydrogen-like atoms such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, and $\ce{B^4+}$. The energy levels are neatly arranged:

$$E_n = - \frac {E_0} {n^2}$$

where:


$E_n$ is the energy of the $n$-th energy level
$n$ is a positive integer ($1$, $2$, $\dots$)
$E_0 = 13.6~\mathrm{eV}$


This is possible because we only need to consider two particles: the nucleus which has a positive charge and the electron which has a negative charge. We only need to consider the interaction between those two particles. Also, the nucleus is not moving, which simplifies some calculations

However, for higher atoms such as helium, where there are more than one electron, the Schrödinger equation contains two terms for the two nucleus-electron attractions and a term for the electron-electron repulsion. The term for the electron-electron repulsion makes things difficult because both electrons can be moved. That term makes the equation impossible to solve analytically.



For hydrogen, the energy of an energy level is determined solely by its principal quantum number. That means, the $2s$ orbital has the same energy as the $2p$ orbital (ignoring the virtual particle effect).

This is not true in higher atoms such as helium. The energy levels of helium can be seen here:



(The energy levels of hydrogen is on the right of the diagram for comparison.)

Here, the azimuthal quantum number has an effect on the energy of the energy levels, which makes for more possible differences between energy levels, which makes for more lines as well as different on the emission spectrum.