Energy levels transitions of an electron? I have an example that tells me that the energy transition between $n=1$ and $n=3$ is $12.1\ \mathrm{eV}$ using the following formula:

$$E= -RH (\frac{1}{m^2} - \frac{1}{n^2})$$

and that the conversion from $12.1\ \mathrm{eV}$ gives a wavelength of $102.6~\mathrm{nm}$.

I understand that $-RH$ is the Rydberg's constant that corresponds to $13.60\ \mathrm{eV}$ and $m$ is the inner orbit and $n$ the outer orbit (but I might be wrong with the latter). So,

$$E= -13.60\ \mathrm{eV}\ (\frac{1}{1^2} - \frac{1}{3^2})$$

$$E= -13.60\ \mathrm{eV} \cdot 0.83$$

$$E= -11.288$$

I don't understand how they get the first value for the electron Volts and then convert these into wavelength. Additionally, my task is to find the energy level for the 4th orbit in the Lyman's series.
If I remember correctly, you can use the equation:

$$E = \frac{hc}{\lambda}$$

and 

$$\nu = \frac{c}{\lambda}$$

to solve for the frequency, wavelength, or energy of the photon. Note that $E$ is energy, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\nu$ is frequency.

The energy, (actually 11.97eV not 11.2eV)converted into joules (2.719 $*10^{-18}$ J) and then put into the first equation:

1.918 $×10^{-18}$ J = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{\lambda}$

then lambda can be solved for:

$\lambda$ = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{1.918×10^{-18} J}$

$\lambda$ = $1.036×10^{-7} m $

= 103.6 nanometers.

This solves for your wavelength like you asked for.