$A$ hydrogen atom initially in the ground level absorbs a photon,which excites it to the $n = 4$ level. Determine the wavelength and frequency of the photon.

For the ground level,$n_{1} = 1$.
The energy of this level is given by $E_{1} = \frac{-13.6}{n_{1}^{2}} \, eV = -13.6 \, eV$.
The atom is excited to the level $n_{2} = 4$.
The energy of this level is $E_{2} = \frac{-13.6}{n_{2}^{2}} \, eV = \frac{-13.6}{16} = -0.85 \, eV$.
The energy of the absorbed photon is $E = E_{2} - E_{1} = -0.85 - (-13.6) = 12.75 \, eV$.
Converting energy to Joules: $E = 12.75 \times 1.6 \times 10^{-19} \, J = 2.04 \times 10^{-18} \, J$.
Using the relation $E = \frac{hc}{\lambda}$,the wavelength is $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}} \approx 9.75 \times 10^{-8} \, m = 97.5 \, nm$.
The frequency is $\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{9.75 \times 10^{-8}} \approx 3.08 \times 10^{15} \, Hz$.