Question

The longest wavelength in the Lyman series is 121.5 nm. Use this wavelength together with the values of c and h to find the ionization energy of hydrogen.

Answer 1

The Lyman series is the result of transition of an electron $n_2=2→n_1=1$:

$$\frac1\lambda=\frac{E_i}{hc}\bigg(\frac1{n_2^2}-\frac1{n_1^2}\bigg),\\\space\\ E_i=\frac{hc}{\lambda}·\frac{n_1^2n_2^2}{n_2^2-n_1^2}=13.6\text{ eV}.$$