In the emission spectrum of a hydrogen atom, which of the following...

For the Paschen series in the hydrogen emission spectrum, the electron transitions from a higher energy level $n_2$ to the lower energy level $n_1 = 3$.

The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:

$\dfrac{1}{\lambda} = R_H$\left( \dfrac{1}{3^2} - \dfrac{1}{n_2^2} \right)$where$n_2 = 4, 5, 6, \dots$<br><br>The longest wavelength corresponds to the minimum energy difference between the two levels. This occurs when the electron transitions from the nearest higher energy level.<br><br>Therefore, for the minimum energy and longest wavelength,$n_2 = 4$.<br><br>The required transition is$n = 4 \rightarrow n = 3$.<br><br>Answer:$n = 4 \rightarrow n = 3