Every time hydrogen emits a photon in the Balmer series, it is followed by another photon in the Lyman series. Find the wavelength corresponding to this latter photon.

Step-by-step solution

The Balmer series corresponds to the transition of an electron from a higher energy level ( n 3 ) to the first excited state ( n = 2 ). After this transition, the electron is in the n = 2 state. It will subsequently transition to the ground state ( n = 1 ), emitting a photon in the Lyman series. The energy difference for the transition from n = 2 to n = 1 is given by: E = 13.6 ( 1 1^2 - 1 2^2 ) eV = 13.6 3 4 eV = 10.2 eV The wavelength of the emitted photon is: = hc E = 1242 eV nm 10.2 eV 121.7 nm Rounding to the n