The electron in a hydrogen atom falls from an excited energy level to the ground state in two steps, causing the emission of photons with wavelengths of 1870 and 102.5 nm. What is the quantum number of the initial excited energy level from which the electron falls?
A)2
B)3
C)4
D)6
E)8

Question

Quizplus · Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy difference between two energy levels is equal to the energy of the photon emitted when the electron falls from the higher energy level to the lower energy level. In this case, the energy difference between the initial excited energy level and the ground state is equal to the sum of the energies of the two photons emitted, which is:E = hc/λ1 + hc/λ2 = (6.626 x 10^-34 J s)(3 x 10^8 m/s)/(1870 x 10^-9 m) + (6.626 x 10^-34 J s)(3 x 10^8 m/s)/(102.5 x 10^-9 m) = 2.18 x 10^-18 JThe energy levels of a hydrogen atom are given by the formula:En = -13.6 eV/n^2where n is the principal quantum number. The ground state energy level is n = 1, and the energy of the ground state is -13.6 eV. The energy of the initial excited energy level is therefore:E = -13.6 eV/n^2 + 2.18 x 10^-18 JSolving for n, we get:n = 3Therefore, the quantum number of the initial excited energy level from which the electron falls is 3.

The Electron in a Hydrogen Atom Falls from an Excited