A hydrogen atom in its ground state is irradiated by light of wavelength 1026 Å . The electron gets excited to higher energy state n . Given the ground state energy is -13.6 eV and hc = 1242 eV nm , then the value of principal quantum number n is

Step-by-step solution

The energy of the incident photon is given by E = hc Substituting the given values with = 102.6 nm: E = 1242 102.6 12.1 eV The energy of the electron in the excited state n is: E_n = E₁ + E E_n = -13.6 + 12.1 = -1.5 eV The energy of the n -th state of a hydrogen atom is given by: E_n = -13.6 n^2 eV -13.6 n^2 = -1.5 n^2 = 13.6 1.5 9 n = 3 Answer: 3