Practice question

Given: An electron in a hydrogen atom is excited to n = 3 and then falls to n = 1 . What is the maximum energy of the emitted photon? (Use E_n = -13.6/n² eV ) These values define the system as per NCERT data. Formula: Maximum energy from n = 3 to n = 1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_3 = -1.51 eV, E_1 = -13.6 eV . Δ E = 12.09 eV . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.