engineering physics MCQ #454

The energy of the photon absorbed when an electron jumps from n = 2 to n = 4 is given by ΔE = E4 - E2. What is ΔE?

engineering physics MCQ #454

  1. Question 1

    Q1. The energy of the photon absorbed when an electron jumps from n = 2 to n = 4 is given by ΔE = E4 - E2. What is ΔE?

    • A) 2.55 eV
    • B) 3.4 eV
    • C) 1.89 eV
    • D) 4.09 eV

    Answer: 2.55 eV

    Explanation: E4 = -13.6 / 4² = -0.85 eV, E2 = -3.4 eV. ΔE = E4 - E2 = -0.85 - (-3.4) = 2.55 eV.