engineering physics MCQ #400

The energy of the photon absorbed when an electron jumps from n = 2 to n = 3 is given by ΔE = 13.6 [1/2² - 1/3²] eV. What is ΔE?

engineering physics MCQ #400

  1. Question 1

    Q1. The energy of the photon absorbed when an electron jumps from n = 2 to n = 3 is given by ΔE = 13.6 [1/2² - 1/3²] eV. What is ΔE?

    • A) 1.89 eV
    • B) 1.51 eV
    • C) 1.13 eV
    • D) 0.66 eV

    Answer: 1.89 eV

    Explanation: ΔE = 13.6 [1/4 - 1/9] = 13.6 * (5/36) = 1.89 eV.