engineering physics MCQ #448

The energy of the photon emitted when an electron jumps from n = 4 to n = 1 is given by ΔE = 13.6 * (1 - 1 / 4²) eV. The value of ΔE is

engineering physics MCQ #448

  1. Question 1

    Q1. The energy of the photon emitted when an electron jumps from n = 4 to n = 1 is given by ΔE = 13.6 * (1 - 1 / 4²) eV. The value of ΔE is

    • A) 12.75 eV
    • B) 13.6 eV
    • C) 10.2 eV
    • D) 1.7 eV

    Answer: 12.75 eV

    Explanation: ΔE = 13.6 * (1 - 1/16) = 13.6 * (15/16) = 12.75 eV.