engineering physics MCQ #359

The energy of a photon emitted when an electron jumps from n = 3 to n = 2 is given by ΔE = E3 - E2. What is ΔE?

engineering physics MCQ #359

  1. Question 1

    Q1. The energy of a photon emitted when an electron jumps from n = 3 to n = 2 is given by ΔE = E3 - E2. What is ΔE?

    • A) 1.89 eV
    • B) 1.51 eV
    • C) 3.4 eV
    • D) 12.09 eV

    Answer: 1.89 eV

    Explanation: Calculate E3 and E2 using En = -13.6 / n², then ΔE = E3 - E2 = -1.51 - (-3.4) = 1.89 eV.