The energy of the photon emitted when an electron jumps from n = 5 to n = 2 is given by ΔE = E5 - E2.
Q1. The energy of the photon emitted when an electron jumps from n = 5 to n = 2 is given by ΔE = E5 - E2.
Answer: 2.86 eV
Explanation: Calculate E5 and E2 using En = -13.6 / n² eV, then find ΔE = E5 - E2 = -0.544 - (-3.4) = 2.856 eV.