Energy E of a hydrogen atom with principal quantum number n is given by E = -13.6/n² eV. The energy of a photo

Question

Energy E of a hydrogen atom with principal quantum number n is given by E = -13.6/n² eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen, is approximately:

Accepted Answer

The energy of the emitted photon corresponds to the difference in energy between the two states: ΔE = E₃ - E₂.

1. Calculate energy of initial state (n=3): E₃ = -13.6 / 3² = -13.6 / 9 ≈ -1.51 eV.
2. Calculate energy of final state (n=2): E₂ = -13.6 / 2² = -13.6 / 4 = -3.4 eV.
3. Calculate the energy difference (energy of ejected photon): ΔE = E₃ - E₂ = -1.51 - (-3.4) = 3.4 - 1.51 = 1.89 eV.

Rounding to one decimal place, the energy is approximately 1.9 eV.

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