Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

Solution

Physics sir · 4 steps

The highest level is n = ∞, where the energy is zero. Compare both photon energies against 13.6 eV.

1. Energy levels. In hydrogen, Eₙ = -13.6/n² eV. The first level is n = 1, the second is n = 2, and the highest permitted level is n → ∞, where E_∞ = 0.

   Eₚₕₒₜₒₙ = 13.6(1/n_f² - 1/nᵢ²) eV

2. Case (i). From n = 2 to n = 1: E₁ = 13.6(1 - 1/4) = 3/4 × 13.6 eV.
3. Case (ii). From n = ∞ to n = 1: E₂ = 13.6(1 - 0) = 13.6 eV.
4. So. E₁ : E₂ = 3/4 : 1 = 3 : 4, which is option (a).

The idea to keep. For hydrogen, a jump to n = 1 releases 13.6(1 - 1/n²) eV. The highest level means n = ∞, so that jump releases the full 13.6 eV, the series limit.

Answer from the NTA final answer key. Solution by PrepWiser.