An electron in the hydrogen atom initially in the fourth excited state makes a transition to nᵗʰ energy state by emitting a photon of energy 2.86 eV . The integer value of n will be__________.

Solution

Physics sir · 4 steps

Fourth excited state means n = 5. Drop 2.86 eV from that level and see which level you land on.

1. Find the start. The ground state is n = 1, so the first excited state is n = 2. Counting up, the fourth excited state is n = 5.

   Eₙ = -13.6/n² eV

2. Energy after emission. E₅ = -13.6/25 = -0.544 eV. Emitting the photon lowers the energy by 2.86 eV, so the final level has Eₙ = -0.544 - 2.86 = -3.404 eV.
3. Solve for n. n² = 13.6/3.404 ≈ 3.995, so n² = 4 and n = 2. As a check, 13.6(1/4 - 1/25) = 2.856 eV ≈ 2.86 eV.
4. So. The electron jumps from n = 5 to n = 2, so n = 2. This is a Balmer line.

The idea to keep. The k-th excited state is level n = k + 1. Emission lowers the energy, so subtract the photon energy and then match the result to -13.6/n² eV.

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