A hydrogen atom in its first excited state absorbs a photon of energy x × 10⁻² eV and excited to a higher energy state where the potential energy of electron is -1.08 eV. The value of x is ______________.

Solution

Physics sir · 3 steps

The given number is potential energy, not total energy, so halve it first and then take the energy gap.

1. PE to total energy. In a Bohr orbit, U = -2K, so the total energy is E = K + U = U/2. With U = -1.08 eV, the final state has E = -0.54 eV.
2. Identify the levels. Eₙ = -13.6/n² eV, and -0.54 eV matches n = 5, since 13.6/25 ≈ 0.54. The first excited state is n = 2, with E₂ = -13.6/4 = -3.4 eV.

   Δ E = E₅ - E₂ = -0.54 - (-3.4) = 2.86 eV

3. So. The photon carries 2.86 eV = 286 × 10⁻² eV, so x = 286.

The common slip: 232. Treating -1.08 eV as the total energy gives 3.4 - 1.08 = 2.32 eV, that is x = 232. The total energy is half the potential energy.

The idea to keep. In Bohr orbits, E = -K = U/2. When a question gives kinetic or potential energy, convert it to total energy before you find any energy gap.

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