A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is x/π × 10⁻¹⁷ eVs. The value of x is ___________ (use h=4.14 × 10⁻¹⁵ eVs, c=3 × 10⁸ ms⁻¹ ).

Solution

Physics sir · 4 steps

First find which level 12.75 eV lands on, then Bohr's rule L = nh/2π gives the answer.

1. Energy after absorbing. The ground state energy is -13.6 eV. After absorbing 12.75 eV, the electron has -13.6 + 12.75 = -0.85 eV.
2. Find n. Use Eₙ = -13.6/n² eV. Then n² = 13.6/0.85 = 16, so the atom is in the n = 4 state.

   L = nh/2π = (4 × 4.14 × 10⁻¹⁵)/2π = (8.28 × 10⁻¹⁵)/π eV s

3. Match the form. Rewrite 8.28 × 10⁻¹⁵ as 828 × 10⁻¹⁷, so L = 828/π × 10⁻¹⁷ eV s. The value of c is not needed here.
4. So. x = 828.

The common slip: 8.28. Stopping at the coefficient of 10⁻¹⁵ gives 8.28. The question writes the answer with 10⁻¹⁷, so the coefficient has to be 828.

The idea to keep. For photon absorption in hydrogen, add the photon energy to -13.6 eV and match the result to -13.6/n². Once you know n, Bohr's rule L = nh/2π gives the angular momentum in one line.

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