The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :

Solution

Physics sir · 3 steps

Find the energy gap between n = 2 and n = 1, then turn that energy into a wavelength using hc/E.

1. Energy levels. For hydrogen, Eₙ = -13.6/n² eV. So E₂ = -3.4 eV and E₁ = -13.6 eV, and the photon carries away the gap Δ E = 10.2 eV.

   λ = hc/(Δ E) = (1240 eV nm)/(10.2 eV) ≈ 121.6 nm

2. Check with Rydberg. 1/λ = R(1/1²-1/2²) = 3R/4, so λ = 4/(3 × 1.097 × 10⁷) m ≈ 1.215 × 10⁻⁷ m. That is the same result, about 121.5 nm.
3. So. The wavelength is about 122 nm, the first Lyman line in the ultraviolet. The small gap from the option comes from rounding the constants, so the answer is (d) 121.8 nm.

The idea to keep. Any jump that ends on n = 1 is a Lyman line, between about 91 nm and 122 nm in the ultraviolet. Remember hc ≈ 1240 eV nm to turn eV into nm fast.

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