For hydrogen, the energy difference between levels n_i=3 and n_f=2 results in an emitted photon with an approximate wavelength of what?

• A. 656 nm
• B. 434 nm
• C. 121.6 nm
• D. 700 nm

Answer: (A)

Energy levels in hydrogen are quantized, so when an electron drops from ni = 3 to nf = 2 the energy difference ΔE is emitted as a photon. The wavelength of that photon is set by λ = hc/ΔE, or equivalently by the hydrogen Rydberg formula 1/λ = R_H(1/n_f^2 − 1/n_i^2). Plugging in nf = 2 and ni = 3 gives 1/λ = R_H(1/4 − 1/9) = R_H(5/36). With R_H ≈ 1.097×10^7 m^−1, λ ≈ 6.56×10^−7 m, which is about 656 nm. This is the red line of the Balmer series, known as the H-alpha line. Other transitions yield different wavelengths (for example, nf = 1 gives the Lyman series around 121.6 nm, and a transition like 4→2 would give a shorter wavelength near 434 nm), so the 3→2 transition specifically lands near 656 nm.