Q.45

The wavelength of a photon emitted during a transition from n = 3 to n = 2 state in the H

atom is _____________ nm. (answer in integer).

[Rydberg energy constant RH = 2.18 × 10-18 J; Planck constant h = 6.626 × 10-34 J s,

c = 3 × 108 m s -1 ]

Hydrogen Atom n=3 to n=2 Transition Wavelength Calculation

The wavelength of the photon emitted during the transition from n=3 to n=2 in a hydrogen atom is 656 nm.

Calculation Steps

The energy difference ΔE between levels is calculated using E_n = -R_H / n², where R_H = 2.18 × 10^{-18} J. For n=3, E_3 = -R_H / 9 = -2.422 × 10^{-19} J; for n=2, E_2 = -R_H / 4 = -5.45 × 10^{-19} J. Thus, ΔE = E_3 – E_2 = 2.178 × 10^{-19} J (emission releases this energy).

Photon energy equals ΔE, so λ = hc / ΔE. Substituting h = 6.626 × 10^{-34} J s and c = 3 × 10^8 m/s gives λ = (6.626 × 10^{-34} × 3 × 10^8) / 2.178 × 10^{-19} = 9.117 × 10^{-7} m = 656 nm (rounded to integer).

Alternatively, the Rydberg formula 1/λ = R_H (1/2² – 1/3²) / (hc) yields the same result, confirming the Balmer series red line at 656 nm.

Common Options Explained

• 656 nm (correct): Matches exact computation with given constants; corresponds to H-alpha line in visible red spectrum.

• 658 nm: Arises from using approximate Rydberg constant R_∞ = 1.097 × 10^7 m^{-1} instead of energy-based R_H, causing slight overestimate.

• 123 nm: Incorrectly assumes n=3 to n=1 (Lyman-alpha); much higher energy UV transition.

• 102 nm: For n=3 to n=1 with minor rounding; not applicable here.