Bohr's model defines the energy of each orbit in hydrogen as E_n = -13.6/n^2 eV, and the Rydberg formula 1/λ = R_H(1/n1^2 - 1/n2^2) relates wavelength to the transition. For the transition from n=3 to n=1 (Lyman series), we substitute n1=1 and n2=3: 1/λ = 1.097 × 10^7 × (1/1 - 1/9) = 1.097 × 10^7 × (8/9) = 9.751 × 10^6 m^-1. Therefore λ = 1/(9.751 × 10^6) = 1.026 × 10^-7 m = 102.6 nm. Option 121.5 nm is incorrect because it corresponds to the n=2 to n=1 transition, not n=3 to n=1. Option 656.3 nm is the Balmer series line for n=3 to n=2, which lies in the visible region. Option 97.2 nm corresponds to n=4 to n=1 in the Lyman series. This question tests the Rydberg formula, a core JEE topic under Bohr's atomic model and spectral line series. Plausibility check: all Lyman transitions produce UV photons below 122 nm, and 102.6 nm falls correctly in the UV range, confirming the answer.