Question

A spectral line of atomic hydrogen has its wave number equal to the difference between the two lines of Balmer series, 486.1 nm and 410.2nm. To which series does the spectral line belong?

Answer 1

Note that wave number is proportional to energy. The wavelength 486.1 nm in the Balmer series to the energy difference of 2.55 eV, and is due to the transition between n = 4(E₄ = −0.85 eV) and n = 2(E₂ = −3.4 eV). ΔE₄₂ = −0.85 − (−3.4) = 2.55 eV. The wavelength 410.2 nm in the Balmer series corresponds to the energy difference of 3.0 eV and is due to the transition between n = 6(E₆ = −0.38 eV) and n = 2(E₂ = −3.4 eV). ΔE6₂ = −0.38 − (−3.4) = 3.02 eV 

Thus ΔE₆₂ − ΔE42 = 3.02 − 2.55 = 0.47 eV 

The difference of 0.47 eV is also equal to difference in E₆ = −0.38 eV (n = 6) and E₄ = −0.85 eV (n = 4). Thus the line arising from the transition n = 6 → n = 4, must belong to Bracket series. 

Note that in the above analysis we have used the well known law of spectroscopy, ˜v_mn − v˜_kn = v˜_mk