Four Lines, Two Series: Checking a Hydrogen Spectrum Record Book
A student examines a low-pressure hydrogen discharge tube with a hand-held spectroscope, then repeats the measurement on a small grating spectrometer. Four lines are entered in the record book at 656.3 nm, 486.1 nm, 434.0 nm and 121.6 nm. The student writes: 'All four lines belong to the Balmer series of hydrogen; the 121.6 nm line is the series limit.' The demonstrator asks for the assignment to be checked against the Rydberg equation (R = 1.097 × 10⁷ m⁻¹) before the report is signed. Later the same student mounts a helium discharge tube and records a line at 164.1 nm, which is written up as 'the same 3 → 2 transition as the red hydrogen line, therefore the same energy gap'. The tube is labelled 'He, low pressure', and the spectrometer calibration is confirmed with a sodium lamp, so the wavelengths themselves are reliable. The only issue is interpretation.
1. Use the Rydberg equation to assign the 656.3 nm and 121.6 nm lines and name the series each belongs to.
Read the worked answer
For 656.3 nm: 1/λ = 1/(656.3 × 10⁻⁹) = 1.5237 × 10⁶ m⁻¹. Dividing by R gives 0.13890 = 1/n₁² − 1/n₂². With n₁ = 2: 0.2500 − 0.13890 = 0.11110, so n₂² = 9.00 and n₂ = 3. The line is 3 → 2, the first member of the Balmer series, in the visible. For 121.6 nm: 1/λ = 8.224 × 10⁶ m⁻¹, and 8.224 × 10⁶/1.097 × 10⁷ = 0.7497 = 1 − 1/n₂², giving n₂² = 3.99 and n₂ = 2. The line is 2 → 1, Lyman-α, in the far ultraviolet. The two lines cannot belong to one series: they fall to different lower levels (n₁ = 2 versus n₁ = 1). Only the three visible wavelengths, 656.3, 486.1 and 434.0 nm, are Balmer lines (3 → 2, 4 → 2, 5 → 2).
2. Show that 121.6 nm cannot be the Balmer limit, calculate the true limit, and explain the student's likely mistakes.
Read the worked answer
The series limit is reached when the upper level tends to infinity with n₁ fixed. For Balmer, n₁ = 2: 1/λ = R(1/4 − 0) = R/4, so λ = 4/R = 4/(1.097 × 10⁷) = 3.646 × 10⁻⁷ m = 364.6 nm. Every Balmer line therefore lies between 656.3 nm and 364.6 nm; no member can be shorter than the limit. A line at 121.6 nm is 243 nm below the limit, so it is outside the series entirely. Likely errors: (i) a series limit is the shortest wavelength of a series, not the longest or most intense, so the student has the concept inverted; (ii) the observed 121.6 nm is Lyman-α, whose own series limit is 1/R = 91.2 nm. There is also an instrumental objection: 121.6 nm is vacuum ultraviolet and is absorbed by air and by glass optics, so a hand-held glass spectroscope cannot have recorded it at all - the entry most probably came from a data table rather than from the instrument.
3. Show that the 164.1 nm line from the helium tube is the He⁺ 3 → 2 transition, and assess the remark about 'the same energy gap'.
Read the worked answer
A hydrogen-like ion obeys 1/λ = RZ²(1/n₁² − 1/n₂²). For He⁺, Z = 2, so Z² = 4: 1/λ = 1.097 × 10⁷ × 4 × (0.2500 − 0.1111) = 1.097 × 10⁷ × 4 × 0.13889 = 6.094 × 10⁶ m⁻¹, and λ = 1.641 × 10⁻⁷ m = 164.1 nm. Agreement with the recorded value is within a tenth of a nanometre, so the assignment stands. The remark is wrong in its conclusion. The quantum numbers are the same, but the energies are not: E ∝ Z², so for the same n values He⁺ has four times the energy gap and one quarter the wavelength (656.3/4 = 164.1 nm). 'Same transition' should mean the same pair of levels only in a bookkeeping sense; the species must be identified before any energy can be quoted. A line position alone cannot identify the element - the whole series pattern is needed.