Question 1: The energy levels for hydrogen-like atoms (e.g., atoms with a single electron) can be written as En = Z2Eo/n2, where Eo is the energy of the ground level for hydrogen and Z is the number of protons in the atom.

The diagram below shows part of the line spectrum for hydrogen (wavelength increases to the right).

Would you expect the corresponding lines for singly ionized helium to be at the same places as the lines for hydrogen, further to the right, or further to the left?

Question

Question 1: The energy levels for hydrogen-like atoms (e.g., atoms with a single electron) can be written as En = Z2Eo/n2, where Eo is the energy of the ground level for hydrogen and Z is the number of protons in the atom.

The diagram below shows part of the line spectrum for hydrogen (wavelength increases to the right).

Would you expect the corresponding lines for singly ionized helium to be at the same places as the lines for hydrogen, further to the right, or further to the left?

BlackTom AI · Accepted Answer

Energy levels scale as Z^2, so for a hydrogen-like ion En = Z^2 E0 / n^2 and transition energies ΔE are larger when Z is larger.

1) Relevant concepts
- For hydrogen E0 (ground state) = -13.6 eV. For a hydrogen-like ion with nuclear charge Z, every energy level is multiplied by Z^2 in magnitude (more negative).
- Photon wavelength λ is related to transition energy by λ = hc / ΔE. Larger ΔE means smaller λ (bluer / to the left on the diagram, since wavelength increases to the right).

2) Step-by-step reasoning
- Compare hydrogen (Z = 1) with singly ionized helium He+ (Z = 2). The energies for He+ are 4 times the magnitude of hydrogen levels because Z^2 = 4.
- Any given electronic transition between the same pair of n levels (for example n = 3 → n = 2) therefore releases 4 times the energy in He+ than in H.

3) Calculations (intermediate steps shown)
- For H (Z=1), n=3 → n=2: E3 = -13.6/9 = -1.511 eV; E2 = -13.6/4 = -3.400 eV; ΔE = E2 - E3 = 1.889 eV → λ = hc/ΔE ≈ 1240 eV·nm / 1.889 eV ≈ 656 nm.
- For He+ (Z=2), same levels: energies multiply by 4: E3 = 4×(-1.511) = -6.044 eV; E2 = 4×(-3.400) = -13.600 eV; ΔE = 7.556 eV = 4×1.889 eV → λ ≈ 1240 / 7.556 ≈ 164 nm, which is one quarter of 656 nm.

4) Why the correct answer is correct
- Because transition energies scale as Z^2, wavelengths scale as 1/Z^2 (shorter by factor 4 for He+). Shorter wavelengths appear to the left on the diagram, so He+ lines are shifted to the left of the hydrogen lines.

5) Why the other options are incorrect
- "At the same places": incorrect because Z changes the energy levels; only identical Z would give identical wavelengths.
- "Farther to the right" (longer λ): incorrect because He+ transitions have larger ΔE and thus shorter λ, not longer.

6) Conclusion
The lines for singly ionized helium would be farther to the left than the lines for hydrogen (matches the provided answer).

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