Q.Using Bohr's energy-level formula, derive the Rydberg formula for the wavelengths of the lines in the hydrogen spectrum, and name the spectral series obtained for transitions ending on $n=1$, $n=2$ and $n=3$.

Atomic Spectra & the Hydrogen Spectrum

The big idea. When an excited electron drops from a higher level n₂ to a lower level n₁, the atom emits a photon whose energy is exactly the gap between the levels. Because the levels are quantized, only certain photon energies — and therefore only certain wavelengths — appear: a line spectrum, the atom's fingerprint. One equation, the Rydberg formula, delivers the wavelength of every line.

The Rydberg formula

For a hydrogen-like species of nuclear charge Z:

ν̄ = 1/λ = R_H · Z² · (1/n₁² − 1/n₂²), with n₁ < n₂ and R_H = 1.097×10⁷ m⁻¹ (= 109677 cm⁻¹).

• ν̄ is the wavenumber (m⁻¹ or cm⁻¹), λ the wavelength, and the frequency is ν = c/λ = c·ν̄.
• n₁ is the lower level (where the electron lands), n₂ the upper level (where it starts). Keep 1/n₁² − 1/n₂² positive.
• Everything scales as Z²: a He⁺ line (Z = 2) sits at ¼ the wavelength of the same hydrogen transition.

The spectral series

Each series is defined by the level the electron falls to (n₁):

Only the Balmer series lies in the visible region — that is why it was discovered first.

Longest and shortest wavelength in a series

Within one series (fixed n₁):

• The first line (n₂ = n₁+1) has the smallest energy gap, hence the longest wavelength.
• The series limit (n₂ → ∞) has the largest gap, hence the shortest wavelength: 1/λ_min = R_H·Z²/n₁².

A frequent trap is swapping these — remember longest λ ↔ smallest 1/λ ↔ smallest energy jump.

Counting the lines

If a single electron is in level n and cascades down to the ground state, or a large sample of atoms is excited to level n, the number of distinct spectral lines is

N = n(n−1)/2.

Between two arbitrary levels n₂ and n₁ the count is (n₂−n₁)(n₂−n₁+1)/2. (The common wrong formula n(n+1)/2 counts a non-existent n→n "line".)

Energies add, wavelengths do not

For consecutive transitions the energies (and wavenumbers) add, never the wavelengths:

ΔE(n₃→n₁) = ΔE(n₃→n₂) + ΔE(n₂→n₁), so 1/λ₃₁ = 1/λ₃₂ + 1/λ₂₁. …