Q.The electron of a hydrogen atom is excited to nth level. How many total possible spectral lines will be found for the transition from this excited state to all other states below n?

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. …