Number of Spectral Lines Formula: N = n(n − 1)/2

The number of spectral lines formula is $N = \frac{n(n-1)}{2}$ : if a sample of hydrogen atoms has electrons excited up to level n, the total number of different lines that can appear in its emission spectrum as the electrons fall back to the ground state is n(n − 1)/2. From n = 4, for example, the count is 4 × 3 / 2 = 6.

When the electrons stop at a lower level n₁ instead of the ground state, the total number of spectral lines formula becomes $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$ . The idea is simple counting: every pair of levels gives one line. It appears in the Class 12 atoms chapter and in Class 11 chemistry under atomic structure, and it is a favourite one mark question in JEE and NEET. See the modern physics formula sheet for the Bohr's model results it builds on.

Quick facts

Formula
$N = \frac{n(n-1)}{2}$ to the ground state
General form
$N = \frac{\Delta n(\Delta n + 1)}{2}$ , $\Delta n = n_2 - n_1$
Single atom
At most n − 1 lines from level n
Unit
None; it is a count
Why it works
Number of pairs from n levels, ${}^{n}C_2$
Energy of a line
$h\nu = E_{n_2} - E_{n_1}$

Why the number of spectral lines formula works

An emission line is produced when an electron jumps from a higher level to a lower one and gives out a photon of energy $h\nu = E_{high} - E_{low}$ . Each different pair of levels gives a different photon energy, so a different line. With n levels in play, the number of ways to pick two of them is

${}^{n}C_2 = \frac{n(n-1)}{2}$

Another way to see it: from level n the electron can fall to n − 1 lower levels, from n − 1 to n − 2 lower levels, and so on down to one jump from level 2. The total is $(n-1) + (n-2) + \ldots + 1 = \frac{n(n-1)}{2}$ . Both arguments assume a large number of atoms, so every possible route is taken by some atom at the same time.

Formula for number of spectral lines between any two levels

If the electrons fall from n₂ but only as far as n₁, only the levels from n₁ to n₂ take part. That is $\Delta n + 1$ levels with $\Delta n = n_2 - n_1$ , so

$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$

Put n₁ = 1 and this reduces to n(n − 1)/2 with n = n₂. The calculator above uses the general form: enter n₂ and n₁ to get N, or enter N and one level to find the other.

Line counts for common cases
From n₂	To n₁	Levels involved	N
3	1	3	3
4	1	4	6
5	1	5	10
6	1	6	15
5	2	4	6
6	3	4	6

Worked examples

Example 1: the basic count

Electrons in a hydrogen sample are excited to n = 5. How many spectral lines can be emitted?

$N = \frac{5 \times 4}{2} = 10$

Ten different lines are possible.

Example 2: stopping at n = 2

Electrons fall from n = 6 and stop at n = 2. How many lines appear, and in which series?

$N = \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 \times 5}{2} = 10$

All ten end at or above n = 2. The four that end on n = 2 belong to the Balmer series.

Example 3: exam style with energy

Ground state hydrogen atoms absorb photons of 12.75 eV. How many lines appear in the emission spectrum that follows?

Energy after absorption: $-13.6 + 12.75 = -0.85\ \text{eV}$ . Since $E_n = -\frac{13.6}{n^2}$ eV, $n^2 = \frac{13.6}{0.85} = 16$ , so n = 4.

$N = \frac{4 \times 3}{2} = 6$

Six lines: three Lyman, two Balmer and one Paschen.

One atom versus many atoms

A single atom makes one jump at a time. Starting from level n, it gives the most lines if it steps down one level at a time, which yields n − 1 lines, and the fewest (one line) if it drops straight to the ground state. The formula n(n − 1)/2 counts lines from a whole sample, where different atoms take different routes at once. Read the question carefully: "a hydrogen atom" and "a sample of hydrogen atoms" have different answers.

One electron in one atom from n = 4: at most 3 lines. A gas of atoms from n = 4: 6 lines.

Spectral series of hydrogen

Where each series ends
Series	Lower level n₁	Region
Lyman	1	Ultraviolet
Balmer	2	Visible (and near ultraviolet)
Paschen	3	Infrared
Brackett	4	Infrared
Pfund	5	Infrared

The wavelength of each line comes from the Rydberg formula, $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ with $R = 1.097 \times 10^7\ \text{m}^{-1}$ for hydrogen. The counting formula tells you how many lines there are; the Rydberg formula tells you where they sit. Photon energy and wavelength are linked by $E = \frac{hc}{\lambda}$ , covered on the wavelength formula page. The angular momentum of each Bohr level is on the orbital angular momentum formula page.

Slips that cost marks

Using n(n + 1)/2. The factor is n − 1, because a level cannot jump to itself.
Using n₂ in place of Δn. When the lowest level is not 1, use the difference n₂ − n₁ in the general formula.
Ignoring the single atom case. One atom gives at most n − 1 lines.
Counting absorption lines the same way. In absorption from the ground state, atoms start at n = 1, so only n − 1 lines (to levels 2 through n) are seen.
Forgetting to find n first. In energy questions, use $E_n = -\frac{13.6}{n^2}$ eV to locate the level, and check that n comes out a whole number.

Frequently asked questions

What is the formula for number of spectral lines?

For electrons falling from level n to the ground state, N = n(n − 1) / 2. Between two levels n₂ and n₁, use N = (n₂ − n₁)(n₂ − n₁ + 1) / 2.

How many spectral lines are produced when an electron falls from n = 4 to n = 1?

In a sample of atoms, 4 × 3 / 2 = 6 lines. For a single atom the most it can give is 3 lines, by stepping 4 to 3 to 2 to 1.

What is the total number of spectral lines formula from n₂ to n₁?

N = Δn(Δn + 1) / 2 with Δn = n₂ − n₁. From n₂ = 5 to n₁ = 3, Δn = 2, so N = 3.

Does the number of spectral lines depend on the element?

The counting formula only uses the number of levels, so it holds for any one electron system such as H, He⁺ or Li²⁺. The wavelengths of the lines do depend on the element through Z².

Why is the number of spectral lines formula n(n − 1)/2?

Each line comes from one pair of levels, and the number of pairs that can be picked from n levels is n(n − 1) / 2.