MHT-CET Chemistry · Structure of Atom

Hydrogen Spectrum and the Rydberg Equation

When an excited electron in a hydrogen atom falls to a lower orbit it emits a photon of a definite wavelength; the Rydberg equation gives the wavenumber of every such line, and the lines group into named series (Lyman, Balmer, Paschen, and so on) by the orbit they land on.

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Why this matters

This is one of the most dependable scoring blocks in MHT-CET Structure of Atom. Most PYQs are one-step plug-ins into the Rydberg equation to get a line's wavenumber, plus a straight recall of which series lands where (Balmer is visible, Lyman is UV). Learn the Rydberg formula with the fraction ordering fixed, memorise the five series and their regions, and every question here becomes an on-sight tick.

Concept 1 of 3

Rydberg equation — wavenumber of a spectral line

Intuition

An electron sitting in an outer orbit is unstable; when it drops to a lower orbit it dumps the energy difference as one photon. The Rydberg equation converts the two orbit numbers straight into the photon's wavenumber (lines per centimetre). The one rule that fixes every calculation: put the SMALLER orbit number as n1 (the first, positive fraction) and the LARGER as n2, so the bracket stays positive.

Definition

The Rydberg equation for a hydrogen-like line:

Wavenumber $\bar{\nu} = \dfrac{1}{\lambda} = R_H\, Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ , with $n_1 < n_2$ .
For hydrogen $Z = 1$ , so $\bar{\nu} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ .
$R_H$ is the Rydberg constant: $1.097\times 10^{7}\ \text{m}^{-1}$ , i.e. $109677\ \text{cm}^{-1}$ (MHT-CET papers almost always give it in $\text{cm}^{-1}$ ).
Wavenumber $\bar{\nu}$ has units of $\text{cm}^{-1}$ (or $\text{m}^{-1}$ ); the wavelength is its reciprocal, $\lambda = 1/\bar{\nu}$ .
$n_1$ is always the lower orbit (the one the electron falls TO); $n_2$ is the upper orbit (the one it falls FROM).

Rydberg equation

\bar{\nu} = \dfrac{1}{\lambda} = R_H\, Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) \qquad (n_1 < n_2)

\bar{\nu}wavenumber of the line (cm^-1 or m^-1)
\lambdawavelength of the line
R_HRydberg constant = 1.097 x 10^7 m^-1 = 109677 cm^-1
Zatomic number (Z = 1 for hydrogen)
n_1lower orbit (fallen TO), the larger positive fraction
n_2upper orbit (fallen FROM)

Worked example

Find the wavenumber of the line emitted when an electron in a hydrogen atom falls from n = 4 to n = 2. (RH = 109677 cm^-1)

The lower orbit is $n_1 = 2$ and the upper is $n_2 = 4$ .
$\bar{\nu} = R_H\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right) = 109677\left(\dfrac{1}{4} - \dfrac{1}{16}\right)$ .
$\dfrac{1}{4} - \dfrac{1}{16} = \dfrac{4 - 1}{16} = \dfrac{3}{16}$ .
$\bar{\nu} = 109677 \times \dfrac{3}{16} = 109677 \times 0.1875$ .

Answer:

\bar{\nu} \approx 20564\ \text{cm}^{-1}

Practice this conceptself-check · 4 quick reps

Try it yourself

Calculate the wavenumber of the photon emitted when the electron drops from n = 3 to n = 1 in a hydrogen atom. (RH = 109677 cm^-1)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Wavenumber of the n = 3 to n = 2 line in hydrogen? (RH = 109677 cm^-1)
2.
Wavenumber of the n = 2 to n = 1 line? (RH = 109677 cm^-1)
3.
In the Rydberg equation, which orbit number is n1 for a fall from n = 6 to n = 3?
4.
Value of the Rydberg constant in cm^-1?

From the bank · past-year question

Example 1Structure of AtomEASY

What is the wavenumber of the photon emitted during transition from the orbit n = 5 to that of n = 2 in hydrogen atom?

[R_H = 109677\text{ cm}^{-1}]

[Shift || · 2025]

n1 is the smaller orbit — keep the bracket positive

\bar{\nu} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)

with

n_1 < n_2

. For a fall from

n = 5

n = 2

, use

n_1 = 2, n_2 = 5

— NOT the reverse. Swapping them flips the sign and gives a negative (wrong) wavenumber.

Wavenumber and wavelength are reciprocals

\bar{\nu} = 1/\lambda

. If a question asks for the wavelength, compute

\bar{\nu}

first, then take its reciprocal. Reporting

\bar{\nu}

as the wavelength (or vice versa) is a classic careless slip.

Drill 6 more on rydberg equation — wavenumber of a spectral line

Concept 2 of 3

The spectral series of hydrogen

Intuition

Every line for which the electron lands on the SAME final orbit belongs to one named series. Falls to n = 1 make the Lyman series, falls to n = 2 make the Balmer series, and so on. The single most-tested fact: Balmer (to n = 2) is the only series in the VISIBLE region; Lyman (to n = 1) is ultraviolet; everything landing on n = 3 or higher is infrared.

Definition

Each series is named by its lower orbit $n_1$ (the orbit the electron falls to):

Lyman — $n_1 = 1$ , ultraviolet (UV).
Balmer — $n_1 = 2$ , visible (the only visible series).
Paschen — $n_1 = 3$ , infrared (IR).
Brackett — $n_1 = 4$ , infrared.
Pfund — $n_1 = 5$ , infrared.

Series	Falls to (n1)	From (n2)	Region
Lyman	1	2, 3, 4, ...	Ultraviolet (UV)Q MHT-CET 2024 — the series for a jump from n2 = infinity to n1 = 1 is the Lyman series.
Balmer	2	3, 4, 5, ...	VisibleQ MHT-CET 2023 + 2021 — Balmer is the ONLY series in the visible region.
Paschen	3	4, 5, 6, ...	Infrared (IR)
Brackett	4	5, 6, 7, ...	Infrared (IR)
Pfund	5	6, 7, 8, ...	Infrared (IR)

Memory aid: La-Ba-Pa-Bra-Pf for n1 = 1, 2, 3, 4, 5. Only Balmer (n1 = 2) is visible; Lyman is UV; the rest are IR.

Practice this conceptself-check · 4 quick reps

Try it yourself

An electron jumps from n = infinity to n = 1 in a hydrogen atom. Which spectral series does the emitted line belong to, and in which region does it lie?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which hydrogen series lies in the visible region?
2.
Which series lands on n = 1 and lies in the UV?
3.
The Paschen series corresponds to falls to which orbit?
4.
Name the series for transitions ending at n = 5.

From the bank · past-year question

Example 2Structure of AtomEASY

Which of the following series of transitions of hydrogen spectrum falls in the visible region?

[Q79 · May Shift 1 · 2021]

Balmer is visible, Lyman is not

The visible series is Balmer (falls to n = 2). Lyman (falls to n = 1) is ultraviolet. A large share of MHT-CET marks are lost by picking Lyman for the 'visible region' question.

The series is named by the LOWER orbit, not the upper one

A jump 'from n = 4 to n = 2' is a Balmer line, because it lands on

n_1 = 2

. Don't name a series by the starting (upper) orbit — always look at where the electron finishes.

Drill 2 more on the spectral series of hydrogen

Concept 3 of 3

Longest wavelength, series limit, and number of spectral lines

Intuition

Within one series the transition with the smallest energy gap gives the LONGEST wavelength (the first line, from the very next orbit); the transition from n2 = infinity gives the SHORTEST wavelength — the series limit. Separately, if an electron is excited to orbit n and can cascade down in every possible way, the total number of distinct lines emitted is a fixed count, n(n-1)/2.

Definition

Two recurring sub-results:

Longest wavelength (first line) of a series: smallest gap, so $n_2 = n_1 + 1$ . Its wavenumber is $\bar{\nu} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{(n_1+1)^2}\right)$ , and $\lambda_{\max} = 1/\bar{\nu}$ .
Series limit (shortest wavelength): $n_2 = \infty$ , so $\dfrac{1}{n_2^2} \to 0$ and $\bar{\nu}_{\max} = \dfrac{R_H}{n_1^2}$ .
Number of spectral lines when an electron de-excites from orbit $n$ down to the ground state (all possible jumps): $\dfrac{n(n-1)}{2}$ .

Number of spectral lines from orbit n

\text{No. of lines} = \dfrac{n(n-1)}{2}

nhighest orbit the electron is excited to

Worked example

Calculate the longest wavelength line of the Balmer series of hydrogen. (RH = 109677 cm^-1)

Longest wavelength = smallest energy gap = the first line, $n_2 = n_1 + 1$ .
For Balmer $n_1 = 2$ , so the first line is $n_2 = 3 \to n_1 = 2$ .
$\bar{\nu} = 109677\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = 109677\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 109677 \times \dfrac{5}{36} \approx 15233\ \text{cm}^{-1}$ .
$\lambda = \dfrac{1}{\bar{\nu}} = \dfrac{1}{15233}\ \text{cm}$ .

Answer:

\lambda \approx 6.565 \times 10^{-5}\ \text{cm}

(about 656 nm, the red H-alpha line).

Practice this conceptself-check · 4 quick reps

Try it yourself

An electron in a hydrogen atom is excited to the n = 4 orbit. How many spectral lines can be emitted as it returns to the ground state?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Series limit (shortest-wavelength) wavenumber of the Lyman series?
2.
Number of spectral lines emitted from the n = 5 orbit to the ground state?
3.
Within a series, which transition gives the longest wavelength?
4.
Series limit of the Balmer series (wavenumber)?

From the bank · past-year question

Example 3Structure of AtomMODERATE

Calculate the longest wavelength in hydrogen emission spectrum of Lyman series. (

R_{H}= 109677{\text{ }cm}^{- 1}

)

[Q96 · 19 April Shift I · 2025]

Longest wavelength = smallest gap, not the biggest jump

The longest wavelength corresponds to the least energy, i.e. the closest pair of orbits (

n_2 = n_1 + 1

). Students often plug in

n_2 = \infty

— that gives the SHORTEST wavelength (the series limit), the exact opposite.

n(n-1)/2 counts every downward jump

The formula

\dfrac{n(n-1)}{2}

is the number of distinct lines when the electron can cascade down in ALL possible ways from orbit

n

. For

n = 4

that is 6, not 3.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Rydberg equation — wavenumber of a spectral line
Rydberg equation
$\bar{\nu} = \dfrac{1}{\lambda} = R_H\, Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) \qquad (n_1 < n_2)$
Longest wavelength, series limit, and number of spectral lines
Number of spectral lines from orbit n
$\text{No. of lines} = \dfrac{n(n-1)}{2}$

Reference tables (1)

The spectral series of hydrogen5 rows

Series	Falls to (n1)	From (n2)	Region
Lyman	1	2, 3, 4, ...	Ultraviolet (UV)Q MHT-CET 2024 — the series for a jump from n2 = infinity to n1 = 1 is the Lyman series.
Balmer	2	3, 4, 5, ...	VisibleQ MHT-CET 2023 + 2021 — Balmer is the ONLY series in the visible region.
Paschen	3	4, 5, 6, ...	Infrared (IR)
Brackett	4	5, 6, 7, ...	Infrared (IR)
Pfund	5	6, 7, 8, ...	Infrared (IR)

Memory aid: La-Ba-Pa-Bra-Pf for n1 = 1, 2, 3, 4, 5. Only Balmer (n1 = 2) is visible; Lyman is UV; the rest are IR.

Watch out for (6)

n1 is the smaller orbit — keep the bracket positive
→ Rydberg equation — wavenumber of a spectral line
Wavenumber and wavelength are reciprocals
→ Rydberg equation — wavenumber of a spectral line
Balmer is visible, Lyman is not
→ The spectral series of hydrogen
The series is named by the LOWER orbit, not the upper one
→ The spectral series of hydrogen
Longest wavelength = smallest gap, not the biggest jump
→ Longest wavelength, series limit, and number of spectral lines
n(n-1)/2 counts every downward jump
→ Longest wavelength, series limit, and number of spectral lines

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Structure of AtomMODERATE

Calculate the wave number of the photon emitted during transition from the orbit of

n=2

n=1

in hydrogen atom (

R_H=109677\,cm^{-1}

)

[Q62 · 9th May Shift 1 · 2024]

Example 2Structure of AtomEASY

Which of the following series of emission spectral lines for hydrogen is observed in the visible region?

[Q66 · Shift 1 · 2023]

Example 3Structure of AtomHARD

What is the wave number of lowest transition in Balmer series?

[Q77 · 3rd May Shift 2 · 2023]

Example 4Structure of AtomEASY

Which emission transition series is obtained when electron jumps from

n_2 = \infty

n_1 = 1

[Q80 · 11th May Shift 1 · 2024]

Example 5Structure of AtomMODERATE

What is the wavenumber of the photon emitted during transition from the orbit

n = 5

to that of

n = 2

in hydrogen atom?

\left\lbrack R_{H}= 109677{\text{ }cm}^{- 1} \right\rbrack

[Q99 · 19 April Shift II · 2025]

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