The Line Spectra of the Hydrogen Atom

Q: Why does hydrogen give a line spectrum instead of a continuous spectrum?

The electron in hydrogen can have only certain energies, -13.6/(n²) eV. Light is emitted only when it jumps between these levels, and each jump gives a photon of one definite energy and wavelength. So only certain sharp wavelengths appear.

Q: What is the Rydberg formula?

It gives the wavelength of any hydrogen line: 1/λ = R(1/(n₁²) - 1/(n₂²)), where R = 1.097 × 10⁷ m⁻¹ and n₂ > n₁. For hydrogen-like ions multiply the right side by Z².

PhysicsAtomsFor NEET aspirants

The line spectra of the hydrogen atom are the sharp wavelengths that hydrogen emits or absorbs when its electron jumps between Bohr energy levels. Each jump gives one photon with $h ν = E_{n_{2}} - E_{n_{1}}$ , which leads to the Rydberg formula $\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ and the Lyman, Balmer, Paschen, Brackett and Pfund series. This page also covers counting lines, excitation, ionisation, recoil and atomic collisions. The line spectra of the hydrogen atom are a favourite topic in JEE Main and NEET.

On this page1Emission and absorption2Transition rule3Spectral series4Counting lines5Excitation and ionisation6Photon vs electron impact7Hydrogen-like ions8Recoil9Atomic collisions

Key Formulas - Quick Reference

★ Must learnPhoton energy in a jump: $h ν = \frac{h c}{λ} = E_{n_{2}} - E_{n_{1}} = 13.6 Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) eV$
★ Must learnRydberg formula: $\overset{ν}{ˉ} = \frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , $R = 1.097 \times 1 0^{7} m^{- 1}$
★ Must learnQuick conversion: $λ (\overset{˚}{A}) = \frac{12400}{Δ E ( eV )}$ , or $λ (nm) = \frac{1240}{Δ E ( eV )}$
Series: Lyman $n_{1} = 1$ (UV), Balmer $n_{1} = 2$ (visible, limit in near UV), Paschen $3$ , Brackett $4$ , Pfund $5$ , Humphreys $6$ (infrared)
First line $n_{2} = n_{1} + 1$ (longest $λ$ ); series limit $n_{2} = \infty$ , $λ_{l imi t} = \frac{n _{1}^{2}}{R Z ^{2}}$ (shortest $λ$ )
★ Must learnNumber of lines from level $n$ : $\frac{n ( n - 1 )}{2}$ for many atoms; at most $n - 1$ from one atom
Ionisation energy from level $n$ : $\frac{13.6 Z ^{2}}{n ^{2}} eV$ ; hydrogen: excitation $10.2$ , $12.09 eV$ , ionisation $13.6 eV$
Recoil speed of the atom: $v = \frac{h}{M λ} = \frac{Δ E}{M c}$
Head-on collision with an atom at rest: maximum energy loss $= \frac{M}{m + M} K$ ( $\frac{K}{2}$ for a neutron on hydrogen)

1. Emission and Absorption Line Spectra

Light from a hot solid (a bulb filament) spreads into a continuous spectrum containing all wavelengths. Light from a gas of free atoms is different: when hydrogen at low pressure is excited by an electric discharge, a prism or grating separates its light into a few sharp, bright lines.

Figure 1: A prism spectroscope. The prism bends every ray towards its base, violet most and red least, so the four visible hydrogen wavelengths land as four sharp lines, not a rainbow. Rays are traced exactly with Snell's law at both faces; the glass's dispersion is exaggerated about three times so that the lines separate.

Emission line spectrum: bright lines on a dark background, produced when excited atoms fall to lower levels. Every element has its own unique set of lines, like a fingerprint, so line spectra identify the composition of unknown samples and of stars.
Absorption line spectrum: when white light passes through a cooler gas, atoms absorb exactly those wavelengths that can lift them to higher levels. The result is dark lines on the continuous spectrum, at the same wavelengths as the emission lines.

Figure 2: Emission and absorption. Hot hydrogen emits bright lines on a dark background; cool hydrogen in front of a white source removes the same wavelengths, leaving dark lines on the continuous spectrum. (Cool hydrogen in the laboratory absorbs Lyman lines; the dark Balmer lines appear when many atoms are already in

n = 2

, as in a hot star's atmosphere.)

Emission spectrum

Bright lines on a dark background. Atoms fall from higher to lower levels and emit photons. All series appear (Lyman, Balmer, Paschen, ...).

Absorption spectrum

Dark lines on a bright continuous background. Atoms rise from lower to higher levels by absorbing photons. At room temperature nearly all hydrogen atoms are in $n = 1$ , so only the Lyman lines are absorbed.

On the energy-level diagram both processes use the same gaps. An absorbed photon lifts the electron; about $1 0^{- 8} s$ later it falls back, often in several steps, so a gas that absorbs one wavelength can re-emit several.

Figure 3: Absorption and emission use the same levels. A

12.09 eV

photon lifts the electron from

n = 1

3

; on the way down the sample gives

\frac{3 \times 2}{2} = 3

lines, one of which (

102.6 nm

) has exactly the absorbed wavelength (levels to scale, energies in eV).

2. The Transition Rule and the Rydberg Formula

Bohr's third postulate says a photon is emitted when the electron jumps from an upper level $n_{2}$ to a lower level $n_{1}$ , and carries the energy difference:

$E_{n} = - \frac{2 π ^{2} k ^{2} Z ^{2} e ^{4} m}{n ^{2} h ^{2}}$ , so $h ν = E_{n_{2}} - E_{n_{1}} = \frac{2 π ^{2} k ^{2} Z ^{2} e ^{4} m}{h ^{2}} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) = 13.6 Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) eV$ .
With $ν = \frac{c}{λ}$ , divide by $h c$ to get the wave number $\overset{ν}{ˉ} = \frac{1}{λ}$ (waves per metre):

★ Must learn

\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), R = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3} c} = 1.097 \times 1 0^{7} m^{- 1}

This is the Rydberg formula. It was found from measurements (Balmer 1885, Rydberg 1888) before Bohr; Bohr's model derived $R$ from $m$ , $e$ , $h$ , $ε_{0}$ and $c$ , a major success.

Figure 4: Every hydrogen line, from any series, lies on one straight line through the origin. The slope is the Rydberg constant

R = 1.097 \times 1 0^{7} m^{- 1}

, which Bohr's theory derives as

\frac{m e ^{4}}{8 ε _{0}^{2} h ^{3} c}

In numericals it is usually quicker to work in electronvolts and use $h c = 12400 eV \overset{˚}{A} = 1240 eV nm$ :

λ (\overset{˚}{A}) = \frac{12400}{Δ E ( eV )}

For example, $n = 3 \to 2$ in hydrogen: $Δ E = 13.6 (\frac{1}{4} - \frac{1}{9}) = 1.89 eV$ , so $λ = \frac{12400}{1.89} = 6563 \overset{˚}{A}$ , the red $H_{α}$ line.

Quick Recall: tap to check

What is the wave number of a line?

\overset{ν}{ˉ} = \frac{1}{λ}

, the number of waves per unit length, in

m^{- 1}

Energy of a photon of wavelength

4000 \overset{˚}{A}

\frac{12400}{4000} = 3.1 eV

How does

λ

for a given jump change from H to

He^{+}

It becomes

\frac{1}{4}

, since

\frac{1}{λ} \propto Z^{2}

3. Spectral Series of Hydrogen

All jumps that end on the same lower level $n_{1}$ form a series. The series are named after their discoverers.

Figure 5: Energy levels of hydrogen,

E_{n} = - \frac{13.6}{n ^{2}} eV

(levels

n \geq 2

to scale;

n = 1

below a scale break). Each series is the set of jumps ending on one level; the dashed arrow from

n = \infty

gives the series limit (shortest wavelength).

Series	$n_{1}$	$n_{2}$	Region	First line ( $λ_{ma x}$ )	Series limit ( $λ_{min}$ )
Lyman	1	$2, 3, 4, \dots$	ultraviolet	$121.5 nm$	$91.2 nm$
Balmer	2	$3, 4, 5, \dots$	visible (limit in near UV)	$656.3 nm$	$364.6 nm$
Paschen	3	$4, 5, 6, \dots$	near infrared	$1875 nm$	$820.4 nm$
Brackett	4	$5, 6, 7, \dots$	infrared	$4051 nm$	$1459 nm$
Pfund	5	$6, 7, 8, \dots$	far infrared	$7458 nm$	$2279 nm$
Humphreys	6	$7, 8, 9, \dots$	far infrared	$12.37 μ m$	$3282 nm$

The first line (the $α$ line) of each series comes from $n_{2} = n_{1} + 1$ : the smallest energy, longest wavelength.
The series limit comes from $n_{2} = \infty$ : the largest energy, shortest wavelength, $λ_{l imi t} = \frac{n _{1}^{2}}{R}$ .
Lines crowd together towards the series limit, because the upper levels crowd together near $E = 0$ .
The infrared series overlap: Brackett starts ( $1459 nm$ ) before Paschen ends ( $1875 nm$ ).

Figure 6: The whole hydrogen spectrum, computed from

\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

(log scale). The first line of each series is drawn taller; lines crowd towards the dashed series limit. Lyman is ultraviolet, Balmer mostly visible, the rest infrared, and the infrared series overlap.

3.1 The Balmer series

Balmer found (1885) that the four visible hydrogen lines fit $λ = 364.6 \frac{n ^{2}}{n ^{2} - 4} nm$ , $n = 3, 4, 5, 6$ . This is the Rydberg formula with $n_{1} = 2$ . Only $H_{α}$ (red), $H_{β}$ (blue-green), $H_{γ}$ and $H_{δ}$ (violet) are visible; the rest lie in the near ultraviolet.

Figure 7: The Balmer series,

\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{n ^{2}})

with

n = 3, 4, 5, \dots

The four bright lines

H_{α}

H_{δ}

are visible; the higher lines crowd together and the series limit (

364.6 nm

) lies in the near ultraviolet.

Exam Trick

Ratios need no constants. $λ \propto \frac{1}{\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}}$ . Learn the brackets: first Lyman $\frac{3}{4}$ , Lyman limit $1$ , first Balmer $\frac{5}{36}$ , Balmer limit $\frac{1}{4}$ , first Paschen $\frac{7}{144}$ . So $\frac{λ _{B a l m er, f i r s t}}{λ _{L y man, f i r s t}} = \frac{3/4}{5/36} = \frac{27}{5}$ , and $\frac{λ _{B a l m er, l imi t}}{λ _{L y man, l imi t}} = 4$ .

Key idea

A series is fixed by its lower level

n_{1}

. The first line is the longest wavelength, the series limit the shortest,

λ_{l imi t} = \frac{n _{1}^{2}}{R}

4. Number of Spectral Lines

Suppose atoms are excited to level $n$ . From $n$ an electron can jump to any of the $n - 1$ lower levels; from $n - 1$ to $n - 2$ levels, and so on. For a sample containing many atoms every path occurs:

N = (n - 1) + (n - 2) + \dots + 2 + 1 = \frac{n ( n - 1 )}{2}

Figure 8: A sample of atoms excited to

n = 4

gives

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

lines (photon energies in eV; level spacing not to scale): three Lyman, two Balmer, one Paschen. One atom cascading down gives at most

n - 1 = 3

of them.

★ Must learn

Many atoms excited to level $n$ : $\frac{n ( n - 1 )}{2}$ emission lines. Between levels $n_{2}$ and $n_{1}$ : $\frac{( n _{2} - n _{1} ) ( n _{2} - n _{1} + 1 )}{2}$ .
A single atom cascading down: at most $n - 1$ photons (one step at a time).
Lines of one series from level $n$ : $n - n_{1}$ (for example, Balmer lines from $n = 5$ : $3$ ).
Absorption from the ground state up to level $n$ : only $n - 1$ lines, all Lyman, because absorption starts from $n = 1$ .

A sample excited to $n = 5$ gives $\frac{5 \times 4}{2} = 10$ lines. Ranking them by photon energy answers the common "which transition has the highest frequency (or longest wavelength)" questions at a glance.

Figure 9: The ten lines from a sample excited to

n = 5

, ranked by photon energy (highest frequency, shortest

λ

at the top). Every jump that ends on

n = 1

beats every other jump, because the gap

E_{2} - E_{1} = 10.2 eV

is larger than all the gaps above it put together.

5. Excitation and Ionisation

Raising the electron to a higher level is excitation; the higher levels are excited states. Removing it completely ( $n \to \infty$ , $E = 0$ ) is ionisation.

Figure 10: Excitation lifts the electron to a higher level; ionisation removes it to

E = 0

. The energy needed depends on the starting level:

13.6 eV

from

n = 1

but only

3.4 eV

from

n = 2

. In volts, the same numbers are the excitation and ionisation potentials.

Term	Meaning	Hydrogen	$He^{+}$
First excitation energy	$E_{2} - E_{1}$	$10.2 eV$	$40.8 eV$
Second excitation energy	$E_{3} - E_{1}$	$12.09 eV$	$48.4 eV$
Ionisation energy (ground state)	$0 - E_{1} = 13.6 Z^{2} eV$	$13.6 eV$	$54.4 eV$
Ionisation energy from level $n$	$\frac{13.6 Z ^{2}}{n ^{2}} eV$	$3.4 eV$ from $n = 2$	$13.6 eV$ from $n = 2$
Excitation / ionisation potential	energy in eV divided by $e$ , in volts	$10.2 V$ / $13.6 V$	$40.8 V$ / $54.4 V$

The binding energy of the electron in level $n$ is the same as the ionisation energy from that level, $∣ E_{n} ∣$ . An excited state lives for only about $1 0^{- 8} s$ before the electron drops back and emits a photon. (Some special metastable states live much longer, around $1 0^{- 3} s$ ; lasers use them.)

Quick Recall: tap to check

What is the ionisation potential of hydrogen in the

n = 3

state?

\frac{13.6}{9} = 1.51 V

Why is the first excitation energy of

He^{+}

40.8 eV

It is

4 \times 10.2 eV

, because energies scale as

Z^{2}

How long does a typical excited state last?

About

1 0^{- 8} s

6. Excitation by Photons and by Particle Impact

Energy can be given to an atom by absorbing a photon or by a collision with an electron or another particle. The two behave very differently.

Figure 11: A photon is absorbed only whole, so it must match a level difference exactly (

10.2

12.09

12.75 eV

...) or exceed the ionisation energy. A colliding electron can hand over just part of its energy: an

11 eV

electron excites

n = 1 \to 2

and keeps

0.8 eV

Excitation by a photon

A photon is absorbed whole or not at all. It is absorbed only if $h ν$ equals a level difference exactly. A ground-state hydrogen atom absorbs $10.2$ , $12.09$ , $12.75 eV$ ... photons, but an $11 eV$ photon passes straight through. Any photon above $13.6 eV$ ionises: the electron leaves with $K = h ν - 13.6 eV$ .

Excitation by an electron

A colliding electron can give up only part of its kinetic energy. An electron with $K \geq 10.2 eV$ can excite $n = 1 \to 2$ and keep the rest. An $11 eV$ electron excites to $n = 2$ and leaves with $0.8 eV$ ; it cannot reach $n = 3$ ( $12.09 eV$ ).

Exam Trick

Photon: exact match; electron: at least. For a photon beam on ground-state hydrogen, find $n$ from $13.6 (1 - \frac{1}{n ^{2}}) = h ν$ ; if $n$ is not a whole number, nothing is absorbed. For an electron beam, find the highest level with excitation energy $\leq K$ . Then count the emitted lines with $\frac{n ( n - 1 )}{2}$ .

Key idea

Photons need an exact energy match (or enough to ionise); colliding particles can give part of their energy. That is why an

11 eV

photon is transmitted but an

11 eV

electron excites hydrogen.

7. Spectra of Hydrogen-like Ions

For $He^{+}$ , $Li^{2 +}$ and other one-electron ions every photon energy is multiplied by $Z^{2}$ and every wavelength divided by $Z^{2}$ . For $He^{+}$ the Lyman series lies around $23$ to $30 nm$ (extreme ultraviolet). Because $E_{n} \propto \frac{Z ^{2}}{n ^{2}}$ , levels with the same $\frac{n}{Z}$ coincide, and so do some lines.

Figure 12:

He^{+}

levels with even

n

coincide with hydrogen levels:

E = - 13.6 \frac{4}{( 2 m ) ^{2}} = - \frac{13.6}{m ^{2}}

. So

He^{+}

(2 m_{2} \to 2 m_{1})

emits exactly the hydrogen line

(m_{2} \to m_{1})

, for example

6 \to 4

gives

H_{α}

656 nm

(Bohr model, ignoring the tiny reduced-mass shift).

This coincidence confused early spectroscopists: a series of $He^{+}$ lines seen in stars (the Pickering series, $n \to 4$ ) contains every Balmer line plus lines halfway between them. Bohr's model explained it at once.

8. Recoil of the Atom

A photon of wavelength $λ$ carries momentum $p = \frac{h}{λ} = \frac{h ν}{c}$ . An atom of mass $M$ at rest that emits it must recoil the other way, with equal momentum.

Figure 13: Momentum conservation on emission. The photon carries momentum

\frac{h ν}{c} = \frac{h}{λ}

, so the atom recoils with

M v = \frac{h}{λ}

. For the

10.2 eV

Lyman-

α

photon,

v \approx 3.3 m s^{- 1}

M v = \frac{h}{λ} = \frac{Δ E}{c} \Rightarrow v = \frac{Δ E}{M c}

Here we used $\frac{h c}{λ} \approx Δ E$ , because the recoil kinetic energy $\frac{( Δ E ) ^{2}}{2 M c ^{2}}$ is tiny (about $5.5 \times 1 0^{- 8} eV$ for hydrogen's $10.2 eV$ line).

JEE Advanced

Exact photon energy with recoil. Energy conservation is $Δ E = h ν + \frac{p ^{2}}{2 M}$ with $p = \frac{h ν}{c}$ , so

h ν = Δ E - \frac{( h ν ) ^{2}}{2 M c ^{2}} \approx Δ E (1 - \frac{Δ E}{2 M c ^{2}})

The emitted photon is slightly less energetic than the level gap, and an absorbed photon must be slightly more energetic. The fractional change $\frac{Δ E}{2 M c ^{2}}$ is only $5 \times 1 0^{- 9}$ for hydrogen.

Isotope shift. With the reduced mass, $R_{M} = R_{\infty} \frac{M}{M + m}$ . Deuterium ( $M \approx 2 m_{p}$ ) has a slightly larger $R$ than hydrogen, so its $H_{α}$ line is shorter by about $0.18 nm$ ( $656.10$ vs $656.28 nm$ in air). Urey discovered deuterium (1932) from this faint companion line.

9. Atomic Collisions: Elastic or Inelastic?

When a particle (mass $m$ , kinetic energy $K$ ) hits an atom (mass $M$ ) at rest, two laws must hold together:

Momentum conservation (Newtonian mechanics) allows any loss of kinetic energy from $0$ (elastic) up to the loss in a perfectly inelastic collision, $Δ E_{ma x} = \frac{M}{m + M} K$ , when both move together.
Quantum mechanics says the lost energy can only go into exciting the atom, so the loss must be $0$ or one of the excitation energies ( $10.2$ , $12.09$ , $12.75 eV$ ... for hydrogen, or $\geq 13.6 eV$ for ionisation).

Figure 14: A neutron with

K = 24.18 eV

hits a free hydrogen atom head-on (Solved Example 2(d)). Newtonian mechanics allows any loss from

0

\frac{K}{2}

; the atom accepts only its excitation energies. Possible outcomes are the overlap: loss

0

(elastic),

10.2 eV

(inelastic) or

12.09 eV

(perfectly inelastic).

Only losses allowed by both rules can happen. The collision is elastic if the loss is $0$ , inelastic if the loss is an excitation energy smaller than $Δ E_{ma x}$ , and perfectly inelastic if an excitation energy equals $Δ E_{ma x}$ exactly. An inelastic collision is possible only if $\frac{M}{m + M} K \geq$ the first excitation energy.

For a neutron on hydrogen ( $M = m$ ) only half of $K$ can be lost, so $K$ must be at least $20.4 eV$ to excite the atom. For an electron ( $m ≪ M$ ) almost all of $K$ is available, which is why electron beams are used to excite gases (the Franck-Hertz experiment).

10. Solving Spectrum Problems and Revision Map

Use the flowchart to choose the method, then the mind map to revise the whole concept.

Figure 15: Solving a spectral-line problem. Most numericals need only

Δ E

in eV and

λ (nm) = \frac{1240}{Δ E ( eV )}

Figure 16: Mind map of this concept. Cover a branch, recall its three points, then check.

11. Solved Examples

Solved Example 1

Hydrogen atoms in the ground state are exposed to photons of energy (a)

12.09 eV

and (b)

11 eV

. What happens in each case?

Solution:

(a) $E_{1} + 12.09 = - 13.6 + 12.09 = - 1.51 eV = E_{3}$ . The energy matches the jump $n = 1 \to 3$ exactly, so the photon is absorbed and the atom goes to $n = 3$ .

(b) $- 13.6 + 11 = - 2.6 eV$ , which lies between $E_{2} = - 3.4$ and $E_{3} = - 1.51 eV$ . There is no level there, and a photon cannot be absorbed in part.

Answer: (a) absorbed, atom excited to $n = 3$ ; (b) not absorbed, the $11 eV$ photons pass through the gas unaffected.

Solved Example 2

A neutron of kinetic energy

K

collides head-on with a hydrogen atom at rest in its ground state (the atom is free to move). Is the collision elastic, inelastic or perfectly inelastic if

K

is (a)

14 eV

(b)

20.4 eV

(c)

22 eV

(d)

24.18 eV

Solution:

Newtonian limit: equal masses, so in a perfectly inelastic collision $m v_{0} = 2 m v_{f}$ , $v_{f} = \frac{v _{0}}{2}$ , and the final KE is $\frac{1}{2} (2 m) \frac{v _{0}^{2}}{4} = \frac{K}{2}$ . The loss can be anything from $0$ to $\frac{K}{2}$ .

Quantum rule: the loss must be $0$ , $10.2$ , $12.09$ , $12.75 eV$ , ...

$K$ (eV)	Newtonian range of loss	Possible losses	Type
14	$0$ to $7 eV$	$0$ only	elastic only
20.4	$0$ to $10.2 eV$	$0$ or $10.2 eV$	elastic, or perfectly inelastic ( $10.2 = \frac{K}{2}$ )
22	$0$ to $11 eV$	$0$ or $10.2 eV$	elastic or inelastic
24.18	$0$ to $12.09 eV$	$0$ , $10.2$ or $12.09 eV$	elastic, inelastic ( $10.2$ ) or perfectly inelastic ( $12.09$ )

Answer: (a) elastic; (b) elastic or perfectly inelastic; (c) elastic or inelastic; (d) elastic, inelastic or perfectly inelastic.

Solved Example 3

He^{+}

ion is at rest in its ground state. A neutron with kinetic energy

K

collides head-on with it. Find the minimum

K

for which the collision can be inelastic. (Take the mass of

He^{+}

4

times that of the neutron.)

Solution:

Quantum: $E_{n} = - \frac{54.4}{n ^{2}} eV$ , so possible losses are $0$ , $40.8$ , $48.4$ , ... , $54.4 eV$ .

Newtonian: perfectly inelastic, $m v_{0} = 5 m v_{f}$ , final KE $= \frac{1}{2} (5 m) \frac{v _{0}^{2}}{25} = \frac{K}{5}$ , so the maximum loss is $\frac{4 K}{5}$ .

An inelastic collision needs $\frac{4 K}{5} \geq 40.8 eV$ .

Answer: $K_{min} = 51 eV$ .

Solved Example 4

How many different wavelengths may be observed in the spectrum of a hydrogen sample if the atoms are excited to states with principal quantum number

n

Solution:

From level $n$ there are $n - 1$ possible jumps; from $n - 1$ there are $n - 2$ ; and so on down to $1$ from level $2$ .

N = (n - 1) + (n - 2) + \dots + 1 = \frac{n ( n - 1 )}{2}

Answer: $\frac{n ( n - 1 )}{2}$ (for example, $6$ lines for $n = 4$ , $10$ for $n = 5$ ).

Solved Example 5

A hydrogen atom at rest in its first excited state drops to the ground state. Find its recoil speed. (Mass of hydrogen atom

M = 1.67 \times 1 0^{- 27} kg

)

Solution:

Photon energy $Δ E = 10.2 eV = 10.2 \times 1.6 \times 1 0^{- 19} = 1.63 \times 1 0^{- 18} J$ (the recoil energy is negligible).

Momentum conservation: $M v = \frac{h}{λ} = \frac{Δ E}{c}$ , so $v = \frac{Δ E}{M c} = \frac{1.63 \times 1 0 ^{- 18}}{1.67 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{8}}$ .

Answer: $v \approx 3.3 m s^{- 1}$ .

Solved Example 6

Find the wavelength of the

H_{α}

line and of the Balmer series limit. (

R = 1.097 \times 1 0^{7} m^{- 1}

)

Solution:

$H_{α}$ : $n_{2} = 3 \to n_{1} = 2$ : $\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ , so $λ = \frac{36}{5 R} = \frac{36}{5 \times 1.097 \times 1 0 ^{7}}$ .

Limit: $n_{2} = \infty$ : $\frac{1}{λ} = \frac{R}{4}$ , so $λ = \frac{4}{R}$ .

Answer: $λ_{H_{α}} = 656.3 nm$ ; Balmer limit $= 364.6 nm$ .

Solved Example 7

Ground-state hydrogen atoms absorb

12.09 eV

photons. Find the wavelengths that are then emitted.

Solution:

The atoms go to $n = 3$ (Example 1). Possible lines: $\frac{3 \times 2}{2} = 3$ .

$3 \to 1$ : $Δ E = 12.09 eV$ , $λ = \frac{1240}{12.09} = 102.6 nm$ (Lyman).
$2 \to 1$ : $Δ E = 10.2 eV$ , $λ = 121.6 nm$ (Lyman).
$3 \to 2$ : $Δ E = 1.89 eV$ , $λ = 656 nm$ (Balmer $H_{α}$ ).

Answer: $102.6 nm$ , $121.6 nm$ and $656 nm$ .

Solved Example 8

A photon of energy

15 eV

is absorbed by a hydrogen atom in its ground state. Find the kinetic energy of the ejected electron.

Solution:

$15 eV > 13.6 eV$ , so the atom is ionised; the extra energy becomes kinetic energy of the free electron (any amount is allowed above $E = 0$ ).

$K = 15 - 13.6$ .

Answer: $K = 1.4 eV$ .

Solved Example 9

The ratio of the longest wavelength of the Lyman series to the longest wavelength of the Balmer series of hydrogen is
(A)

5/27

(B)

27/5

(C)

4/9

(D)

1/4

Solution:

Answer: (A). $\frac{1}{λ _{L}} = R (1 - \frac{1}{4}) = \frac{3 R}{4}$ and $\frac{1}{λ _{B}} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ . So $\frac{λ _{L}}{λ _{B}} = \frac{5/36}{3/4} = \frac{5}{27}$ .

Solved Example 10

Which transition in

He^{+}

emits the same wavelength as the

H_{α}

line (

3 \to 2

) of hydrogen?
(A)

3 \to 2

(B)

4 \to 2

(C)

6 \to 4

(D)

5 \to 3

Solution:

Answer: (C). For $He^{+}$ , $\frac{1}{λ} = 4 R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) = R (\frac{1}{( n _{1} /2 ) ^{2}} - \frac{1}{( n _{2} /2 ) ^{2}})$ . We need $\frac{n _{1}}{2} = 2$ and $\frac{n _{2}}{2} = 3$ , so $n_{1} = 4$ , $n_{2} = 6$ .

Solved Example 11

A sample of hydrogen atoms emits

6

different wavelengths after excitation. Find the level they were excited to, and the largest and smallest photon energies emitted.

Solution:

$\frac{n ( n - 1 )}{2} = 6 \Rightarrow n = 4$ .

Largest: $4 \to 1$ : $13.6 (1 - \frac{1}{16}) = 12.75 eV$ ( $97.3 nm$ ). Smallest: $4 \to 3$ : $13.6 (\frac{1}{9} - \frac{1}{16}) = 0.66 eV$ ( $1875 nm$ ).

Answer: $n = 4$ ; $12.75 eV$ and $0.66 eV$ .

Solved Example 12

Find the first excitation potential and the ionisation potential of

He^{+}

Solution:

$E_{n} = - \frac{54.4}{n ^{2}} eV$ : $E_{1} = - 54.4 eV$ , $E_{2} = - 13.6 eV$ .

First excitation energy $= E_{2} - E_{1} = 40.8 eV$ ; ionisation energy $= 54.4 eV$ . Dividing by $e$ gives potentials in volts.

Answer: $40.8 V$ and $54.4 V$ .

Solved Example 13

The ratio of the longest to the shortest wavelength in the Balmer series of hydrogen is
(A)

9/5

(B)

5/9

(C)

4/3

(D)

27/5

Solution:

Answer: (A). Longest: $n = 3 \to 2$ , $\frac{1}{λ _{ma x}} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ . Shortest (series limit): $\frac{1}{λ _{min}} = \frac{R}{4}$ . So $\frac{λ _{ma x}}{λ _{min}} = \frac{36/5}{4} = \frac{9}{5}$ , that is $656.3 nm$ against $364.6 nm$ .

Solved Example 14

Light from the

n = 2 \to 1

transition of

He^{+}

falls on hydrogen atoms in the ground state. Find the kinetic energy of the electrons ejected from the hydrogen atoms.

Solution:

Photon energy: $Δ E = 13.6 \times 2^{2} (1 - \frac{1}{4}) = 40.8 eV$ ( $λ = \frac{1240}{40.8} = 30.4 nm$ , extreme ultraviolet).

$40.8 eV > 13.6 eV$ , so the photon ionises the hydrogen atom and the rest becomes kinetic energy: $K = 40.8 - 13.6$ .

Answer: $K = 27.2 eV$ .

Solved Example 15

Which transition in the hydrogen atom emits the photon of the highest frequency?
(A)

n = 2 \to 1

(B)

n = 5 \to 2

(C)

n = 4 \to 1

(D)

n = 5 \to 4

Solution:

Answer: (C). Photon energies: $2 \to 1$ : $10.2 eV$ ; $5 \to 2$ : $2.86 eV$ ; $4 \to 1$ : $12.75 eV$ ; $5 \to 4$ : $0.31 eV$ . The largest energy means the highest frequency (Figure 9). Option (D) has the lowest frequency and longest wavelength, $4051 nm$ .

Solved Example 16

The

H_{α}

and

H_{β}

lines of hydrogen fall on a caesium surface (work function

2.14 eV

). Which of them can eject photoelectrons, and with what maximum kinetic energy?

Solution:

$H_{α}$ ( $3 \to 2$ ): $h ν = 13.6 (\frac{1}{4} - \frac{1}{9}) = 1.89 eV < 2.14 eV$ , so no electrons are emitted, however bright the light.

$H_{β}$ ( $4 \to 2$ ): $h ν = 13.6 (\frac{1}{4} - \frac{1}{16}) = 2.55 eV > 2.14 eV$ , so $K_{ma x} = 2.55 - 2.14$ .

Answer: only $H_{β}$ ejects electrons, with $K_{ma x} = 0.41 eV$ (stopping potential $0.41 V$ ).

Practice Questions

Find the wavelength of the second line of the Lyman series.Answer: $102.6 nm$
Find the series limit of the Paschen series.Answer: $820 nm$
How much energy is needed to ionise a hydrogen atom in the $n = 3$ state?Answer: $1.51 eV$
How many spectral lines can a sample of hydrogen atoms excited to $n = 5$ emit?Answer: $10$
Find the minimum photon energy that can excite $He^{+}$ from its ground state.Answer: $40.8 eV$
Find the ratio of the shortest wavelength of the Lyman series to the shortest wavelength of the Balmer series.Answer: $1 : 4$
Electrons of energy $12.5 eV$ bombard ground-state hydrogen. Which wavelengths are emitted?Answer: atoms reach $n = 3$ : $102.6$ , $121.6$ and $656 nm$

Common Mistakes to Avoid

Watch out

Thinking a photon of energy between two level differences is partly absorbed. A photon is absorbed whole or not at all; an $11 eV$ photon passes through ground-state hydrogen.
Applying the photon rule to electrons. A colliding electron can give part of its energy, so any $K \geq 10.2 eV$ can excite hydrogen to $n = 2$ .
Mixing up first line and series limit. The first line ( $n_{2} = n_{1} + 1$ ) has the longest wavelength; the limit ( $n_{2} = \infty$ ) the shortest.
Using $n (n - 1) /2$ for a single atom. One atom gives at most $n - 1$ photons; the formula counts different lines from many atoms.
Counting absorption lines with $n (n - 1) /2$ . Absorption from the ground state gives only $n - 1$ lines, all in the Lyman series.
Forgetting $Z^{2}$ for $He^{+}$ and $Li^{2 +}$ . Energies are multiplied, and wavelengths divided, by $Z^{2}$ .
Writing Balmer as entirely visible. Only four lines are visible; the series limit, $364.6 nm$ , is ultraviolet.
Assuming a neutron can transfer all its energy to a hydrogen atom. Momentum conservation limits the loss to $\frac{K}{2}$ for equal masses.

Frequently Asked Questions

Why does hydrogen give a line spectrum instead of a continuous spectrum?

The electron in hydrogen can have only certain energies, $- \frac{13.6}{n ^{2}} eV$ . Light is emitted only when it jumps between these levels, and each jump gives a photon of one definite energy and wavelength. So only certain sharp wavelengths appear.

What are the spectral series of hydrogen?

Lines ending on the same lower level form a series: Lyman ends on n equal to 1 and lies in the ultraviolet, Balmer ends on 2 and is mostly visible, and Paschen, Brackett and Pfund end on 3, 4 and 5 and lie in the infrared.

What is the Rydberg formula?

It gives the wavelength of any hydrogen line: $\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , where $R = 1.097 \times 1 0^{7} m^{- 1}$ and $n_{2} > n_{1}$ . For hydrogen-like ions multiply the right side by $Z^{2}$ .

What is meant by the series limit?

It is the shortest wavelength in a series, from a jump that starts at $n = \infty$ . For the Lyman series it is $91.2 nm$ and for the Balmer series $364.6 nm$ . The lines crowd closer together as they approach the limit.

What is the difference between emission and absorption spectra?

An emission spectrum shows bright lines on a dark background from excited atoms falling to lower levels. An absorption spectrum shows dark lines on a continuous background where cooler atoms absorb exactly the same wavelengths to reach higher levels.

Why is only the Lyman series seen in the absorption spectrum of hydrogen at room temperature?

At room temperature almost every hydrogen atom is in the ground state, $n = 1$ . Absorption must start from the level the atom is in, so every absorbed line is a jump upward from $n = 1$ , which belongs to the Lyman series.

Which hydrogen spectrum questions come in NEET?

NEET often asks which series lies in the visible or ultraviolet region, the ratio of wavelengths of two lines, the number of spectral lines from a given level, and the energy of a photon for a given transition. Remember that Lyman is ultraviolet, Balmer is mostly visible with its limit in the near ultraviolet, and Paschen is infrared.

How are hydrogen spectra tested in JEE Main and Advanced?

JEE Main tests the Rydberg formula, series limits, number of lines, excitation and ionisation energies and matching lines of $He^{+}$ . JEE Advanced adds recoil of the atom, reduced-mass isotope shifts and collisions where both momentum conservation and allowed energy levels decide the outcome.

Previous year questions on The Line Spectra of the Hydrogen Atom

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

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