Bohr's Atomic Model

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Niels Bohr (1913) fixed the fatal defects of Rutherford's model by borrowing Planck's quantum idea. He proposed that electrons revolve only in fixed, quantised orbits (angular momentum = $nh /2 π$ ), and they emit or absorb energy only when jumping between orbits. This single postulate explains why hydrogen shows a discrete line spectrum instead of a continuous one, why atoms are stable, and gives us the famous formula $E_{n} = - 13.6 Z^{2} / n^{2}$ eV. Understanding waves, EM spectrum, and Planck's theory is essential background.

Key Formulas - Quick Reference

Wave: $c = ν λ$ ; wave number $\overset{ν}{ˉ} = 1/ λ = ν / c$
Planck: $E = h ν = h c / λ$ ; $h = 6.626 \times 1 0^{- 34}$ J s
Bohr quantisation: $m v r = nh /2 π$
Radius of nth orbit: $r_{n} = 0.529 \times n^{2} / Z$ Å
Velocity in nth orbit: $v_{n} = 2.18 \times 1 0^{6} \times Z / n$ m/s
Energy in nth orbit: $E_{n} = - 13.6 \times Z^{2} / n^{2}$ eV/atom $= - 21.8 \times 1 0^{- 19} \times Z^{2} / n^{2}$ J/atom
Rydberg formula: $\overset{ν}{ˉ} = 1/ λ = R Z^{2} (1/ n_{1}^{2} - 1/ n_{2}^{2})$ , $R = 109678$ cm $^{- 1}$
Spectral lines from level $n$ to ground: $= n (n - 1) /2$

1. Characteristics of a Wave

A wave is a periodic disturbance travelling through space (or a medium). Every wave has five key characteristics:

Figure 1: Characteristics of a wave - wavelength (

λ

), amplitude (

a

), crest and trough.

Property	Symbol	Definition	Units
Wavelength	$λ$	Distance between two consecutive crests/troughs	m, nm, Å (1 Å = $1 0^{- 10}$ m)
Frequency	$ν$ (nu)	Number of waves passing per second	Hz (s $^{- 1}$ ) = cps
Velocity	$c$	Distance travelled per second; $c = ν λ$	m/s ( $3 \times 1 0^{8}$ for EM waves)
Wave number	$\overset{ν}{ˉ}$	Waves per unit length; $\overset{ν}{ˉ} = 1/ λ$	cm $^{- 1}$ or m $^{- 1}$
Amplitude	$a$	Maximum displacement from equilibrium	controls intensity/brightness

Electromagnetic radiation and the EM spectrum

Visible light, X-rays, $γ$ -rays, radio waves etc. are all electromagnetic radiations - oscillating electric + magnetic fields travelling at $c = 3 \times 1 0^{8}$ m/s in vacuum. They do not need a medium.

Figure 2: Electromagnetic spectrum - wavelength (

λ

) increases from left (

γ

-rays) to right (radio waves).

2. Atomic Spectrum

When atoms are excited (heat, electric discharge), electrons jump to higher energy levels, and when they fall back they emit light of specific wavelengths - producing an emission spectrum (bright lines on dark background). Conversely, atoms absorb specific wavelengths, giving an absorption spectrum (dark lines on continuous background).

Line spectrum: Discrete, characteristic wavelengths - a "fingerprint" of the atom. Each element has a unique line spectrum.

3. Planck's Quantum Theory (1900)

Max Planck proposed that energy is not continuous but quantised - emitted or absorbed only in discrete packets called quanta (or photons for light).

Energy of one photon:

E = h ν = \frac{h c}{λ}

where

h = 6.626 \times 1 0^{- 34}

J s is Planck's constant.

For $n$ photons of frequency $ν$ : total energy $= nh ν$ .

Solved Example 1

The wave number of a radiation is 400 cm

^{- 1}

. Find its (a) wavelength, (b) frequency, (c) energy per photon.

Solution:

(a) $λ = 1/ \overset{ν}{ˉ} = 1/400$ cm $= 2.5 \times 1 0^{- 3}$ cm $= 2.5 \times 1 0^{- 5}$ m
(b) $ν = c / λ = (3 \times 1 0^{8}) / (2.5 \times 1 0^{- 5}) = 1.2 \times 1 0^{13}$ Hz
(c) $E = h ν = (6.626 \times 1 0^{- 34}) (1.2 \times 1 0^{13}) = 7.95 \times 1 0^{- 21}$ J

Solved Example 2

Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt.

Solution:

Energy per electron $= e V = (1.6 \times 1 0^{- 19}) (1) = 1.6 \times 1 0^{- 19}$ J.
Per mole: $E = (1.6 \times 1 0^{- 19}) (6.022 \times 1 0^{23}) = 96500$ J/mol $= 96.5 kJ/mol$ .

Solved Example 3

A near-UV photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red at 760 nm. What is the wavelength of the second photon?

Solution:

By energy conservation: $\frac{h c}{300} = \frac{h c}{760} + \frac{h c}{λ _{2}}$

$\frac{1}{λ _{2}} = \frac{1}{300} - \frac{1}{760} = \frac{760 - 300}{300 \times 760} = \frac{460}{228000}$

$λ_{2} = \frac{228000}{460} \approx 495.6 nm$ (green-blue).

4. Bohr's Atomic Model - The Postulates

Figure 3: Bohr's shell model - electrons occupy stationary orbits (K, L, M, N) at fixed radii

r_{n}

and energies

E_{n}

Stationary states: Electrons revolve around the nucleus in fixed circular orbits (called stationary states or shells) without radiating energy.
Quantised angular momentum: Only orbits for which $m v r = \frac{nh}{2 π}$ (where $n = 1, 2, 3, ...$ ) are allowed. This is the quantisation condition.
Energy jumps: Energy is absorbed or emitted only when an electron jumps from one stationary state to another. The photon energy equals the energy gap: $Δ E = E_{2} - E_{1} = h ν$ .
Force balance: The centripetal force is supplied by electrostatic attraction between the electron and nucleus.

Derivation - Radius of the nth orbit

From force balance: $\frac{m v ^{2}}{r} = \frac{K Z e ^{2}}{r ^{2}}$ ... (1)
From quantisation: $m v r = \frac{nh}{2 π}$ , so $v = \frac{nh}{2 π m r}$ ... (2)
Substituting (2) into (1) and solving for $r$ :

r_{n} = \frac{n ^{2} h ^{2}}{4 π ^{2} m K Z e ^{2}} = 0.529 \times \frac{n ^{2}}{Z}

where $K = 1/ (4 π ε_{0})$ . For hydrogen ( $Z = 1$ , $n = 1$ ): $r_{0} = 0.529$ Å (Bohr radius).

Velocity in the nth orbit

Substituting $r$ back into (2):

v_{n} = \frac{2 π K Z e ^{2}}{nh} = 2.18 \times 1 0^{6} \times \frac{Z}{n}

m/s

Energy in the nth orbit

Total energy $E = K E + P E = \frac{1}{2} m v^{2} - \frac{K Z e ^{2}}{r}$ . Using (1), $\frac{1}{2} m v^{2} = \frac{K Z e ^{2}}{2 r}$ , so:

$E = \frac{K Z e ^{2}}{2 r} - \frac{K Z e ^{2}}{r} = - \frac{K Z e ^{2}}{2 r}$

Substituting for $r$ :

E_{n} = - \frac{2 π ^{2} m K ^{2} Z ^{2} e ^{4}}{n ^{2} h ^{2}} = - 13.6 \times \frac{Z ^{2}}{n ^{2}}

eV/atom

= - 21.8 \times 1 0^{- 19} \times \frac{Z ^{2}}{n ^{2}}

J/atom

= - 313.6 \times \frac{Z ^{2}}{n ^{2}}

kcal/mol

Energy is negative because a bound electron has less energy than a free one (defined as zero). As

n \to \infty

E \to 0

(ionisation). Larger

n

= less negative = higher energy level.

Solved Example 4

Calculate the velocity of an electron in Bohr's first orbit of hydrogen (given

r = 0.53 \times 1 0^{- 10}

m).

Solution:

$v = \frac{nh}{2 π m r} = \frac{( 1 ) ( 6.62 \times 1 0 ^{- 34} )}{2 π ( 9.1 \times 1 0 ^{- 31} ) ( 0.53 \times 1 0 ^{- 10} )} = 2.18 \times 1 0^{6} m/s$

Solved Example 5

The velocity of

e^{-}

in the first Bohr orbit is

2.17 \times 1 0^{6}

m/s. Calculate the velocity in the 3rd orbit of He

^{2 +}

(

Z = 2

Solution:

$v_{n} = v_{0} \times Z / n = 2.17 \times 1 0^{6} \times (2/3) = 1.45 \times 1 0^{6} m/s$ .

5. Hydrogen Spectrum - Spectral Series

When excited electrons in hydrogen fall from higher to lower orbits, they emit photons of specific wavelengths, forming spectral series. All wavelengths follow the Rydberg formula:

\frac{1}{λ} = \overset{ν}{ˉ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

where

R = 109678

^{- 1}

(Rydberg constant),

n_{2} > n_{1}

Figure 4: Hydrogen atom energy levels and spectral series - Lyman (UV), Balmer (visible), Paschen/Brackett/Pfund (IR).

Series	$n_{1}$ (final)	$n_{2}$ (initial)	Region	Discovered by
Lyman	1	2, 3, 4, ...	Ultraviolet	Lyman (1906)
Balmer	2	3, 4, 5, ...	Visible	Balmer (1885)
Paschen	3	4, 5, 6, ...	Infrared	Paschen (1908)
Brackett	4	5, 6, 7, ...	Infrared	Brackett (1922)
Pfund	5	6, 7, 8, ...	Infrared	Pfund (1924)

Figure 5: Balmer series - visible emission lines of hydrogen. H

α

= 656 nm (red), H

β

= 486 nm, H

γ

= 434 nm, H

δ

= 410 nm.

Important: All visible-region lines belong to the Balmer series, but not all Balmer lines are visible (higher transitions fall in UV).

Number of spectral lines emitted when $e^{-}$ jumps from level $n_{2}$ to $n_{1}$ :

No. of lines = \frac{( n _{2} - n _{1} ) ( n _{2} - n _{1} + 1 )}{2}

For jumps to ground state (

n_{1} = 1

= \frac{n ( n - 1 )}{2}

Solved Example 6

Find the longest wavelength absorption line for hydrogen gas in the ground state.

Solution:

Absorption from ground state ( $n_{1} = 1$ ). Longest $λ$ = smallest $Δ E$ = smallest jump ( $n = 1 \to n = 2$ ).

$\frac{1}{λ} = 109678 \times 1 \times (\frac{1}{1} - \frac{1}{4}) = 109678 \times \frac{3}{4}$

$λ = \frac{4}{3 \times 109678}$ cm $\approx 1.215 \times 1 0^{- 5}$ cm $= 1215 \overset{˚}{A}$ (Lyman- $α$ , UV).

Solved Example 7

Find the wavelength of a spectral line when an electron in the H-atom jumps from the 4th level to the 2nd level.

Solution:

$\frac{1}{λ} = 109678 (\frac{1}{4} - \frac{1}{16}) = 109678 \times \frac{3}{16}$ cm $^{- 1}$

$λ = \frac{16}{3 \times 109678}$ cm $\approx 4863 \overset{˚}{A}$ (H $β$ line, Balmer).

Solved Example 8

The radii of two of the first four Bohr orbits of hydrogen are in the ratio 1 : 4. What is the ratio of their energies?

Solution:

$r_{n} \propto n^{2}$ , so ratio $1 : 4$ implies $n_{1} : n_{2} = 1 : 2$ .
Energy $E \propto - 1/ n^{2}$ , so ratio $E_{1} : E_{2} = (1/1) : (1/4) = 4 : 1$ in magnitude.

Solved Example 9

An H-atom in ground state absorbs 1.50 times as much energy as the minimum required for its escape. Find the wavelength of the emitted electron.

Solution:

Ionisation energy of H = 13.6 eV. Total absorbed = $1.5 \times 13.6 = 20.4$ eV.
KE of ejected electron $= 20.4 - 13.6 = 6.8$ eV $= 6.8 \times 1.6 \times 1 0^{- 19}$ J $= 1.088 \times 1 0^{- 18}$ J.
Momentum $p = 2 m E = 2 (9.11 \times 1 0^{- 31}) (1.088 \times 1 0^{- 18}) = 1.41 \times 1 0^{- 24}$ kg m/s.
$λ = h / p = (6.626 \times 1 0^{- 34}) / (1.41 \times 1 0^{- 24}) = 4.7 \overset{˚}{A}$ .

Solved Example 10

A series of lines in the H-spectrum lies at 656.46, 482.7, 434.17, 410.29 nm. What is the wavelength of the next line?

Solution:

These are Balmer lines ( $n_{1} = 2$ ): H $α$ (3 $\to$ 2), H $β$ (4 $\to$ 2), H $γ$ (5 $\to$ 2), H $δ$ (6 $\to$ 2). Next: H $ϵ$ (7 $\to$ 2).

$\frac{1}{λ} = 109678 (\frac{1}{4} - \frac{1}{49}) = 109678 \times \frac{45}{196}$

$λ \approx 397 nm$ .

Solved Example 11

Which hydrogen-like species has wavelength difference between first line of Balmer and first line of Lyman equal to

59.3 \times 1 0^{- 9}

Solution:

First Balmer (3 $\to$ 2): $\frac{1}{λ _{B}} = R Z^{2} (1/4 - 1/9) = 5 R Z^{2} /36$
First Lyman (2 $\to$ 1): $\frac{1}{λ _{L}} = R Z^{2} (1 - 1/4) = 3 R Z^{2} /4$
$λ_{B} - λ_{L} = \frac{36}{5 R Z ^{2}} - \frac{4}{3 R Z ^{2}} = \frac{1}{R Z ^{2}} (\frac{36}{5} - \frac{4}{3}) = \frac{88}{15 R Z ^{2}}$

Setting $= 59.3 \times 1 0^{- 9}$ m $= 5.93 \times 1 0^{- 6}$ cm with $R = 1.097 \times 1 0^{5}$ cm $^{- 1}$ :
$Z^{2} = \frac{88}{15 \times 1.097 \times 1 0 ^{5} \times 5.93 \times 1 0 ^{- 6}} \approx 9$ , so $Z = 3$ $\Rightarrow$ Li $^{2 +}$ .

6. Merits and Limitations of Bohr's Model

Merits

Explains the stability of atoms
Correctly predicts the hydrogen line spectrum
Gives correct radii, velocities, and energies for hydrogen-like species (He $^{+}$ , Li $^{2 +}$ )
Explains the concept of stationary states and quantised orbits

Limitations

Fails for multi-electron atoms (only works for H and H-like ions)
Cannot explain the Zeeman effect (splitting of spectral lines in a magnetic field) or the Stark effect (splitting in an electric field)
Assumes electrons are only particles, not waves - contradicts de Broglie's wave-particle duality
Violates Heisenberg's uncertainty principle by fixing both position (orbit radius) and momentum
Cannot explain fine structure (splitting of lines) or intensities of spectral lines

Bohr's model was a brilliant halfway step. It kept the classical picture of orbits but bolted on quantum jumps. Full quantum mechanics (concepts 3 and 4) replaces orbits with orbitals and probabilities.

Common Mistakes to Avoid

Watch out

Sign of energy. $E_{n}$ is negative for bound electrons - a common sign error. As $n$ increases, $E$ becomes less negative (higher, closer to zero).
Rydberg formula needs $n_{2} > n_{1}$ . Always put the larger quantum number as $n_{2}$ , else you'll get negative wave number.
Bohr works only for H-like (single-electron) species. For He, Li, Be with more electrons, Bohr's numbers are wrong.
$Z^{2}$ in energy but $Z$ in velocity. Do not mix - $E_{n} \propto Z^{2} / n^{2}$ , $v_{n} \propto Z / n$ , $r_{n} \propto n^{2} / Z$ .
Absorption vs emission. Absorption = jump up (energy in). Emission = jump down (energy out). Signs and directions matter.
Wavelength vs wave number. $λ$ is a length; $\overset{ν}{ˉ} = 1/ λ$ is inverse length. Don't confuse cm and cm $^{- 1}$ .

Frequently Asked Questions

What are Bohr's postulates in simple terms?

Bohr proposed four ideas: (1) electrons revolve in fixed stationary orbits without radiating energy; (2) only orbits where angular momentum is a multiple of $h /2 π$ are allowed ( $m v r = nh /2 π$ ); (3) energy is emitted or absorbed only when electrons jump between orbits, with $Δ E = h ν$ ; (4) the electrostatic attraction supplies the centripetal force. This fixed the stability problem of Rutherford's model.

What is the value of the Bohr radius and why is it important?

The Bohr radius is $r_{0} = 0.529$ Å $= 5.29 \times 1 0^{- 11}$ m. It's the radius of the first (n=1, ground state) orbit of the hydrogen atom in Bohr's model. It sets the atomic scale of length and is the natural unit for measuring atomic sizes. For any hydrogen-like ion, $r_{n} = 0.529 n^{2} / Z$ Å.

Why is the energy of an electron in an atom negative?

The reference (zero) is a free electron infinitely far from the nucleus. A bound electron has lower energy than a free one, so its energy is negative. The more tightly bound (smaller n), the more negative. As n increases, $E$ becomes less negative and approaches zero at $n = \infty$ (ionisation).

How many spectral lines are emitted when an electron falls from the nth level to the ground state?

The number of spectral lines is $n (n - 1) /2$ . For example, from n=4 to n=1, we get $4 \times 3/2 = 6$ lines. In general, for a jump from $n_{2}$ to $n_{1}$ , the number of lines is $(n_{2} - n_{1}) (n_{2} - n_{1} + 1) /2$ .

What is the Rydberg formula?

The Rydberg formula gives the wavelength of any line in the hydrogen spectrum: $\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ where $R = 109678$ cm $^{- 1}$ is the Rydberg constant, $Z$ is atomic number, and $n_{1} < n_{2}$ . It reproduces every observed line in the H spectrum with remarkable accuracy.

Which spectral series of hydrogen falls in the visible region?

The Balmer series ( $n_{1} = 2$ ) lies in the visible region. Its four main lines are H $α$ (656 nm, red), H $β$ (486 nm, blue-green), H $γ$ (434 nm, blue-violet), and H $δ$ (410 nm, violet). Note: all visible lines are Balmer, but not all Balmer lines are visible - higher members fall in UV.

What are the Zeeman and Stark effects?

Both are splittings of atomic spectral lines caused by external fields. The Zeeman effect is the splitting in a magnetic field. The Stark effect is the splitting in an electric field. Bohr's model cannot explain either, because it does not account for the orientation of orbits in space - one of its important failures.

Why does Bohr's model fail for multi-electron atoms?

Bohr assumes only the electron-nucleus attraction, but in multi-electron atoms there are also electron-electron repulsions and screening effects. Also, Bohr treats electrons as classical particles in circular orbits, ignoring their wave nature and the uncertainty principle. Multi-electron atoms need the full quantum mechanical (Schrödinger) treatment.

Previous year questions on Bohr's Atomic Model

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