Bohr Model of the Hydrogen Atom

Q: How does de Broglie's hypothesis explain Bohr's quantisation rule?

If the electron is a wave of wavelength h/mv, a stable orbit must hold a whole number of wavelengths, 2πr = nλ. Substituting the wavelength gives mvr = nh/(2π), exactly Bohr's second postulate.

PhysicsAtomsFor NEET aspirants

The Bohr model of the hydrogen atom explains why atoms are stable and why hydrogen emits sharp spectral lines. Starting from Rutherford's nuclear atom, Bohr let the electron move only in orbits whose angular momentum is a whole multiple of $\frac{h}{2 π}$ , with light emitted only in a jump between orbits. This gives $r_{n} = 0.529 \frac{n ^{2}}{Z} \overset{˚}{A}$ , $v_{n} = 2.18 \times 1 0^{6} \frac{Z}{n} m s^{- 1}$ and $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$ . The Bohr model of the hydrogen atom is tested every year in JEE Main and NEET.

On this page1Early models2α-scattering3Why Rutherford fails4Bohr's postulates5Radius, speed, energy6Period, current, magnetic moment7Hydrogen-like ions8de Broglie view9Nuclear motion10Limitations

Key Formulas - Quick Reference

Distance of closest approach (head-on): $r_{0} = \frac{k q _{1} q _{2}}{K} = \frac{k ( 2 e ) ( Z e )}{K}$ ; impact parameter $b = \frac{k Z e ^{2} cot ( θ /2 )}{K}$ for an $α$ -particle
★ Must learnBohr's postulates: $\frac{m v ^{2}}{r} = \frac{k Z e ^{2}}{r ^{2}}$ , $m v r = \frac{nh}{2 π}$ , $h ν = E_{n_{2}} - E_{n_{1}}$
★ Must learnRadius: $r_{n} = \frac{ε _{0} h ^{2}}{π m e ^{2}} \frac{n ^{2}}{Z} = 0.529 \frac{n ^{2}}{Z} \overset{˚}{A}$
Speed: $v_{n} = \frac{e ^{2}}{2 ε _{0} h} \frac{Z}{n} = 2.18 \times 1 0^{6} \frac{Z}{n} m s^{- 1}$ $(v_{1} = \frac{c}{137})$
★ Must learnEnergy: $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2}} \frac{Z ^{2}}{n ^{2}} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$
★ Must learnEnergy relations on any orbit: $K = - E$ , $U = 2 E$ , $K = - \frac{U}{2}$
Period and current: $T_{n} = 1.52 \times 1 0^{- 16} \frac{n ^{3}}{Z ^{2}} s$ , $I_{n} = \frac{e}{T _{n}} \propto \frac{Z ^{2}}{n ^{3}}$
Magnetic moment: $M_{n} = \frac{e}{2 m} L = n \frac{e h}{4 π m} = n μ_{B}$ , $μ_{B} = 9.27 \times 1 0^{- 24} A m^{2}$ ; field at nucleus $B_{n} \propto \frac{Z ^{3}}{n ^{5}}$
Finite nuclear mass: replace $m$ by $μ = \frac{m M}{m + M}$ ; $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} \frac{μ}{m} eV$

1. Early Atomic Models

By 1900 it was known that atoms contain negatively charged electrons (J. J. Thomson, 1897) and are electrically neutral, so they must also contain positive charge. The question was how the positive charge and the mass are arranged.

Dalton (1808): atoms are tiny indivisible spheres; atoms of one element are identical. The discovery of the electron showed that atoms can be divided.
Thomson (1898), plum pudding model: the atom is a sphere of radius about $1 0^{- 10} m$ in which positive charge is spread uniformly, with electrons embedded like seeds in a watermelon. The atom as a whole is neutral.

Figure 1: Two early atomic models. Thomson spread the positive charge over the whole atom. Rutherford put all of it, and almost all the mass, in a nucleus about

1 0^{5}

times smaller than the atom.

Thomson's model could not explain the line spectrum of hydrogen, and it failed completely when Rutherford's scattering experiment showed that $α$ -particles are sometimes turned right back by a gold foil. A spread-out positive charge can never exert such a large force.

2. Rutherford's α-Particle Scattering Experiment

In 1911 Geiger and Marsden, working with Rutherford, fired a narrow beam of fast $α$ -particles (helium nuclei, charge $+ 2 e$ ) at a very thin gold foil and counted how many were scattered through each angle $θ$ .

Figure 2: The Geiger-Marsden experiment (1911). A narrow beam of

5.5 MeV

α

-particles hits a thin gold foil. The ZnS screen and microscope rotate round the foil and count the flashes (scintillations) at each scattering angle

θ

in turn.

2.1 Observations and conclusions

Observation	Conclusion
Most $α$ -particles go straight through (only about $0.14%$ are scattered by more than $1^{\circ}$ ).	The atom is mostly empty space.
A few are deflected through large angles; about 1 in 8000 turns back through more than $9 0^{\circ}$ .	The positive charge and almost all the mass sit in a tiny, massive core: the nucleus.
The number scattered falls steeply with angle, as $N (θ) \propto \frac{1}{sin ^{4} ( θ /2 )}$ .	The force is Coulomb repulsion from a point-like nucleus, down to distances of about $1 0^{- 14} m$ .

Figure 3: Rutherford's scattering law,

N (θ) \propto \frac{1}{sin ^{4} ( θ /2 )}

, plotted on a log scale. Going from

θ = 2 0^{\circ}

13 5^{\circ}

the count falls by a factor of about

801

. Most

α

-particles are deflected by less than

1^{\circ}

(off the left edge), and about 1 in 8000 turns back through more than

9 0^{\circ}

2.2 Impact parameter and scattering angle

The impact parameter $b$ is the perpendicular distance of the initial velocity of the $α$ -particle from the centre of the nucleus. A small $b$ means the particle passes close to the nucleus, feels a large force and is scattered through a large angle. Under the Coulomb force the path is a hyperbola with the nucleus at one focus.

Figure 4: Paths computed from Coulomb's law (hyperbolas) for

b = 0.25 d

0.5 d

d

2 d

and

4 d

, where

d

is the head-on distance of closest approach. Each is labelled with its angle from

tan \frac{θ}{2} = \frac{d}{2 b}

: halving

b

roughly doubles a small

θ

, and only

b < 0.5 d

sends the particle backwards (

θ > 9 0^{\circ}

For an $α$ -particle of kinetic energy $K$ scattered by a nucleus of charge $Z e$ :

b = \frac{Z e ^{2} cot ( θ /2 )}{4 π ε _{0} K} = \frac{r _{0}}{2} cot \frac{θ}{2}

Head-on collision: $b = 0$ and $θ = 18 0^{\circ}$ , the particle retraces its path. Large $b$ gives $θ \to 0$ .

2.3 Distance of closest approach

In a head-on approach the $α$ -particle slows down as it climbs the Coulomb "hill", stops momentarily at a distance $r_{0}$ and returns. The nucleus is much heavier and is treated as fixed. Energy conservation between infinity and the turning point:

K + 0 = 0 + \frac{k ( z _{1} e ) ( z _{2} e )}{r _{0}} \Rightarrow r_{0} = \frac{k z _{1} z _{2} e ^{2}}{K}

Figure 5: Head-on approach of a

10 MeV

α

-particle to a nucleus with

Z = 50

(Solved Example 1). The total energy stays at

10 MeV

while KE turns into PE; the particle stops where

\frac{k q _{1} q _{2}}{r} = K

, at

r_{0} = 14.4 fm = 1.44 \times 1 0^{- 14} m

$r_{0}$ is an upper limit on the size of the nucleus: the Coulomb law still held at this distance. Rutherford's data put the nuclear radius at about $1 0^{- 15}$ to $1 0^{- 14} m$ , some $1 0^{4}$ to $1 0^{5}$ times smaller than the atom.

Exam Trick

Use $k e^{2} = 1.44 eV nm = 1.44 MeV fm$ . Then $r_{0} = \frac{1.44 z _{1} z _{2}}{K ( MeV )} fm$ with no powers of ten. A $5 MeV$ $α$ on gold: $r_{0} = \frac{1.44 \times 2 \times 79}{5} = 45.5 fm = 4.55 \times 1 0^{- 14} m$ .

Quick Recall: tap to check

What does the impact parameter measure?

The perpendicular distance between the nucleus and the line of the incoming velocity. Smaller

b

means larger deflection.

What happens to

r_{0}

if the

α

-particle's kinetic energy is doubled?

It halves, since

r_{0} \propto \frac{1}{K}

Why do so few

α

-particles bounce back?

The nucleus is tiny, so very few particles have a small enough impact parameter.

3. Rutherford's Nuclear Model and Its Failures

Rutherford proposed that the atom is a tiny, dense, positive nucleus with the electrons revolving around it in the surrounding empty space, held in orbit by the Coulomb attraction, like planets around the sun. It explained the scattering data perfectly but ran into two serious problems with classical physics.

Stability. An electron in a circular orbit is accelerating. By Maxwell's theory an accelerated charge radiates electromagnetic energy, so the electron should lose energy and spiral into the nucleus. The classical estimate for hydrogen is only about $1 0^{- 11} s$ . Real atoms are stable.
Line spectrum. As the electron spirals in, its frequency of revolution rises smoothly, so it should emit a continuous range of frequencies. Hydrogen actually emits only certain sharp wavelengths.

Figure 6: Failure of the Rutherford model. An accelerating electron radiates, loses energy and spirals in (radius drawn from the classical result

r^{3/2} \propto

angle left to turn). Atoms would be unstable and would emit a continuous spectrum; real atoms are stable and give line spectra.

Rutherford atom

Nucleus plus electrons in orbits, all governed by classical mechanics and electromagnetism. Any orbit allowed; orbiting electrons radiate. Predicts collapse and a continuous spectrum.

Bohr atom

Same nucleus, but only certain orbits are allowed and they do not radiate. Light is emitted only in a jump between orbits. Predicts stable atoms and sharp lines.

Key idea

Rutherford found the nucleus; classical physics could not keep the electron in orbit. Bohr's fix was to add quantum rules to the nuclear atom.

4. Bohr's Postulates

In 1913 Niels Bohr kept Rutherford's nucleus and Newton's laws for the orbit, but added quantum rules. The model applies to hydrogen and hydrogen-like ions (one electron only): $H$ , $He^{+}$ , $Li^{2 +}$ , $Be^{3 +}$ , with nuclear charge $+ Z e$ . Because it mixes classical and quantum ideas it is called a semi-classical model.

Figure 7: The Bohr picture. The Coulomb attraction supplies the centripetal force, the angular momentum is quantised in steps of

ℏ = \frac{h}{2 π}

, and light is emitted only in a jump between orbits.

Circular orbits under the Coulomb force. The electron revolves in a circle; the electrostatic attraction provides the centripetal force:
$\frac{m v ^{2}}{r} = \frac{k Z e ^{2}}{r ^{2}}, k = \frac{1}{4 π ε _{0}}$
Quantised angular momentum (stationary orbits). Only those orbits are allowed in which the angular momentum is an integral multiple of $ℏ = \frac{h}{2 π}$ . In these orbits the electron does not radiate:
$L = m v r = \frac{nh}{2 π}, n = 1, 2, 3, \dots$
Frequency condition. The atom emits (or absorbs) a single photon only when the electron jumps between two stationary orbits; the photon carries the energy difference:
$h ν = E_{n_{2}} - E_{n_{1}}$

$n$ is the principal quantum number. $ℏ = \frac{h}{2 π} = 1.055 \times 1 0^{- 34} J s$ . The first postulate is Newton's second law seen from the ground (inertial) frame; in the rotating frame the same equation reads "centrifugal force balances the Coulomb force".

5. Radius, Speed and Energy of the nth Orbit

5.1 Radius and speed

From postulate 2: $v = \frac{nh}{2 π m r}$ .
Put it in postulate 1, written as $m v^{2} r = k Z e^{2}$ : $m \frac{n ^{2} h ^{2}}{4 π ^{2} m ^{2} r ^{2}} r = k Z e^{2}$ .
Solve for $r$ , then substitute back to get $v$ :
$r_{n} = \frac{n ^{2} h ^{2}}{4 π ^{2} k Z e ^{2} m}, v_{n} = \frac{2 π k Z e ^{2}}{nh}$

r_{n} = \frac{ε _{0} h ^{2}}{π m e ^{2}} \cdot \frac{n ^{2}}{Z} = 0.529 \frac{n ^{2}}{Z} \overset{˚}{A}, v_{n} = \frac{e ^{2}}{2 ε _{0} h} \cdot \frac{Z}{n} = 2.18 \times 1 0^{6} \frac{Z}{n} m s^{- 1}

$a_{0} = 0.529 \overset{˚}{A}$ is the Bohr radius, the radius of the ground state of hydrogen. $\frac{v _{1}}{c} = \frac{1}{137}$ is the fine-structure constant, so the speed is small enough for non-relativistic mechanics.

Figure 8: Hydrogen orbits to scale (

r_{n} = 0.529 n^{2} \overset{˚}{A}

). The jump

n = 3 \to 2

emits the red

H_{α}

photon. Outer orbits are much larger but the electron moves more slowly there.

5.2 Energy of the electron

Kinetic energy, using postulate 1 ( $m v^{2} = \frac{k Z e ^{2}}{r}$ ): $K = \frac{1}{2} m v^{2} = \frac{k Z e ^{2}}{2 r}$ .
Potential energy of the electron-nucleus pair: $U = - \frac{k Z e ^{2}}{r}$ .
Total energy: $E = K + U = - \frac{k Z e ^{2}}{2 r}$ . Put in $r_{n}$ :

★ Must learn

E_{n} = - \frac{2 π ^{2} k ^{2} m e ^{4}}{h ^{2}} \cdot \frac{Z ^{2}}{n ^{2}} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2}} \cdot \frac{Z ^{2}}{n ^{2}} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV

At every orbit $K = - E$ and $U = 2 E$ . The negative total energy means the electron is bound; $E = 0$ corresponds to $n \to \infty$ , a free electron at rest.

Figure 9: Energies on a Bohr orbit of hydrogen against radius. At every radius

K = - E

and

U = 2 E

(inverse-square force). Only the marked radii

r = n^{2} a_{0}

(

n = 1, 2, 3

) are allowed, so

E

takes only the values

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

Writing $E_{n} = - \frac{R h c Z ^{2}}{n ^{2}}$ defines the Rydberg constant $R = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3} c} = 1.097 \times 1 0^{7} m^{- 1}$ . The energy $R h c = 13.6 eV = 2.18 \times 1 0^{- 18} J$ is called one rydberg. For hydrogen: $E_{1} = - 13.6$ , $E_{2} = - 3.40$ , $E_{3} = - 1.51$ , $E_{4} = - 0.85$ , $E_{5} = - 0.54$ , $E_{6} = - 0.38 eV$ .

Exam Trick

One energy gives all three. If the total energy is $E$ , then $K = - E$ and $U = 2 E$ . An electron with $E = - 3.4 eV$ has $K = + 3.4 eV$ and $U = - 6.8 eV$ . When an electron moves to a lower orbit, $K$ rises while $U$ and $E$ fall, and $U$ falls twice as much as $K$ rises. This holds for any inverse-square force, including satellites.

Quick Recall: tap to check

How does the radius of the orbit change from

n = 1

n = 3

in hydrogen?

r \propto n^{2}

: it becomes 9 times,

4.76 \overset{˚}{A}

The kinetic energy of an electron in an orbit is

1.51 eV

. Find its potential and total energy.

E = - 1.51 eV

U = - 3.02 eV

(this is

n = 3

of hydrogen).

Why is the energy of a bound electron negative?

Zero energy is chosen for the electron at rest far away; energy must be supplied to free it, so bound states lie below zero.

6. Period, Frequency, Current and Magnetic Moment

All other orbit quantities follow from $r_{n}$ and $v_{n}$ :

Quantity	Formula	Proportional to	Hydrogen, $n = 1$
Radius $r_{n}$	$0.529 \frac{n ^{2}}{Z} \overset{˚}{A}$	$\frac{n ^{2}}{Z}$	$0.529 \overset{˚}{A}$
Speed $v_{n}$	$2.18 \times 1 0^{6} \frac{Z}{n} m s^{- 1}$	$\frac{Z}{n}$	$2.18 \times 1 0^{6} m s^{- 1}$
Momentum $p_{n} = m v_{n}$	$\frac{2 π mk Z e ^{2}}{nh}$	$\frac{Z}{n}$	$1.99 \times 1 0^{- 24} kg m s^{- 1}$
Angular momentum $L_{n}$	$\frac{nh}{2 π}$	$n$	$1.055 \times 1 0^{- 34} J s$
Period $T_{n} = \frac{2 π r _{n}}{v _{n}}$	$\frac{n ^{3} h ^{3}}{4 π ^{2} k ^{2} Z ^{2} e ^{4} m}$	$\frac{n ^{3}}{Z ^{2}}$	$1.52 \times 1 0^{- 16} s$
Frequency $f_{n}$ , angular speed $ω_{n}$	$f_{n} = \frac{1}{T _{n}}$ , $ω_{n} = \frac{v _{n}}{r _{n}}$	$\frac{Z ^{2}}{n ^{3}}$	$6.56 \times 1 0^{15} Hz$ ; $4.12 \times 1 0^{16} rad s^{- 1}$
Current $I_{n} = e f_{n}$	$\frac{e v _{n}}{2 π r _{n}}$	$\frac{Z ^{2}}{n ^{3}}$	$1.05 mA$
Magnetic moment $M_{n} = I_{n} π r_{n}^{2}$	$\frac{e v _{n} r _{n}}{2} = n \frac{e h}{4 π m}$	$n$ (no $Z$ )	$9.27 \times 1 0^{- 24} A m^{2}$
Field at nucleus $B_{n} = \frac{μ _{0} I _{n}}{2 r _{n}}$	$\frac{μ _{0} e v _{n}}{4 π r _{n}^{2}}$	$\frac{Z ^{3}}{n ^{5}}$	$12.5 T$
Kinetic energy $K_{n}$ , $∣ E_{n} ∣$	$13.6 \frac{Z ^{2}}{n ^{2}} eV$	$\frac{Z ^{2}}{n ^{2}}$	$13.6 eV$
Centripetal acceleration $a_{n}$	$\frac{v _{n}^{2}}{r _{n}}$	$\frac{Z ^{3}}{n ^{4}}$	$9.0 \times 1 0^{22} m s^{- 2}$

Each quantity in the table is a power of $n$ (for fixed $Z$ ). Taking logarithms turns every power law into a straight line whose slope is the power, which is how graph questions test these results.

Figure 10: Each Bohr quantity is a power of

n

, so

lo g \frac{Q _{n}}{Q _{1}}

against

lo g n

is a straight line whose slope is the power:

3

for

T

2

for

r

1

for

L

and

M

- 1

for

v

and

- 2

for

∣ E ∣

. Graph questions ask for exactly these slopes.

Slope = power of $n$ . A graph of $lo g r_{n}$ against $lo g n$ has slope $2$ , of $lo g T_{n}$ slope $3$ , of $lo g v_{n}$ slope $- 1$ and of $lo g ∣ E_{n} ∣$ slope $- 2$ . For $Z$ at fixed $n$ use the powers of $Z$ instead: $r \propto Z^{- 1}$ , $v \propto Z$ , $E \propto Z^{2}$ , $T \propto Z^{- 2}$ .

The revolving electron is a small current loop. Because its charge is negative, the conventional current flows opposite to its velocity, so the magnetic moment points opposite to the angular momentum.

Figure 11: The orbiting electron is a tiny current loop. Because its charge is negative, the current

I

runs opposite to

v

, so the magnetic moment

M

(and the field

B

at the nucleus) points opposite to

L

, with

\frac{M}{L} = \frac{e}{2 m}

for every orbit.

★ Must learn

The ratio of the orbital magnetic moment to the angular momentum is the same for every orbit:

\frac{M}{L} = \frac{e}{2 m}

(the gyromagnetic ratio). So $M_{n} = n μ_{B}$ , where $μ_{B} = \frac{e h}{4 π m} = 9.27 \times 1 0^{- 24} A m^{2}$ is the Bohr magneton. The moment points opposite to $L$ because the electron is negative.

Key idea

Memorise three scalings:

r \propto \frac{n ^{2}}{Z}

v \propto \frac{Z}{n}

E \propto \frac{Z ^{2}}{n ^{2}}

. Every other quantity is built from these.

7. Energy Levels of Hydrogen-like Ions

For $He^{+}$ ( $Z = 2$ ) every energy is $4$ times, and for $Li^{2 +}$ ( $Z = 3$ ) $9$ times, the hydrogen value at the same $n$ . Radii shrink by $\frac{1}{Z}$ and speeds grow by $Z$ . The ground-state energy is $- 13.6 Z^{2} eV$ : $- 54.4 eV$ for $He^{+}$ and $- 122.4 eV$ for $Li^{2 +}$ .

Figure 12: Hydrogen-like ions to scale,

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

. Levels with the same

\frac{n}{Z}

have the same energy: H

(n = 1)

He^{+}

(n = 2)

and

Li^{2 +}

(n = 3)

all sit on the orange line at

- 13.6 eV

Exam Trick

Same $\frac{n}{Z}$ means same energy, same speed. $E \propto (\frac{Z}{n})^{2}$ and $v \propto \frac{Z}{n}$ , so level $n = 2$ of $He^{+}$ matches $n = 1$ of hydrogen, and $n = 4$ of $He^{+}$ matches $n = 2$ of hydrogen. The radius, however, is not the same: $r \propto \frac{n ^{2}}{Z}$ is $2 a_{0}$ for $He^{+}$ ( $n = 2$ ) but $a_{0}$ for $H$ ( $n = 1$ ).

The transitions between these levels, which produce the hydrogen spectrum, are treated in the next concept, The Line Spectra of the Hydrogen Atom.

8. de Broglie's Explanation of the Quantisation Rule

Bohr gave no reason why $L$ should be a multiple of $\frac{h}{2 π}$ . In 1924 de Broglie supplied one: the electron has a wavelength $λ = \frac{h}{m v}$ . A wave running round a circular orbit reinforces itself only if a whole number of wavelengths fits the circumference, forming a standing wave. Otherwise it cancels itself.

Figure 13: de Broglie's reason for Bohr's second postulate. A stationary orbit must hold a whole number of electron wavelengths:

2 π r = nλ = \frac{nh}{m v}

, which is exactly

m v r = \frac{nh}{2 π}

2 π r_{n} = nλ = n \frac{h}{m v _{n}} \Rightarrow m v_{n} r_{n} = \frac{nh}{2 π}

This is exactly Bohr's second postulate. The quantum number $n$ is the number of electron wavelengths in the orbit. In hydrogen $λ_{n} = \frac{2 π r _{n}}{n} = 2 π a_{0} n = 3.32 n \overset{˚}{A}$ .

Quick Recall: tap to check

How many de Broglie wavelengths fit in the third orbit?

Three.

Find the de Broglie wavelength of the electron in the ground state of hydrogen.

λ = 2 π a_{0} = 3.32 \overset{˚}{A}

Which postulate does the standing-wave idea explain?

The second one,

m v r = \frac{nh}{2 π}

9. Effect of the Motion of the Nucleus

The nucleus is not infinitely heavy. The electron and nucleus both revolve about their common centre of mass with the same angular speed.

Figure 14: Finite nuclear mass (not to scale). Both bodies circle the centre of mass with the same

ω

. The problem becomes one body of reduced mass

μ = \frac{M m}{M + m}

at distance

r

, so

m

is replaced by

μ

in every Bohr formula.

JEE Advanced

Reduced mass. With $M r_{1} = m r_{2}$ and $r = r_{1} + r_{2}$ , the force equation and the quantisation of the total angular momentum $I ω = \frac{nh}{2 π}$ (with $I = μ r^{2}$ ) are those of a single particle of mass

μ = \frac{m M}{m + M}

at distance $r$ . So in every Bohr formula $m \to μ$ :

r_{n} = 0.529 \frac{n ^{2}}{Z} \frac{m}{μ} \overset{˚}{A}, E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} \frac{μ}{m} eV

Hydrogen: $\frac{μ}{m} = \frac{1836}{1837} = 0.99946$ , so the correction is tiny but measurable. Deuterium lines are shifted slightly from hydrogen lines, which is how deuterium was discovered (1932).
Positronium ( $e^{-} e^{+}$ , $M = m$ ): $μ = \frac{m}{2}$ , so $E_{1} = - 6.8 eV$ and the separation is $2 a_{0} = 1.06 \overset{˚}{A}$ .
Muonic hydrogen (muon of mass $207 m$ around a proton): $μ \approx 186 m$ , so $E_{1} \approx - 2.53 keV$ and $r_{1} \approx 2.8 \times 1 0^{- 13} m$ .

JEE Advanced

Bohr quantisation with other forces. Keep $m v r = \frac{nh}{2 π}$ and change the force law. For a force towards the centre $F = k r$ (potential $\frac{1}{2} k r^{2}$ ): $m v^{2} = k r^{2}$ gives $v = r \frac{k}{m}$ , so $m r^{2} \frac{k}{m} = \frac{nh}{2 π}$ and

r_{n} \propto n, v_{n} \propto n, E_{n} = k r_{n}^{2} = \frac{nh}{2 π} \frac{k}{m} \propto n

For gravity (satellite) the algebra is the same as for the atom with $k Z e^{2} \to GM m$ , so $r \propto n^{2}$ and $E \propto - \frac{1}{n ^{2}}$ .

10. Successes and Limitations of the Bohr Model

Successes: explains the stability of the atom, gives the size of hydrogen ( $a_{0} = 0.529 \overset{˚}{A}$ ), the ionisation energy $13.6 eV$ , the Rydberg constant from fundamental constants and every line of the hydrogen spectrum, and works for all one-electron ions.
Limitations: it fails for atoms with two or more electrons (even helium), since it ignores electron-electron repulsion.
It cannot explain the relative intensities of spectral lines, their fine structure (each line is really a group of close lines), or the splitting of lines in a magnetic field (Zeeman effect) and an electric field (Stark effect).
It assumes definite circular orbits with an exactly known radius and speed, which violates the uncertainty principle. Quantum mechanics replaces orbits with orbitals (probability clouds), but keeps Bohr's energy formula for hydrogen.
It gives no reason for the quantisation rule itself; de Broglie's waves supplied one later.

Key idea

Bohr's energies

- 13.6 \frac{Z ^{2}}{n ^{2}} eV

are exactly right for one-electron atoms, but his picture of definite orbits is not. Use the model for numbers, not for the shape of the atom.

11. Solving Bohr-Model Problems and Revision Map

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

Figure 15: Solving a Bohr-model problem. Check that there is one electron, then decide: ratio questions never need constants; absolute values use the three numerical forms, the energy relations and, for positronium or muonic atoms, the reduced mass.

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

12. Solved Examples

Solved Example 1

α

-particle with kinetic energy

10 MeV

is heading towards a stationary point nucleus of atomic number

50

. Calculate the distance of closest approach.

Solution:

Given: $K = 10 MeV = 1.6 \times 1 0^{- 12} J$ , $z_{1} = 2$ , $z_{2} = 50$ .

At closest approach all the KE has become PE: $K = \frac{k ( 2 e ) ( 50 e )}{r _{0}}$ .

r_{0} = \frac{9 \times 1 0 ^{9} \times 100 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1.6 \times 1 0 ^{- 12}} = 1.44 \times 1 0^{- 14} m

Shortcut: $r_{0} = \frac{1.44 \times 2 \times 50}{10} fm = 14.4 fm$ .

Answer: $r_{0} = 1.44 \times 1 0^{- 14} m = 1.44 \times 1 0^{- 4} \overset{˚}{A}$ .

Solved Example 2

A beam of

α

-particles of speed

2.1 \times 1 0^{7} m s^{- 1}

is scattered by a gold foil (

Z = 79

). Find the distance of closest approach of an

α

-particle to a gold nucleus. Take the charge-to-mass ratio of the

α

-particle as

4.8 \times 1 0^{7} C kg^{- 1}

Solution:

Energy conservation: $\frac{1}{2} m v^{2} = \frac{k ( 2 e ) ( Z e )}{r _{0}}$ , so $r_{0} = \frac{2 k ( Z e ) ( 2 e / m )}{v ^{2}}$ , where $\frac{2 e}{m} = \frac{q}{m} = 4.8 \times 1 0^{7} C kg^{- 1}$ .

r_{0} = \frac{2 \times 9 \times 1 0 ^{9} \times 79 \times 1.6 \times 1 0 ^{- 19} \times 4.8 \times 1 0 ^{7}}{( 2.1 \times 1 0 ^{7} ) ^{2}}

Answer: $r_{0} \approx 2.5 \times 1 0^{- 14} m$ .

Solved Example 3

A proton moving with speed

7.45 \times 1 0^{5} m s^{- 1}

heads directly towards another proton that is free and initially at rest. Find the distance of closest approach. (Mass of proton

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

)

Solution:

The second proton is free, so it also moves. At closest approach both have the same velocity $v_{1}$ (no relative motion).

Momentum: $m v = 2 m v_{1} \Rightarrow v_{1} = \frac{v}{2}$ .

Energy: $\frac{1}{2} m v^{2} = \frac{k e ^{2}}{r _{0}} + 2 \times \frac{1}{2} m (\frac{v}{2})^{2} \Rightarrow \frac{k e ^{2}}{r _{0}} = \frac{1}{4} m v^{2}$ .

r_{0} = \frac{4 k e ^{2}}{m v ^{2}} = \frac{4 \times 9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1.67 \times 1 0 ^{- 27} \times ( 7.45 \times 1 0 ^{5} ) ^{2}}

Answer: $r_{0} \approx 1.0 \times 1 0^{- 12} m$ , twice the value for a fixed target, because only half the KE (the KE relative to the centre of mass) can be converted into PE.

Solved Example 4

What is the angular momentum of an electron in a Bohr hydrogen atom whose energy is

- 3.4 eV

Solution:

$E_{n} = - \frac{13.6}{n ^{2}} eV = - 3.4 eV \Rightarrow n^{2} = 4$ , $n = 2$ .

$L = \frac{nh}{2 π} = \frac{2 h}{2 π} = \frac{h}{π} = \frac{6.626 \times 1 0 ^{- 34}}{3.1416}$ .

Answer: $L = \frac{h}{π} \approx 2.11 \times 1 0^{- 34} J s$ .

Solved Example 5

In a hydrogen atom the period of revolution of the electron in orbit

n_{1}

is 8 times the period in orbit

n_{2}

. Find the possible values of

n_{1}

and

n_{2}

Solution:

$T_{n} = \frac{n ^{3} h ^{3}}{4 π ^{2} k ^{2} Z ^{2} e ^{4} m} \propto n^{3}$ for a given atom, so $\frac{T _{1}}{T _{2}} = (\frac{n _{1}}{n _{2}})^{3} = 8 \Rightarrow n_{1} = 2 n_{2}$ .

Answer: $(n_{1}, n_{2}) = (2, 1), (4, 2), (6, 3), \dots$

Solved Example 6

An electron in the ground state of hydrogen revolves anticlockwise in a circular orbit of radius

R

. (a) Obtain an expression for its orbital magnetic dipole moment. (b) The atom is placed in a uniform magnetic field

B

such that the normal to the orbit makes

3 0^{\circ}

with the field. Find the torque on the orbit.

Solution:

(a) For $n = 1$ : $m v R = \frac{h}{2 π} \Rightarrow v = \frac{h}{2 π m R}$ . Current $i = \frac{e}{T} = \frac{e v}{2 π R} = \frac{e h}{4 π ^{2} m R ^{2}}$ .

$M = iπ R^{2} = \frac{e h}{4 π ^{2} m R ^{2}} π R^{2} = \frac{e h}{4 π m}$ , one Bohr magneton, independent of $R$ .

(b) $τ = M \times B$ , so $τ = M B sin 3 0^{\circ} = \frac{e h}{4 π m} \cdot \frac{B}{2}$ .

Answer: (a) $M = \frac{e h}{4 π m} = 9.27 \times 1 0^{- 24} A m^{2}$ ; (b) $τ = \frac{e h B}{8 π m}$ .

Solved Example 7

α

-particles of kinetic energy

7.7 MeV

are scattered by gold nuclei (

Z = 79

). Find (a) the distance of closest approach and (b) the impact parameter for scattering through

6 0^{\circ}

Solution:

(a) $r_{0} = \frac{k ( 2 e ) ( 79 e )}{K} = \frac{1.44 \times 158}{7.7} fm = 29.5 fm$ .

(b) $b = \frac{r _{0}}{2} cot \frac{θ}{2} = \frac{29.5}{2} cot 3 0^{\circ} = 14.8 \times 1.732 fm$ .

Answer: (a) $r_{0} \approx 2.95 \times 1 0^{- 14} m$ ; (b) $b \approx 2.56 \times 1 0^{- 14} m$ .

Solved Example 8

For the

He^{+}

ion in its first excited state (

n = 2

), find the orbit radius, the electron's speed, and its total, kinetic and potential energies.

Solution:

$Z = 2$ , $n = 2$ , so $\frac{n ^{2}}{Z} = 2$ and $\frac{Z}{n} = 1$ .

$r = 0.529 \times 2 = 1.058 \overset{˚}{A}$ ; $v = 2.18 \times 1 0^{6} \times 1 m s^{- 1}$ ; $E = - 13.6 \times \frac{4}{4} = - 13.6 eV$ ; $K = - E$ ; $U = 2 E$ .

Answer: $r = 1.06 \overset{˚}{A}$ , $v = 2.18 \times 1 0^{6} m s^{- 1}$ , $E = - 13.6 eV$ , $K = 13.6 eV$ , $U = - 27.2 eV$ .

Solved Example 9

The ratio of the speed of the electron in the ground state of hydrogen to the speed of light is about
(A)

1/2

(B)

1/137

(C)

1/1836

(D)

1/237

Solution:

Answer: (B). $\frac{v _{1}}{c} = \frac{2.18 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} = \frac{1}{137}$ , the fine-structure constant $\frac{e ^{2}}{2 ε _{0} h c}$ .

Solved Example 10

When the electron of a hydrogen atom jumps from

n = 3

n = 1

:
(A) KE, PE and total energy all decrease
(B) KE increases, PE and total energy decrease
(C) KE decreases, PE increases
(D) all three increase

Solution:

Answer: (B). $E$ falls from $- 1.51$ to $- 13.6 eV$ (the difference leaves as a photon). $K = - E$ rises from $1.51$ to $13.6 eV$ ; $U = 2 E$ falls from $- 3.02$ to $- 27.2 eV$ .

Solved Example 11

Find the current due to the electron and the magnetic field it produces at the nucleus, for the ground state of hydrogen.

Solution:

$T_{1} = \frac{2 π r _{1}}{v _{1}} = \frac{2 π \times 0.529 \times 1 0 ^{- 10}}{2.18 \times 1 0 ^{6}} = 1.52 \times 1 0^{- 16} s$ .

$I = \frac{e}{T _{1}} = \frac{1.6 \times 1 0 ^{- 19}}{1.52 \times 1 0 ^{- 16}} = 1.05 \times 1 0^{- 3} A$ .

$B = \frac{μ _{0} I}{2 r _{1}} = \frac{4 π \times 1 0 ^{- 7} \times 1.05 \times 1 0 ^{- 3}}{2 \times 0.529 \times 1 0 ^{- 10}}$ .

Answer: $I \approx 1.05 mA$ , $B \approx 12.5 T$ , a very strong field.

Solved Example 12

Find the de Broglie wavelength of the electron in the

n = 2

orbit of hydrogen in two ways.

Solution:

Method 1: $2 π r_{2} = 2 λ \Rightarrow λ = π r_{2} = π \times 2.116 \overset{˚}{A} = 6.65 \overset{˚}{A}$ .

Method 2: $v_{2} = 1.09 \times 1 0^{6} m s^{- 1}$ , $λ = \frac{h}{m v _{2}} = \frac{6.626 \times 1 0 ^{- 34}}{9.11 \times 1 0 ^{- 31} \times 1.09 \times 1 0 ^{6}} = 6.67 \times 1 0^{- 10} m$ .

Answer: $λ \approx 6.65 \overset{˚}{A}$ (the small difference is rounding).

Solved Example 13

Positronium is a bound state of an electron and a positron. Find its ground-state energy and the electron-positron separation.

Solution:

$M = m$ , so $μ = \frac{m \cdot m}{2 m} = \frac{m}{2}$ .

$E_{1} = - 13.6 \times \frac{μ}{m} = - 13.6 \times \frac{1}{2}$ ; $r_{1} = 0.529 \times \frac{m}{μ} = 0.529 \times 2$ .

Answer: $E_{1} = - 6.8 eV$ , separation $= 1.06 \overset{˚}{A}$ .

Solved Example 14

The ratio of the radius of the second orbit of

He^{+}

to the radius of the third orbit of

Li^{2 +}

is
(A)

2 : 3

(B)

3 : 2

(C)

4 : 9

(D)

8 : 27

Solution:

Answer: (A). $r \propto \frac{n ^{2}}{Z}$ . For $He^{+}$ ( $n = 2$ , $Z = 2$ ): $\frac{4}{2} = 2$ . For $Li^{2 +}$ ( $n = 3$ , $Z = 3$ ): $\frac{9}{3} = 3$ . So the ratio is $2 : 3$ ( $1.06 \overset{˚}{A}$ and $1.59 \overset{˚}{A}$ ).

Solved Example 15

For a hydrogen atom, the graph of

ln (\frac{T _{n}}{T _{1}})

against

ln n

, where

T_{n}

is the period of revolution in the

n

th orbit, is a straight line of slope
(A)

1

(B)

2

(C)

3

(D)

- 2

Solution:

Answer: (C). $T_{n} = \frac{2 π r _{n}}{v _{n}} \propto \frac{n ^{2}}{1/ n} = n^{3}$ , so $ln \frac{T _{n}}{T _{1}} = 3 ln n$ : a straight line through the origin with slope $3$ (Figure 10).

Solved Example 16

A particle of mass

m

moves in a circular orbit under an attractive central force of magnitude

F = \frac{k}{r}

(potential energy

U = k ln r

). Using Bohr's quantisation rule, find how the radius, speed and energy of the

n

th orbit depend on

n

Solution:

Force equation: $\frac{m v ^{2}}{r} = \frac{k}{r} \Rightarrow v = \frac{k}{m}$ , the same for every orbit.

Quantisation: $m v r = \frac{nh}{2 π} \Rightarrow r_{n} = \frac{nh}{2 π mk} \propto n$ .

Energy: $E_{n} = \frac{1}{2} m v^{2} + k ln r_{n} = \frac{k}{2} + k ln r_{n}$ , so the gap between neighbouring levels is $E_{n + 1} - E_{n} = k ln \frac{n + 1}{n}$ .

Answer: $r_{n} \propto n$ , $v$ does not depend on $n$ , and $E_{n} = \frac{k}{2} + k ln r_{n}$ (the levels crowd together slowly, as $ln n$ ).

Solved Example 17

In muonic hydrogen a muon (charge

- e

, mass

207 m_{e}

) orbits a proton (mass

1836 m_{e}

). Find the radius and energy of the ground state and the wavelength of the photon emitted in the jump

n = 2 \to 1

Solution:

The muon is not light compared with the proton, so use the reduced mass: $μ = \frac{207 \times 1836}{207 + 1836} m_{e} = 186 m_{e}$ .

$r_{1} = 0.529 \times \frac{m _{e}}{μ} \overset{˚}{A} = \frac{0.529}{186} \overset{˚}{A} = 2.84 \times 1 0^{- 13} m$ .

$E_{1} = - 13.6 \times \frac{μ}{m _{e}} eV = - 13.6 \times 186 = - 2530 eV = - 2.53 keV$ .

$Δ E_{2 \to 1} = \frac{3}{4} \times 2.53 keV = 1.90 keV$ , so $λ = \frac{1240}{1897} nm = 0.654 nm$ .

Answer: $r_{1} = 2.84 \times 1 0^{- 13} m$ , $E_{1} = - 2.53 keV$ , $λ = 0.654 nm$ (an X-ray, because the muon orbit is 186 times smaller and 186 times more tightly bound).

Practice Questions

Find the radius of the $n = 3$ orbit of $Li^{2 +}$ .Answer: $1.59 \overset{˚}{A}$
Find the ratio of the period of revolution in the ground state of hydrogen to that in the $n = 2$ state of $He^{+}$ .Answer: $1 : 2$
Find the total energy of the electron in the $n = 3$ state of $He^{+}$ .Answer: $- 6.04 eV$
Find the kinetic and potential energies of the electron in the $n = 2$ orbit of hydrogen.Answer: $K = 3.4 eV$ , $U = - 6.8 eV$
Find the angular momentum of an electron in the third orbit.Answer: $\frac{3 h}{2 π} = 3.16 \times 1 0^{- 34} J s$
Find the distance of closest approach of a $5 MeV$ $α$ -particle to a gold nucleus ( $Z = 79$ ).Answer: $4.55 \times 1 0^{- 14} m$
Find the orbital magnetic moment of the electron in the $n = 3$ state of any hydrogen-like ion.Answer: $3 μ_{B} = 2.78 \times 1 0^{- 23} A m^{2}$

Common Mistakes to Avoid

Watch out

Writing the energy of hydrogen-like ions as $- \frac{13.6 Z}{n ^{2}}$ . The factor is $Z^{2}$ : $He^{+}$ ground state is $- 54.4 eV$ , not $- 27.2 eV$ .
Using $r \propto \frac{n ^{2}}{Z ^{2}}$ or $v \propto \frac{Z}{n ^{2}}$ . The correct scalings are $r \propto \frac{n ^{2}}{Z}$ and $v \propto \frac{Z}{n}$ .
Thinking kinetic energy decreases when the electron moves to a lower orbit. It increases ( $K = - E$ ); only $U$ and $E$ decrease.
Assuming $∣ U ∣ = ∣ K ∣$ . On a Bohr orbit $∣ U ∣ = 2 K$ , so $U = 2 E$ and $K = - E$ .
Treating a free target as fixed in closest-approach problems. If the target can move, use momentum conservation: only the KE relative to the centre of mass turns into PE.
Believing the magnetic moment depends on $Z$ . $M_{n} = n μ_{B}$ is the same for H, $He^{+}$ and $Li^{2 +}$ in the same $n$ .
Applying the Bohr formulas to neutral helium or lithium. They hold only for one-electron systems.
Saying the Rutherford atom fails because electrons would fly off. It fails because a radiating electron would spiral in, and the spectrum would be continuous.

Frequently Asked Questions

What are the postulates of the Bohr model of the hydrogen atom?

The electron moves in circular orbits in which the Coulomb attraction provides the centripetal force. Only orbits with angular momentum equal to a whole number times h over 2 pi are allowed, and in them the electron does not radiate. A photon is emitted or absorbed only when the electron jumps between two such orbits.

What is the radius of the first Bohr orbit?

The radius of the ground state of hydrogen, called the Bohr radius, is $0.529 \overset{˚}{A}$ ( $5.29 \times 1 0^{- 11} m$ ). In general $r_{n} = 0.529 \frac{n ^{2}}{Z} \overset{˚}{A}$ , so the radius grows as the square of $n$ and shrinks with atomic number.

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

Energy is measured from zero for an electron at rest far from the nucleus. A bound electron has less energy than that, because energy must be supplied to pull it free, so its energy is negative. For hydrogen $E_{n} = - \frac{13.6}{n ^{2}} eV$ .

What did Rutherford's alpha scattering experiment show?

Most alpha particles passed straight through a thin gold foil, but a few were deflected through large angles and about one in eight thousand bounced back. This showed that the atom is mostly empty space with its positive charge and mass concentrated in a tiny nucleus about $1 0^{- 15} m$ across.

Why did Rutherford's model fail?

An electron circling the nucleus is accelerating, so classical electromagnetism says it must radiate energy and spiral into the nucleus in about ten to the minus eleven seconds, giving a continuous spectrum. Real atoms are stable and emit sharp line spectra, which the model could not explain.

How does de Broglie's hypothesis explain Bohr's quantisation rule?

If the electron is a wave of wavelength $\frac{h}{m v}$ , a stable orbit must hold a whole number of wavelengths, $2 π r = nλ$ . Substituting the wavelength gives $m v r = \frac{nh}{2 π}$ , exactly Bohr's second postulate.

Which Bohr model questions come in NEET?

NEET usually asks the ratio of radii, speeds or energies for different $n$ or $Z$ , the relations $K = - E$ and $U = 2 E$ , and the energy or angular momentum in a given orbit. Remember $r \propto \frac{n ^{2}}{Z}$ , $v \propto \frac{Z}{n}$ and $E \propto \frac{Z ^{2}}{n ^{2}}$ .

How is the Bohr model tested in JEE Main and Advanced?

JEE Main tests Bohr formulas for hydrogen-like ions, current and magnetic moment of an orbit, distance of closest approach and impact parameter. JEE Advanced adds reduced-mass corrections, positronium and muonic atoms, and Bohr quantisation applied to other force laws.

Previous year questions on Bohr Model of the Hydrogen Atom

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

Show all 23 questions

Ready to master Atoms?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test