CBSE Class 12 Physics Revision Notes Chapter 12: Atoms

Rutherford’s scattering experiment showed that nearly all atomic mass and positive charge are concentrated in a tiny nucleus.
Bohr explained atomic stability and hydrogen line spectra by introducing stationary orbits and quantised energy levels.

The chapter Atoms traces the development of atomic models from Thomson’s model to Rutherford’s nuclear model and Bohr’s model of the hydrogen atom. It explains how alpha-particle scattering revealed the nucleus and why classical physics could not explain atomic stability.

Use these CBSE Class 12 Physics Revision Notes Chapter 12 for the 2026–27 session. Begin with Rutherford’s experiment and atomic spectra. Then revise Bohr’s postulates, orbit radius, energy levels, spectral transitions and the wave explanation of quantisation.

Key Takeaways

• Atomic structure: Most of an atom is empty space, with positive charge and mass concentrated in the nucleus.
• Bohr quantisation: The electron’s angular momentum is L = nh/2π.
• Hydrogen energy: En = −13.6/n² eV for the nth stationary state.
• Electron transition: A photon with energy hν = Ei − Ef is emitted during a downward transition.

Access Class 12 Physics Chapter 12 Atoms Notes in 30 Minutes

Divide the chapter into three revision blocks:

• First 10 minutes: Thomson’s model, Rutherford scattering and the nuclear model
• Next 10 minutes: Bohr’s postulates, orbit radius and energy levels
• Final 10 minutes: Hydrogen spectra, electron transitions and de Broglie’s explanation

While revising numericals, remember that the electron’s total energy is negative in a bound state. Also distinguish excitation energy from ionisation energy.

Need help revising Rutherford scattering, energy-level diagrams and hydrogen-atom numericals?
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Atomic Models in Class 12 Physics Chapter 12 Notes

Experiments in the nineteenth century established that matter is made of atoms.

The discovery of the electron by J. J. Thomson showed that atoms contain negatively charged particles. Since atoms are electrically neutral, they must also contain positive charge.

This raised an important question: how are positive charge and electrons arranged inside an atom?

Thomson’s Atomic Model in Class 12 Atoms Revision Notes

1898. J. Thomson proposed the first detailed atomic model in 1898.

According to Thomson’s atomic model:

• The atom is a positively charged sphere.
• Positive charge is spread uniformly throughout its volume.
• Electrons are embedded inside the positive charge.
• The total positive and negative charges are equal.
• The atom is electrically neutral.

The model is also called the:

• Plum pudding model
• Watermelon model

The positive charge was compared with the red part of a watermelon, while electrons were compared with seeds.

Limitations of Thomson’s Model

Thomson’s model could explain overall electrical neutrality, but it could not explain:

• Results of alpha-particle scattering
• Presence of a small atomic nucleus
• Distribution of atomic mass
• Atomic line spectra
• Actual arrangement of electrons

Rutherford’s experiment later showed that positive charge is not spread throughout the atom.

Rutherford’s Alpha-Particle Scattering Experiment in CBSE Class 12 Physics Chapter 12 Notes

Rutherford suggested using alpha particles to investigate atomic structure.

Hans Geiger and Ernest Marsden performed the experiment around 1911.

Experimental Arrangement

The setup contained:

• A radioactive source producing alpha particles
• Lead blocks to form a narrow alpha-particle beam
• A very thin gold foil
• A movable zinc sulphide screen
• A microscope to observe scintillations
• A vacuum chamber

The gold foil had a thickness of about:

2.1 × 10⁻⁷ m

Alpha particles striking the zinc sulphide screen produced small flashes of light called scintillations.

Why Alpha Particles Were Used

Alpha particles are:

• Positively charged
• Relatively heavy
• Fast-moving
• Capable of penetrating thin metal foils

Their large mass means that atomic electrons cannot significantly change their paths.

Observations of Rutherford’s Scattering Experiment

The main observations were:

• Most alpha particles passed through the foil without noticeable deflection.
• Some particles were deflected through small angles.
• A very small number were deflected through large angles.
• Around 1 in 8000 particles suffered a deflection greater than 90°.
• A few particles appeared to rebound.

Only about 0.14% of the incident alpha particles were scattered by more than 1°.

Conclusions from the Observations

Rutherford concluded that:

• Most of the atom is empty space.
• Positive charge occupies a very small central region.
• Most of the atomic mass is concentrated in this region.
• The central region is called the nucleus.
• Electrons occupy the surrounding space.

A large deflection can occur only when an alpha particle passes very close to the positively charged nucleus.

Rutherford’s Nuclear Model in Class 12 Physics Atoms Notes

According to Rutherford’s nuclear model:

• The atom has a very small, dense and positively charged nucleus.
• Nearly all the atomic mass is concentrated in the nucleus.
• Electrons revolve around the nucleus.
• Electrostatic attraction keeps electrons in their orbits.
• The atom is mostly empty space.

The atomic radius is of the order of:

10⁻¹⁰ m

The nuclear radius is approximately:

10⁻¹⁵ m to 10⁻¹⁴ m

Thus, the atom is about 10,000 to 100,000 times larger than its nucleus.

Atomic and Nuclear Size Comparison

This large difference explains why most alpha particles pass through the foil.

Alpha-Particle Trajectory in Physics Chapter 12 Revision Notes

An alpha particle and a nucleus are both positively charged.

Therefore, an alpha particle experiences an electrostatic repulsive force as it approaches the nucleus.

For an alpha particle of charge +2e and a nucleus of charge +Ze:

F = (1/4πε₀)(2Ze²/r²)

Here:

• Z is the atomic number of the target atom.
• r is the separation between the alpha particle and nucleus.

The force changes continuously as the particle approaches and moves away from the nucleus.

Impact Parameter

The impact parameter b is the perpendicular distance between:

• The initial direction of the alpha particle
• The centre of the target nucleus

It determines the scattering angle.

A smaller impact parameter means that the particle comes closer to the nucleus and experiences a stronger repulsive force.

Scattering Angle

The scattering angle is represented by θ.

• For large impact parameter, θ is close to 0°.
• For a head-on collision, θ approaches 180°.

The small number of backward-scattered particles shows that the nucleus occupies a tiny volume.

Distance of Closest Approach in Chapter 12 Physics Notes

For a head-on collision, an alpha particle approaches the nucleus, slows down and momentarily stops before reversing its direction.

At the closest point:

• The alpha particle’s kinetic energy becomes zero.
• Its initial kinetic energy is converted into electrostatic potential energy.

For an alpha particle and a nucleus of charge +Ze:

K = (1/4πε₀)(2Ze²/d)

Therefore:

d = (1/4πε₀)(2Ze²/K)

Here:

• d is the distance of closest approach.
• K is the initial kinetic energy.

This equation gives an upper limit to the size of the nucleus.

A particle with greater kinetic energy comes closer to the nucleus.

Electron Orbits in Rutherford’s Atomic Model Revision Notes

In Rutherford’s model, an electron revolves around the positively charged nucleus.

For a hydrogen atom, the electrostatic force provides the centripetal force.

Therefore:

(1/4πε₀)(e²/r²) = mv²/r

This gives:

mv² = e²/(4πε₀r)

Kinetic Energy of the Electron

The kinetic energy is:

K = ½mv²

Using the force relation:

K = e²/(8πε₀r)

Potential Energy of the Electron

The electrostatic potential energy is:

U = −e²/(4πε₀r)

The negative sign indicates an attractive interaction.

Total Energy

The total energy is:

E = K + U

Therefore:

E = −e²/(8πε₀r)

The total energy is negative because the electron is bound to the nucleus.

The relations are:

U = −2K

E = −K

E = U/2

Limitations of Rutherford’s Model in CBSE Class 12 Atoms Notes

Rutherford’s model successfully explained the nuclear structure of the atom, but it had two major problems.

Atomic Stability Problem

An electron moving in a circular orbit is accelerating.

According to classical electromagnetic theory, an accelerating charged particle should continuously radiate energy.

Therefore:

• The electron should lose energy.
• Its orbital radius should decrease.
• It should spiral towards the nucleus.
• The atom should collapse.

Actual atoms are stable, so this prediction is incorrect.

Atomic Spectrum Problem

As the electron spirals inward, its speed and orbital frequency should change continuously.

It should therefore emit radiation with continuously changing frequencies.

This would produce a continuous spectrum.

However, atoms produce discrete line spectra. Rutherford’s model could not explain this observation.

Atomic Spectra in Class 12 Physics Chapter 12 Notes

Each element produces a characteristic spectrum.

When a low-pressure atomic gas is excited, it emits light only at certain wavelengths.

These wavelengths appear as separate lines.

The spectrum acts like a fingerprint for identifying an element.

Continuous Spectrum

A continuous spectrum contains a continuous range of wavelengths.

It is commonly produced by:

• Solids
• Liquids
• Dense gases

Line Spectrum

A line spectrum contains only certain specific wavelengths.

It is commonly produced by:

• Rarefied gases
• Electrically excited gases
• Individual atoms

Emission Line Spectrum

An emission line spectrum consists of bright lines on a dark background.

It is produced when excited atoms emit radiation of specific frequencies.

Absorption Spectrum

An absorption spectrum consists of dark lines on a continuous bright background.

It is produced when atoms absorb specific wavelengths from incident light.

The absorption lines of an element occur at the same wavelengths as its emission lines.

Hydrogen Atom Spectrum in Atoms Class 12 Notes

Hydrogen is the simplest atom because it contains:

• One proton
• One electron

Its emission spectrum consists of several discrete lines.

In 1885, Johann Balmer found an empirical relation for a group of visible hydrogen lines.

The fixed wavelengths in the hydrogen spectrum suggested that electrons can have only certain allowed energies.

Rutherford’s classical model could not explain these discrete wavelengths.

Bohr’s model provided the first successful explanation.

Bohr Model of the Hydrogen Atom in Class 12 Physics Atoms Notes

Niels Bohr modified Rutherford’s nuclear model by adding quantum ideas.

He proposed three postulates to explain:

• Atomic stability
• Quantised electron orbits
• Hydrogen line spectra

Bohr’s First Postulate in Physics Chapter 12 Revision Notes

An electron can revolve around the nucleus only in certain stable orbits.

These orbits are called:

• Stationary orbits
• Stationary states
• Allowed orbits

While moving in a stationary orbit:

• The electron does not radiate energy.
• Its total energy remains constant.
• The atom remains stable.

Each stationary state has a definite energy.

This postulate contradicts classical electromagnetic theory but explains atomic stability.

Bohr’s Second Postulate in CBSE Class 12 Physics Chapter 12 Notes

Only those electron orbits are allowed for which angular momentum is an integral multiple of h/2π.

Therefore:

L = mvr = nh/2π

Here:

• m is electron mass.
• v is orbital speed.
• r is orbit radius.
• n is the principal quantum number.
• h is Planck’s constant.

The allowed values of n are:

n = 1, 2, 3, ...

Angular momentum is therefore quantised.

An electron cannot revolve in an orbit with an arbitrary angular momentum.

Bohr’s Third Postulate in Class 12 Atoms Revision Notes

An electron emits or absorbs radiation only when it moves from one stationary state to another.

For a transition from a higher energy state Ei to a lower energy state Ef:

hν = Ei − Ef

A photon is emitted when:

Ei > Ef

For absorption:

Ef = Ei + hν

A photon is absorbed when an electron moves from a lower level to a higher level.

The frequency of emitted light is determined by the energy difference between the two states. It is not equal to the electron’s orbital frequency.

Radius of Bohr Orbits in Chapter 12 Physics Notes

For a hydrogen atom:

Electrostatic force = Centripetal force

Therefore:

e²/(4πε₀r²) = mv²/r

Bohr’s quantisation condition is:

mvr = nh/2π

Using both equations, the radius of the nth orbit is:

rn = n²h²ε₀/(πme²)

It may also be written as:

rn = n²a₀

Here, a₀ is the Bohr radius.

Bohr Radius

For n = 1:

a₀ = 5.3 × 10⁻¹¹ m

or:

a₀ = 0.53 Å

Therefore:

rn = 0.53n² Å

Radius Dependence on Quantum Number

The orbit radius varies as:

rn ∝ n²

Therefore:

r₁ : r₂ : r₃ = 1 : 4 : 9

The spacing between successive orbits increases with n.

Electron Speed in Bohr Orbits Revision Notes

The speed of an electron in the nth orbit of hydrogen is:

vn = e²/(2ε₀hn)

Therefore:

vn ∝ 1/n

For the first orbit:

v₁ ≈ 2.2 × 10⁶ m/s

Hence:

vn = v₁/n

The electron’s speed decreases as it moves to a higher orbit.

Hydrogen Energy Levels in Class 12 Physics Chapter 12 Notes

The total energy of the electron in the nth stationary state is:

En = −me⁴/(8ε₀²h²n²)

For hydrogen:

En = −2.18 × 10⁻¹⁸/n² J

In electron volts:

En = −13.6/n² eV

Energy Dependence on Quantum Number

The energy varies as:

En ∝ −1/n²

The negative sign shows that the electron is bound to the nucleus.

As n increases:

• Energy becomes less negative.
• The electron is less tightly bound.
• Energy levels move closer together.
• The orbit radius increases.

Ground State and Excited States in Atoms Revision Notes

The state with the lowest energy is called the ground state.

For hydrogen:

n = 1

E₁ = −13.6 eV

The higher-energy states are called excited states.

Examples:

• n = 2 is the first excited state.
• n = 3 is the second excited state.
• n = 4 is the third excited state.

Excitation Energy

The energy required to move an electron from a lower state to a higher state is called excitation energy.

For excitation from n = 1 to n = 2:

ΔE = E₂ − E₁

ΔE = −3.4 − (−13.6)

ΔE = 10.2 eV

For excitation from n = 1 to n = 3:

ΔE = −1.51 − (−13.6)

ΔE = 12.09 eV

Ionisation Energy of Hydrogen in CBSE Class 12 Physics Chapter 12 Notes

Ionisation means removing the electron completely from the atom.

For complete removal:

n = ∞

E∞ = 0

The ionisation energy from a state n is:

Eionisation = 0 − En

Therefore:

Eionisation = 13.6/n² eV

For the ground state:

Eionisation = 13.6 eV

This means that 13.6 eV must be supplied to remove the electron from a ground-state hydrogen atom.

The ionisation energy decreases for excited states.

Line Spectrum of the Hydrogen Atom in Class 12 Physics Chapter 12 Notes

According to Bohr’s third postulate, light is emitted when an electron moves from a higher energy state to a lower state.

For a transition from ni to nf:

hν = Eni − Enf

where:

ni > nf

Using the hydrogen energy formula:

hν = 13.6[1/nf² − 1/ni²] eV

The wavelength is:

λ = hc/(Eni − Enf)

Since only certain energy levels exist, only certain photon frequencies are emitted.

This produces a line spectrum.

Emission and Absorption Transitions

Emission

During emission:

• Electron moves from higher n to lower n.
• Atom loses energy.
• A photon is emitted.

Absorption

During absorption:

• Electron moves from lower n to higher n.
• Atom gains energy.
• A photon is absorbed.

The absorbed photon must have exactly the energy difference between the two levels.

Hydrogen Spectral Series in Class 12 Atoms Notes

Transitions ending at a common lower energy level form a spectral series.

The Balmer series contains the important visible lines of hydrogen.

Rydberg Formula

The wavelengths of hydrogen spectral lines can be represented as:

1/λ = R[1/nf² − 1/ni²]

Here:

• R is the Rydberg constant.
• ni is the initial quantum number.
• nf is the final quantum number.
• ni > nf for emission.

The approximate value of R is:

R = 1.097 × 10⁷ m⁻¹

Energy-Level Diagram in Physics Chapter 12 Revision Notes

An energy-level diagram shows the allowed energies of the hydrogen atom.

Important features are:

• Horizontal lines represent allowed energy states.
• The lowest line represents n = 1.
• Higher lines represent excited states.
• The levels become closer as n increases.
• The level n = ∞ corresponds to E = 0.
• Energies above zero form a continuous range for a free electron.

An upward arrow represents absorption.

A downward arrow represents emission.

The photon energy equals the vertical energy gap between two levels.

De Broglie Explanation of Bohr’s Second Postulate in CBSE Class 12 Atoms Notes

Bohr introduced angular-momentum quantisation as a postulate.

Louis de Broglie later explained it using the wave nature of electrons.

According to de Broglie:

λ = h/p

For a non-relativistic electron:

λ = h/(mv)

Standing-Wave Condition

A stable electron orbit must contain a whole number of de Broglie wavelengths around its circumference.

Therefore:

2πrn = nλ

where:

n = 1, 2, 3, ...

Substituting:

λ = h/(mvn)

we get:

2πrn = nh/(mvn)

Therefore:

mvnrn = nh/2π

This is Bohr’s quantisation condition.

Meaning of the Standing-Wave Condition

Only those electron waves that fit completely around the orbit can persist.

If the circumference does not contain a whole number of wavelengths:

• The electron wave interferes destructively with itself.
• A stable standing wave cannot form.
• That orbit is not allowed.

Thus, quantised electron orbits arise from the wave nature of the electron.

Bohr Model for Hydrogen-Like Atoms Revision Notes

Bohr’s model also applies to one-electron species called hydrogenic atoms.

Examples include:

• Hydrogen, H
• Singly ionised helium, He⁺
• Doubly ionised lithium, Li²⁺

For a hydrogen-like atom with nuclear charge +Ze:

Orbit Radius

rn = n²a₀/Z

Therefore:

rn ∝ n²/Z

Electron Energy

En = −13.6Z²/n² eV

Therefore:

En ∝ −Z²/n²

A larger nuclear charge:

• Reduces orbit radius
• Increases binding energy
• Increases ionisation energy

Limitations of Bohr Model in Class 12 Physics Chapter 12 Notes

Bohr’s model successfully explains the main features of hydrogen and hydrogen-like atoms.

However, it has several limitations.

Limited to One-Electron Atoms

The model works only for hydrogenic atoms.

It cannot explain atoms containing two or more electrons because it does not include electron-electron interactions properly.

Cannot Explain Spectral Intensities

Bohr’s model predicts the frequencies of hydrogen spectral lines.

It does not explain:

• Why some lines are brighter than others
• Why some transitions are more probable
• Relative intensities of spectral lines

Does Not Give a Complete Quantum Picture

The model treats the electron as moving in a definite circular orbit.

Modern quantum mechanics describes electrons using probability distributions rather than fixed classical paths.

Other Limitations

The basic Bohr model does not fully explain:

• Fine structure of spectral lines
• Splitting in magnetic fields
• Splitting in electric fields
• Spectra of multi-electron atoms
• Chemical bonding

Despite these limitations, Bohr’s model remains useful for understanding atomic energy levels.

Thomson, Rutherford and Bohr Atomic Models Comparison

Atoms Formula Notes

Important Terms in Class 12 Physics Chapter 12

Useful Links for Class 12 Physics