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Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF

Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF

Updated on Jul 08, 2025 01:22 AM IST

Atoms make everything around us. Since the late nineteenth century, scientists have put forth more and more accurate explanations of the structure of atoms. This has been a result of countless thought-provoking experiments which has revolutionised the field of physics. The understanding of atoms has led to advancements in all other fields of science such as molecular biology, chemistry, and engineering. In the chapter Atoms of Class 12 NCERT Notes, we understand the process, the merits, as well as the limitations of the first discoveries made in this field.

This Story also Contains
  1. NCERT Notes for Class 12 Physics Chapter 12: Download PDF
  2. NCERT Notes for Class 12 Physics Chapter 12 Atoms
  3. Atoms Previous Year Question and Answer
  4. CBSE Class 12 Chapterwise Notes
Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF
Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF

Chapter 12 of the NCERT Notes for Class 12 Physics provides a brief introduction to the basic concepts of atomic structure. The Rutherford's Nuclear Model, Atomic Spectra and Bohr Model of the Hydrogen Atom are the major topics this chapter discusses. The concept of the De Broglie Hypothesis, along with its explanation of Bohr's postulates, is discussed in greater detail in this chapter.

This chapter is a vital part of modern physics and is frequently tested in CBSE board exams and entrance exams like JEE Main and NEET. It is essential for students as both numerical and theoretical questions are common from its topics. The NCERT Notes of Chapter Atoms provide an easy-to-understand explanation of the concepts which are essential for understanding modern physics. These notes have been prepared by experts at Careers360 and give students detailed explanations where required.

Also, students can refer,

NCERT Notes for Class 12 Physics Chapter 12: Download PDF

Download the NCERT Notes for Atoms using the button. Using the PDF, students can access the notes offline at their convenience and is useful for revision and studying on the go.

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NCERT Notes for Class 12 Physics Chapter 12 Atoms

The NCERT Notes of Atoms contain short derivations, simple clarifications, along with diagrams, for better learning and are suitable for exam preparation. It builds a strong conceptual base for atomic structure, quantum physics, and light-matter interaction, all of which are critical for higher studies in physics and chemistry.

Thomson’s Atomic Model (Plum Pudding Model)

J. J. Thomson, who discovered the electron in 1897, proposed the plum pudding model of the atom in 1904 before the discovery of the atomic nucleus in order to include the electron in the atomic model. In Thomson’s model, the atom is composed of electrons surrounded by a soup of positive charge to balance the electrons’ negative charges, like negatively charged “plums” surrounded by positively charged “pudding”.

Drawbacks
  • Failed to explain results from the alpha-particle scattering experiment.
  • Failed to explain atomic spectra.
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Alpha Particle Scattering and Rutherford’s Nuclear Model of the Atom

Rutherford and his colleagues Geiger and Marsden bombarded a thin gold foil of thickness approximately 8.6×106 cm with a beam of alpha particles in a vacuum. They used gold since it is highly malleable, producing sheets that can be only a few atoms thick, thereby ensuring smooth passage of the alpha particles. A circular screen coated with zinc sulphide surrounded the foil. Since the positively charged alpha particles possess mass and move very fast, it was hypothesized that they would penetrate the thin gold foil and land themselves on the screen, producing fluorescence in the part they struck.

Observations:
  • Most of the alpha particles behaved as expected, but there was a noticeable fraction of particles that got scattered by angles greater than 90 degrees.
  • There were about 1 in every 2000 particles that got scattered by a full 180 degrees.
  • The distribution of scattered α-particles was analyzed and plotted as a graph as a function of scattering angle (θ).

For a detector at a specific angle (θ) with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula:

N(θ)1sin4(θ2)

Conclusion:
  • Most of the atom is empty space, since most particles pass undeflected.

  • The positive charge and mass of the atom are concentrated in a very small, dense region, which repelled the few alpha particles that were deflected. This region was named the nucleus.

  • Electrons revolve around the nucleus in circular orbits at very high speeds, like planets orbiting the Sun.

  • The atom remains electrically neutral as the positive charge of protons is balanced by the negative charge of electrons.

  • The electrostatic force of attraction between electrons and protons provides the centripetal force needed to keep the electrons in their orbits.

Limitations:
  • Could not explain the stability of atom (electrons should spiral into nucleus).

  • Did not explain discrete spectral lines of hydrogen.

  • Could not describe atoms with more than one electron.

Alpha-particle Trajectory

We can understand the trajectory of alpha-particles by calculating the distance of closest approach and impact parameter.

Distance of Closest Approach

The distance of the closest approach in Rutherford scattering refers to the minimum distance between an incoming alpha particle and the nucleus of the target atom during a scattering event. At this distance, the entire initial kinetic energy has been converted into potential energy.

( K.E. )initial =12mv2=14πε0(Ze)2er0r0=Ze2mv2πε0=4kZe2mv2=2kZe2K

Impact Parameter

Impact parameter is defined as the perpendicular distance of the velocity of the alpha particle from the centre of the nucleus when it is far away from the atom. The shape of the trajectory of the scattered alpha particle depends on the impact parameter 'b' and the nature of the potential field. Rutherford deduced the following relationship between the impact parameter 'b' and the scattering angle θ. It is given as

b=Ze2cot(θ/2)4πε0(12mv2)bcot(θ/2)

Electron Orbits

According to Rutherford, the electrostatic force of attraction, provided the required centripetal force for orbit.

Fc=Femv2r=14πε0Ze2r2r=Ze24πε0mv2

Here,
r=radius of orbit,
v = velocity of orbiting electron,
e = charge of an electron,
m = mass of an electron,
Z = atomic mass of the atom,
ε0= permittivity of free space

Kinetic Energy(K):

K=12mv2=Ze28πε0r

Potential Energy(U):

Using the electrostatic potential between 2 charged bodies, we get:

U=14πε0eX+Zer=Ze24πε0r

Here, negative signs show that there is a force of attraction, and energy has to be given to the system to overcome this force of attraction.

Total Energy(T):

T = U + K

Some important relations to note:

Kinetic energy (K) = -(1/2)Potential energy(U)
Kinetic energy (K) = -Total energy(T)
Potential energy(U) = 2×Total energy(T)

Atomic Spectra

  • When electrons in an atom absorb energy, they jump to higher energy levels (excited state).
  • When they return to lower energy levels (ground state), they emit energy in the form of light.
  • This emitted light, when passed through a prism, shows discrete lines known as the line spectrum or atomic spectrum.
  • Each element has a unique atomic spectrum.
  • These spectral lines correspond to specific wavelengths (energies) of emitted radiation.
  • The atomic spectrum of hydrogen is divided into series based on the final energy level.

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Bohr Model of the Hydrogen Atom

Proposed by Niels Bohr in 1913, this model explained the stability of the atom and the line spectra of hydrogen, improving upon Rutherford’s nuclear model. By combining classical and early quantum concepts, Bohr's Model of the hydrogen atom has three postulates.

(1) Atoms have certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom. In these states, electrons don't radiate energy.

For electrons revolving in a stable orbit, the necessary centripetal force is provided by the coulomb's force

mvn2rn=kze2rn2

(2) This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised.

L=mvnrn=nh2π;n=1,2,3

(3). Bohr’s third postulate states that when an electron makes a transition from a higher to lower energy orbit, it emits a photon with energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by hν=EiEf

Using the first and second postulates, we can find the radius of electron orbit.

rn=n2h24π2kZme2=n2h2ε0πmZe2=0.53n2Z (k=14πε0)rnn2Z

We can find the energy of an electron in stable orbit using Rutherford's expression for total energy.

T=Ze28πε0r

By putting the value of Bohr’s radius, we get

T=me28πε02n2h2

By putting the values of electron mass (m), charge (e), the permittivity of free space (ε0), and Planck’s constant (h), we get,

T=13.6n2eV

Energy Levels

In Bohr's atomic model, the energy levels of an electron in an atom are quantized. This means that electrons can only occupy specific, discrete energy levels and cannot exist in between these levels. The potential energy is always negative, indicating that the electron is bound to the nucleus.

The energy of nth  level of hydrogen atom (z=1) is given as :
En=z213.6n2eV=13.6n2eV(z=1)

  • Energy of ground state (n=1)
    E1=13.61eV=13.6eV
  • Energy of first excited state (n=2)
    E2=13.64eV=3.4eV
  • Energy of second excited state (n=3)
    E3=13.69eV=1.51eV
  • Energy of third excited state (n=4)
    E4=13.616eV=0.85eV

Line Spectra of Hydrogen Atom

According to Bohr, when an atom makes a transition from a higher energy level to a lower energy level, it emits a photon with energy equal to the energy difference between the initial and final levels. If Ei, is the initial energy of the atom before such a transition, Ef is its final energy after the transition, then conservation of energy gives the energy of the emitted photon.

hv=hcλ=EiEfhcλ=13.6ni2eV13.6nf2eV=13.6eV(1nf21ni2)Rch=13.6eV=1 Rydberg energy 1λ=R(1nf21ni2)
where R= Rydberg's constant =1.097×107 m1

The Line Spectra Of the Hydrogen Atom and its different series can be seen from the diagram below.

De-Broglie's Explanation of Bohr's Second Postulate

De-Broglie explaination of Bohr's Second Postulate show us clearly as to why the angular momentum of the revolving electron is the integral multiple of the h/2π. In his experiment he proved that the electron revolving in a circular orbit has a wave nature. Davison and Germer also proved that the electron shows the wave nature. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. This occurs when the total distance travelled by a wave down the string and back is any integral number of wavelengths.

For an electron moving in nth  circular orbit of radius rn, the total distance is the circumference of the orbit, 2πrn.

2πrn=nλ,n=1,2,3

We also know, λ=hp=hmvn

From the above equations, we have,

2πrn=nhmvnmvnrn=nh2π

This is the quantum condition proposed by Bohr for the angular momentum of the electron. Thus, De-Broglie's hypothesis confirms Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron.

Limitations of Bohr's Model

  • The model works well only for hydrogen and other single-electron systems (like He⁺, Li²⁺). It fails for multi-electron atoms (like helium, lithium, etc.).

  • Could not explain the fine structure or small splitting of spectral lines seen in high-resolution spectra.

  • Failed to explain: Zeeman and Stark effect – splitting of spectral lines in a magnetic and electric fields.

Atoms Previous Year Question and Answer

Q.1 Write shortcomings of Rutherford atomic model. Explain how these were overcome by the postulates of Bohr’s atomic model.

Answer:

Rutherford’s Model

1. While the electrons are moving around the nucleus they'll lose energy constantly. After losing energy, they should fall into the nucleus ultimately. Hence atom should collapse, but in reality, this does not happen.
2. He could not explain the spectrum of an atom.

Bohr’s Atomic Model postulates

(i) Explained the stability of the nucleus by postulating that the electron can orbit the nucleus in only those special orbits in which it does not radiate energy.
(ii) Explained the line spectrum by postulating that the electron emits photons of well-defined unique energies when it undergoes a transition from one permitted orbit to another.

Q.2 Use the Bohr model of the hydrogen atom to calculate the speed of the electron in the first excited state.

Answer:

The speed of the electron

Vn=1ne24πε0×2πh

For first existed a state n=2

Vn=12e24πε0×2πh=14×(1.6×1019)28.85×1012×6.6×1034=0.01096×108=1.096×106 ms1

Q.3 How is the stability of the hydrogen atom in the Bohr model explained by de Broglie's hypothesis?

Answer:

The quantised electron orbits and energy states are due to the wave nature of the electron, and only resonant standing waves can persist. Thus, the De Broglie hypothesis gives an explanation for Bohr’s second postulate.

According to de Broglie's hypothesis

2πr=nλ=nhp=nhmvmvr=nh2π

CBSE Class 12 Chapterwise Notes

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Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Class 12 Physics Chapter 12?

The important topics of Class 12 Physics Chapter 12 Atoms include Rutherford's gold foil experiment, postulates of Bohr's model of the hydrogen atom, and the line spectra of the hydrogen atom. 

2. What is the significance of energy levels in an atom?

Energy levels denote the stable orbits of an electron in an atom. It is the total energy of an electron in that state. 

3. What are the limitations of Bohr’s atomic model?

The major limitation of Bohr's atomic model is that it is not applicable to multi-electron atoms. It is also unable to explain phenomena such as Zeeman and Stark effect.

4. Why did Rutherford’s model fail to explain atomic stability?

According to classical theory, accelerating electrons should radiate energy as electromagnetic radiation. If this were true, electrons would fall into the nucleus and hence, atoms would be unstable. Rutherford's model failed to explain why electron orbits are stable in nature.

5. How to calculate the ionization energy of hydrogen from Bohr’s theory?

To ionize an atom, the electron must be taken out of the atom completely. A free electron has a minimum total energy equal to 0. The magnitude of ionization energy is thus equivalent to the energy of an electron in the ground state. For a hydrogen atom, this is equal to 13.6 eV. 

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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