Atoms Class 12 NCERT Solutions Physics Chapter 12 Explained

Q: 12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × × 10 11 m with orbital speed 3 × × 10 4 m/s. (Mass of earth = 6.0 × × 10 24 (kg.)

12.10 Radius of the Earth's orbit around the Sun, r = 1.5 * 10 11 m Orbital speed of Earth, v = 3 * 10 4 m/s Mass of the Earth, m = 6 * 10 24 kg According to Bohr's model, angular momentum is given as: m v r = n h 2 ? , where h = Planck's constant = 6.626 * 10 - 34 Js n = Quantum number Hence, n= 2 ? m v r h = 2 * ? * 6 * 10 24 * 3 * 10 4 * 1.5 * 10 11 6.626 * 10 - 34 = 2.56 * 10 74 Hence, the quantum number that characterizes earth's revolution is 2.6 * 10 74

Q: 12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

12.9 It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is – 13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e. -1.1 eV Orbital energy is related to orbit level(n) as: E = - 13.6 n 2 eV For n = 3, E = - 13.6 9 eV = - 1.5 eV This energy is approximately equal to the energy of the gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level. During the de-excitation, the electron can jump from n=3 to n=1 directly, which forms a line of the Lyman series of hydrogen spectrum. We have the relation for wave number for Lyman series as: 1 ? = R y 1 1 2 - 1 n 2 where R y = Rydberg constant = 1.097 × 10 7 m - 1 ? = Wavelength of radiation emitted by the transition of electron For n = 3, we get 1 ? = 1.097 × 10 7 × 1 1 2 - 1 3 2 ? = 1.215 × 10 - 7 m = 102.6 nm If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as: 1 ? = 1.097 × 10 7 × 1 1 2 - 1 2 2 ? = 1.215 × 10 - 7 m = 121.5 nm If the transition takes place from n = 3 to n = 2, then the wavelength is given as: 1 ? = 1.097 × 10 7 × 1 2 2 - 1 3 2 ? = 6.563 × 10 - 7 m = 656.3 nm This radiation corresponds to the Balmer series of the hydrogen spectrum. Hence, in Lyman series, two wavelengths i.e. 102.6 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e. 656.3 nm is emitted.

Q: 12.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

12.8 The radius of the innermost orbit of a hydrogen atom, r 1 = 5.3 * 10 - 11 m Let r 2 be the radius of the orbit at n = 2. The relation between the radius of the orbit is r 2 = n 2 * r 1 =4 * 5.3 * 10 - 11 m =21.2 * 10 - 11 m Let r 3 be the radius of the orbit at n = 3. The relation between the radius of the orbit is r 3 = n 2 * r 1 =9 * 5.3 * 10 - 11 m =47.7 * 10 - 11 m

Q: 12.1 Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than) (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model) (c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model) (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model) (e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models)

12.1 The size of the atom in Thomson's model is no different from the atomic size in Rutherford's model. In the ground state of Thomson's model, electrons are in stable equilibrium. While in Rutherford's model, electrons always experience a net force. A classical atom based on Rutherford's model, is doomed to collapse. An atom has a nearly continuous mass distribution in a Thomson's model, but has a highly non-uniform mass distribution in Rutherford's model. The positively charged part of the atom possesses most of the mass in both the models.

Q: 12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

12.5 Ground state energy of hydrogen atom, E = -13.6 eV Kinetic energy is equal to the negative of the total energy (ground state energy) = 13.6 eV Potential energy is equal to the negative two times of kinetic energy = -13.6 * 2 = - 27.2 e V

Updated on Jul 7, 2025 14:07 IST

By Pallavi Pathak, Assistant Manager Content

NCERT Solutions Class 12 Physics Chapter 12 offers effective study material for Science stream students who are preparing for the Board examination. By practicing these solutions, the students can understand all concepts thoroughly and learn how to solve the numerical problems of the Chapter. The Class 12 Physics Chapter 12 – Atoms NCERT solutions is designed as per the latest CBSE curriculum and NCERT textbook. Students can rely on these solutions. It introduces students to key concepts that are crucial for exam preparation. Some of these are:

Energy Levels and Spectra of Hydrogen Atom
Limitations of Classical Physics in Atomic Structure
de Broglie’s Hypothesis and Electron Orbits
Bohr’s Model and Its Postulates
Rutherford’s Nuclear Model of Atom

Also read:

Table of content

NCERT Class 12 Atoms: Key Topics , Weightage and Formulae
Class 12 Physics Chapter 12 – Atoms at a Glimpse
NCERT Physics Class 12th Solution PDF - Atoms Chapter Download
Atoms Chapter 12 Important Formulas & Concepts
Class 12 Physics Chapter 12 Atoms NCERT Solutions
Benefits of Using Chapter 12 Atoms Class 12 Physics NCERT Solutions
Chapter 12 Atoms Class 12 Physics NCERT Solutions- FAQs

NCERT Class 12 Atoms: Key Topics , Weightage and Formulae

The Class 12 Physics Ch 12 Atoms play an important role in forming the basis for advanced topics of modern physics like quantum mechanics and atomic structure. Before starting the preparation of the chapter, it is always a smart strategy to know the topics covered in advance. See below the topics covered in the NCERT Solutions Atoms Class 12:

Exercise	Topics Covered
12.1	Introduction To Atoms
12.2	Alpha-Particle Scattering And Rutherford’s Nuclear Model Of Atom
12.3	Atomic Spectra
12.4	Bohr Model Of The Hydrogen Atom
12.5	The Line Spectra Of The Hydrogen Atom
12.6	De Broglie’s Explanation Of Bohr’s Second Postulate Of Quantisation

For a comprehensive preparation of Class 12 Physics, here you will get all the important topics and weightage details - NCERT Solutions for Class 12 Physics.

Atoms Weightage in JEE Main, NEET Exams

Exam Name	No. of Questions	Percentage
NEET	2-3 questions	5-7%
JEE Main	3-4 questions	6-8%

Class 12 Physics Chapter 12 – Atoms at a Glimpse

See below the quick summary of Atoms:

The chapter introduces you to Rutherford’s Experiment, which concludes that atoms have a positively charged, small, and dense nucleus. The atoms also have electrons orbiting them.
Stability of atoms and the line spectra of hydrogen is understood by Bohr's Model through quantized orbits for electrons with fixed energy levels.
There are also limitations of Bohr's Model as it can't explains the complex atomic behavior of multi-electron atoms, but it works well for hydrogen.
Using Bohr's model, Hydrogen Spectral Series explains the emission spectra of hydrogen are classified into Balmer, Lyman, Paschen, etc.
Energy Levels explains that the electrons exist in specific energy states. The transitions between these states involve discrete energy changes.
Hydrogen Atom used to study atomic structure and spectral patterns. The hydrogen atom is the simplest atom with one electron and one proton.
Atomic Spectra explains the transitions of electrons absorb or produce light at distinct wavelengths, which creates unique spectral lines for each element.

Try these practice questions

Q1:

Choose the correct option from the following options given below :

In the ground state of Rutherford’s model electrons are in stable equilibrium. While in Thomson’s model electrons always

An atom has a nearly continuous mass distribution in a Rutherford’s model but has a highly non-uniform mass distribution

A classical atom based on Rutherford’s model is doomed to collapse.

The positively charged part of the atom possesses most of the mass in Rutherford’s model but not in Thomson’s model.

Q2:

Find the ratio of energies of photos product due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level. and (ii) the highest permitted energy level to the first permitted level.

3 : 4

4 : 3

1 : 4

4 : 1

NCERT Physics Class 12th Solution PDF - Atoms Chapter Download

The Atoms NCERT PDF link is given below. Students must download this PDF. It will help students to work on their conceptual understanding, improve their problem-solving skills, and develop revision strategies.

Students can download the NCERT Class 12 Physics Chapter 12 Solution PDF by clicking on the link given below to boost their exam preparation and get good marks in the exams.

NCERT Solution Class 12 Physics Atoms PDF: Download Free PDF

Atoms Chapter 12 Important Formulas & Concepts

Important Formulae of Class 12 Physics Chapter 12 Atoms

Students can access the important formulae from Class 12 Physics Chapter 12 Atoms:

1. Radius of nth orbit (Bohr’s Model)

$r_n = \frac{n^2 h^2}{4 \pi^2 m k e^2} = 0.529 \times n^2 \, \text{Å}$

Where:

$n$ = orbit number
$h$ = Planck’s constant
$m$ = mass of electron
$k = \frac{1}{4\pi\varepsilon_0}$
$e$ = charge of electron

2. Velocity of electron in nth orbit

$v_n = \frac{k e^2}{n h} \cdot \frac{1}{2\pi \varepsilon_0 r_n} = \frac{2.18 \times 10^6}{n} \, \text{m/s}$

3. Energy of electron in nth orbit

$E_n = - \frac{13.6}{n^2} \, \text{eV}$

4. Energy difference between two levels

$\Delta E = E_i - E_f = h\nu = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \, \text{eV}$

5. Frequency of emitted/absorbed radiation

$\nu = \frac{\Delta E}{h}$

6. Wavelength of spectral line (Hydrogen atom – Balmer, Lyman series etc.)

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Where:

$R_H$ = Rydberg constant $= 1.097 \times 10^7 \, \text{m}^{-1}$
$n_2 > n_1$

7. Angular momentum (Bohr’s quantization condition)

$L = m v r = \frac{n h}{2 \pi}$

8. Centripetal force equals electrostatic force

$\frac{m v^2}{r} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r^2}$

Class 12 Physics Chapter 12 Atoms NCERT Solutions

Q.12.1 Choose the correct alternative from the clues given at the end of the each statement:

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

Ans.12.1 The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

In the ground state of Thomson’s model, electrons are in stable equilibrium. While in Rutherford’s model, electrons always experience a net force.

A classical atom based on Rutherford’s model, is doomed to collapse.

An atom has a nearly continuous mass distribution in a Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.

The positively charged part of the atom possesses most of the mass in both the models.

Q.12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Ans.12.2 In the alpha-particle scattering experiment, if a thin sheet of hydrogen is used in place of a gold film, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67

\times 10^{- 27}

kg) is less than the mass of incident

α -

particles (6.64

\times 10^{- 27}

kg). Thus, the mass of the scattering particles is more than the target nucleus (hydrogen). As a result, the

α -

particles would not bounce back if solid hydrogen is used in the

α -

particle scattering experiment.

Q.12.3 What is the shortest wavelength present in the Paschen series of spectral lines?

Ans.12.3 Rydberg’s formula is given as:

$\frac{h c}{λ}$ = 21.76 $\times 10^{- 19} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

Where, h = Planck’s constant = 6.6 $\times 10^{- 34}$ Js

c = speed of light = 3 $\times 10^{8}$ m/s

$λ$ = Wavelength

$n_{1}$ and $n_{2}$ are integers.

The shortest wavelength present in the Paschen series of the spectral lines is given for the values $n_{1} = 3$ and $n_{2} = \infty$

Therefore, $\frac{h c}{λ}$ = 21.76 $\times 10^{- 19} [\frac{1}{3^{2}} - \frac{1}{\infty^{2}}]$

$\frac{h c}{λ} =$ 2.42 $\times 10^{- 19}$

$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2.42 \times 10^{- 19}}$ = 8.189 $\times 10^{- 7}$ m= 818.9 nm

Q.12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Ans.12.4 Separation of two energy level of atom, E = 2.3 eV = 2.3 $\times 1.6 \times 10^{- 19}$ J = 3.68 $\times 10^{- 19}$

Let $ν$ be the frequency of radiation emitted when the atom transits from upper level to lower level.

We have the relation for energy as $E = h ν$ , where

h = Planck’s constant = 6.626 $\times 10^{- 34}$ Js

Then $ν = \frac{E}{h}$ = $\frac{3.68 \times 10^{- 19}}{6.626 \times 10^{- 34}}$ Hz = 5.55 $\times 10^{14}$ Hz

Hence the frequency is 5.55 $\times 10^{14}$ Hz

Commonly asked questions

12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × $\times 10^{11}$ m with orbital speed 3 × $\times 10^{4}$ m/s. (Mass of earth = 6.0 × $\times 10^{24}$ (kg.)

12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

12.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

12.1 Choose the correct alternative from the clues given at the end of the each statement:

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than)

(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model)

(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model)

(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model)

(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models)

12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

12.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

12.16 If Bohr’s quantization postulate (angular momentum = nh/2n) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?

12.15 The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

(a) What is the kinetic energy of the electron in this state?

(b) What is the potential energy of the electron in this state?

(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

12.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

2.11 Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

(a) Is the average angle of deflection of $α$ particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(b) Is the probability of backward scattering (i.e., scattering of $α$ -particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of $α$ -particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of $α$ -particles by a thin foil?

12.14 Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.

(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognizing that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ–) of mass about 207me orbits around a proton].

12.3 What is the shortest wavelength present in the Paschen series of spectral lines?

Benefits of Using Chapter 12 Atoms Class 12 Physics NCERT Solutions

Here are the benefits of using the Atoms Class 12 Physics Solutions;

NCERT Solutions provided by Shiksha perfectly align with the CBSE exam format, which are based on the NCERT Textbooks. Students can pick our NCERT Solutions to study on the lines of CBSE Board exams and score well.
Shiksha has prepared NCERT Solutions to provide conceptual explanations for all NCERT problems with lucid language and shortcut techniques.
We have covered all the concepts in that chapter, such as Rutherford’s atomic model, Bohr’s theory, hydrogen spectral lines, and energy level transitions in our NCERT Solutions.
This chapter is a part of the Modern Physics unit, Our NCERT Solutions also provide related chapters solutions also which can give them a complete picture of the unit.